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Why is there more room in a square room than there is in a rectangular room when the perimeter is the same in...
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Why is there an extra square foot in a square room with dimensions of $13×13$ and one less square foot in a room with dimensions of $14×12$? The perimeter for both rooms is the same (52 foot). I'm not asking to see formulas giving me the same results I already know - I'd prefer a more intuitive explanation.
area square-numbers
edited Feb 4 '15 at 3:32
Milo Brandt
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show 6 more comments
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Why is there an extra square foot in a square room with dimensions of $13×13$ and one less square foot in a room with dimensions of $14×12$? The perimeter for both rooms is the same (52 foot). I'm not asking to see formulas giving me the same results I already know - I'd prefer a more intuitive explanation.
area square-numbers
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
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The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
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– Timbuc
Dec 27 '14 at 15:12
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@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
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– Timbuc
Dec 27 '14 at 15:18
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It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
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– guest
Dec 28 '14 at 3:58
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@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
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– Pedro Tamaroff♦
Dec 28 '14 at 4:39
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@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
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– hobbs
Dec 28 '14 at 6:42
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show 6 more comments
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Why is there an extra square foot in a square room with dimensions of $13×13$ and one less square foot in a room with dimensions of $14×12$? The perimeter for both rooms is the same (52 foot). I'm not asking to see formulas giving me the same results I already know - I'd prefer a more intuitive explanation.
area square-numbers
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
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Why is there an extra square foot in a square room with dimensions of $13×13$ and one less square foot in a room with dimensions of $14×12$? The perimeter for both rooms is the same (52 foot). I'm not asking to see formulas giving me the same results I already know - I'd prefer a more intuitive explanation.
area square-numbers
area square-numbers
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
edited Feb 4 '15 at 3:32
Milo Brandt
39.5k475139
39.5k475139
asked Dec 27 '14 at 15:04
user203450
4
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The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
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– Timbuc
Dec 27 '14 at 15:12
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@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
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– Timbuc
Dec 27 '14 at 15:18
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It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
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– guest
Dec 28 '14 at 3:58
5
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@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
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– Pedro Tamaroff♦
Dec 28 '14 at 4:39
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@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
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– hobbs
Dec 28 '14 at 6:42
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show 6 more comments
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The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
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– Timbuc
Dec 27 '14 at 15:12
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@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
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– Timbuc
Dec 27 '14 at 15:18
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It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
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– guest
Dec 28 '14 at 3:58
5
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@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
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– Pedro Tamaroff♦
Dec 28 '14 at 4:39
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@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
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– hobbs
Dec 28 '14 at 6:42
4
4
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The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
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– Timbuc
Dec 27 '14 at 15:12
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The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
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– Timbuc
Dec 27 '14 at 15:12
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9
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@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
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– Timbuc
Dec 27 '14 at 15:18
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@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
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– Timbuc
Dec 27 '14 at 15:18
5
5
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It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
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– guest
Dec 28 '14 at 3:58
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It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
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– guest
Dec 28 '14 at 3:58
5
5
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@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
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– Pedro Tamaroff♦
Dec 28 '14 at 4:39
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@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
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– Pedro Tamaroff♦
Dec 28 '14 at 4:39
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7
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@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
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– hobbs
Dec 28 '14 at 6:42
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@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
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– hobbs
Dec 28 '14 at 6:42
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13 Answers
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From the comments:
I can take that rectangle, cut the length off, add it to the width, and get the same area as the square.
Well, not quite.
While I wait to properly understand the question, here are some more "ridiculous algebraic formulas". Suppose the size of the square is $a$. Then the dimensions of the rectangle must be $a+b$ and $a-b$ for some $b$. Its area is $(a+b)times(a-b)=a^2-b^2$, which is less than the area of the square, $a^2$.
From the later comments:
Mathematics is a man made concept. Everything has a reason for being the way it is. . . There was a reason the inventors of mathematics created it the way they did. . . A human created the formula to determine the area and the perimeter, etc of something.
I see! You're asking the question "Why does the square have larger area than the rectangle?" in the sense of something like "Why is the speed limit only 20 miles per hour in this podunk town?" Someone was responsible for that speed limit. They could have chosen otherwise, but they chose this. Why?
But here I don't think people had that much choice in the matter. Area is just a name for the number of little squares that can fit inside a given shape. Perimeter is a name for the number of little line segments that can go along the boundary of the shape. Those would still be there even if we hadn't given those concepts a name, just as a lemon would have a colour and a taste even if we didn't have words for yellow and sour. I mean, you can draw a $4times2$ rectangle and a $3times3$ square on a piece of graph paper and count the number of little squares and the number of little lengths on the boundary by hand without using any formulas. Go ahead, try it right now! What part of that feels invented, or man-made, or arbitrary? Did someone else choose those numbers, when you counted them yourself?
It's kind of implicit in your original framing of the question, too, isn't it? You've built a $14times12$ foot room and a $13times13$ foot room, and you find that the $13times13$ room has more space in it. And you want to know why someone defined the formulas so that it works out this way. The idea that some mathematicians in antiquity are responsible, that if only they had defined area and perimeter differently it would somehow change the amount of stuff you can fit into your rooms, well, it doesn't make any sense to me.
answered Dec 27 '14 at 15:55
RahulRahul
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I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
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– André 3000
Dec 27 '14 at 20:28
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@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
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– JMoravitz
Dec 27 '14 at 21:22
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What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
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– JMoravitz
Dec 27 '14 at 21:41
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@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
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– JMoravitz
Dec 28 '14 at 0:21
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@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
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– Rahul
Dec 28 '14 at 3:19
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If I may rephrase the problem, what is the maximum area $A$ of a rectangle of given perimeter $P$ ?
Let us call $a$ and $b$ the sizes of the rectangle. We then have $$A=a times b$$ $$P=2times(a+b)$$ Extract $b$ from $P$; this gives $b=frac P2 -a$; then $$A=a timesBig(frac P2 -aBig)$$ Now, compute the derivative with respect to $a$; so $$frac{dA}{da}=frac{P}{2}-2 a$$ So, an extremum is obtained for $a=frac P4=b$ (this means a square). The second derivative test confirms that this is a maximum.
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
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This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
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– lcv
Dec 27 '14 at 16:01
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Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
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– Claude Leibovici
Dec 27 '14 at 16:06
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I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
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– lcv
Dec 27 '14 at 16:51
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The perimeter and the area of a shape have little or no relation to one another.
Cut yourself a length of string and tie it together to make a loop. You can lay this out on the ground so that it surrounds many different sized areas.
There will be a maximum possible area. This is a famous problem known as the Isoperimetric Problem. It turns out that the maximum enclosed area is given by making a circle.
But what about the smallest possible enclosed area? Well, if you use "mathematical string", i.e. a one dimensional object with no thickness and with perfect "bendability", then you can contrive a shape that encloses an area as small as you like. Think about a long, thin rectangle. If you want the enclosed area to be less then make the rectangle longer and thinner.
Mathematically: Let the length of string be $p$ (for perimeter), and make a rectangle with width $w$ and thickness $t$. The perimeter is, by definition, given by $p = 2w+2t$. The area will, by definition, be given by $a=w t$. The perimeter is a constant and so we can rearrange $p=2w+2t$ to give
$$w=tfrac{1}{2}p -t $$
It follows that the area of the rectangle is given by
$$begin{eqnarray*}
a &=& wt \ \
&=& left(tfrac{1}{2}p -tright)t \ \
&=& tfrac{1}{2}pt - t^2
end{eqnarray*}$$
The area $a=tfrac{1}{2}pt - t^2$ has no minimum. No matter what the perimeter $p$ is, as the thickness $t to 0$, the area $a to 0$. As the thickness $t to 0$, the width $w to tfrac{1}{2}p$.
What about the maximum possible area for a rectangle? (A circle encloses the maximum possible area.) Completing the square on $a=tfrac{1}{2}pt - t^2$ gives
$$a = tfrac{1}{16}p^2-left(t-tfrac{1}{4}pright)^{! 2}$$
This shows that, in the case of the rectangle, the maximum enclosed area is given by making by $t-tfrac{1}{4}p=0$, i.e. by making the thickness one quarter of the perimeter. This means we have made a square. In such a case, the maximum area is $tfrac{1}{16}p^2$.
TLDR: There is no general connection between the perimeter and the enclosed area. You can make an enclosed area as small as you like for any given perimeter. There is, however, a maximum possible area for any given perimeter; this is formed by a circle.
edited Jan 5 at 15:05
greedoid
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answered Dec 28 '14 at 0:32
Fly by NightFly by Night
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I'll try to give an intuitive answer. For the rigorous mathematical argument you have Claude's answer.
Let's imagine we build our rectangles and squares from small $1times 1$ blocks.
To build a rectangle with area $4$, we need to join $4$ small squares consecutively. You'll have two internal squares, in which two sides are belong to the perimeter, and two external squares, in which three sides belong to the perimeter.
So you'll have a perimeter of: $2 times 2 + 2times 3 = 10$
Now, to build a square with area $4$, you put $2$ small squares on top of another two small squares. All of these will have $2$ sides that contribute to the perimeter.
The perimeter will be: $4 times 2 = 8$.
This same argument can be increased to any size square/rectangle. Any rectangle will always have more of these blocks exposed to the outside than a square of the same area.
This proves that a rectangle will always have a larger perimeter than a square with the same area. This implies that if a rectangle and a square have the same perimeter, the rectangle must have a smaller area.
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
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But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
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– user203450
Dec 27 '14 at 21:25
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For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
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– Sciencertobe
Dec 27 '14 at 21:34
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I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
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– GWilliams
Dec 28 '14 at 7:11
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Let's says the rectangle has sides of length $A$ and $B$ and the square has sides of length $C$. Because they have the same perimeter we can say
$$
A + B = 2C , .
$$
Now let's assume that in the rectangle, $A$ is the bigger side and $B$ is the smaller side.
Which means $A$ will be bigger than $C$ and $B$ will be smaller than $C$. That means for some non-negative number $x$
begin{align*}
A = C + x\
B = C - x
end{align*}
so the area of the rectangle will be
$$
AB = (C+x)(C-x) = C^2 - x^2
$$
so the area of the square will be greater than that of the rectangle by the square of difference of the sides length.
edited Dec 28 '14 at 18:45
André 3000
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answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
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Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
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– André 3000
Dec 28 '14 at 18:42
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Also, I really like this answer. It goes well with the picture in Rahul's answer.
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– André 3000
Dec 28 '14 at 18:47
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@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
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– Ritesh Srivastava
Dec 29 '14 at 12:09
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My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.
answered Dec 28 '14 at 6:35
hobbshobbs
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@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
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– hobbs
Dec 28 '14 at 6:48
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Let $;R;$ be a rectange with sides $;x,y;$ s.t $;x+y=k;,;;kinBbb N;$ fixed, let us find the maximum of $;f(x,y):=xy$:
$$x+y=kimplies y=k-ximplies g(x)=f(x,k-x)=x(k-x)$$
From here:
$$g'(x)=k-2x=0iff x=frac k2implies y=frac k2$$
and the rectangle is in fact a square. you now prove that in the above we really get a maximum point.
Another more basic way with the same notation:
$$g(x)=-x^2+kx , ;;text{and this is a parabola that open downwards}$$
and thus its vertex is a maximal point. But the vertex is given by
$$left(;-frac k{-2};,;;-fracDelta{-4};right)=left(;frac k2;,;;frac{k^2}4;right)$$
and again $;x=frac k2;$ and etc.
answered Dec 27 '14 at 15:18
TimbucTimbuc
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This is age !! Thanks for pointing it.
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– Claude Leibovici
Dec 27 '14 at 15:30
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@ClaudeLeibovici Hehe...any time, Claude.
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– Timbuc
Dec 27 '14 at 15:31
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This is both in response to the comments made in Rahul's answer, and also as an additional graphical proof.
In addition, let me emphasize that to mathematicians, "Why" something is true is explained in mathematical proof. Explaining things via formula and pictures and such is not a "How" but is indeed a "Why". Also let me emphasize that when beginning anything it is important to remember the definitions. Area of a shape is loosely defined as "how many unit squares can fit in a shape", and for rectangles in particular it is explicitly defined as $Atimes B$ where $A$ and $B$ are the lengths of the sides.
Claim: All rectangles of a given perimeter have less than or equal area to that of the square of the same perimeter. Equivalently, a square of a given perimeter is of greater area than all rectangles of equal perimeter where the rectangle is not a square.
Without loss of generality let the rectangle we are comparing to be longer than it is tall. Having started with a square of side length $B$, choose some positive value of $xleq B$ and relabel the side length as $B=A+x$ where $A = B-x$
Make a horizontal slice a distance of $x$ away from the top extending across the entire square and a vertical slice a distance of $x$ away from the right which extends from the top to the previous slice. By removing the upper-right piece you effectively reduce the area. You may then move the upper-left piece and rotate it to fit nicely with the lower piece. The resulting formed rectangle is of the same perimeter as the original square, however the area is smaller.
As such the area of the blue square is $A^2 + 2Ax + x^2$ which is strictly greater than the area of the red square which is $A^2 + 2Ax$ while the perimeter remain the same.
By changing the value of $x$, you are able to form any rectangle of given perimeter, and if $x=B$ then you get a line segment which has zero area. $square$
If you are looking for a simpler answer, the "why" is because area and perimeter measure two completely different things. If you are looking for a more philosophical answer, then I would direct you to read Plato and Pythagoras.
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
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Take a 13x13 square, and for convenience's sake, draw it on a 1x1 grid, so it takes up 13x13 squares.
Pick one corner of the square (let's say top left), and remove the 1x1 square in that corner from the big square. The shape you have now (big-square-minus-little-square) has the same perimeter as the big square that you started with, but obviously its area is 1 square less than the area of the big square that you started with.
Now remove another square, adjacent to the one that you first removed (let's say top almost-left). You now have a 13x13 square from which you removed a 2x1 rectangle. The perimeter is still the same, but the area has been diminished by two.
Keep removing 1x1 squares along the same edge (top edge), until there is only one 1x1 square left on that edge. You now have a 13x13 square from which you removed a 12x1 rectangle, without changing its perimeter. You can also think of it as a 13x12 rectangle with a single 1x1 square hanging off one corner (at the top of the top right corner).
Now take that one extra 1x1 square and move it from the edge it's on (top edge) to its adjacent edge (right edge), so that now you have a 13x12 rectangle with a single 1x1 square hanging off the same corner, but in the other direction (at the right of the top right corner). You can also think of it as a 14x12 rectangle with a 11x1 rectangle missing. In this step, you once again did not change the perimeter, and you also did not change the area.
Now, starting from the 1x1 square that's hanging off the corner, add one missing 1x1 square back next to it (just below it). You now have a 14x12 rectangle with a 10x1 square missing. In this step, you did not change the perimeter, but you increased the area by 1.
Keep adding 1x1 squares back until you have a full 14x12 rectangle. Every time you add a 1x1 square back, you are increasing the area by 1, without changing perimeter.
Now look at what you did over all: you removed 13-1=12 squares off the top, but you added 12-1=11 squares on the right. Therefore, the area decreased overall from start to finish.
If you think about how this process works, you will see that if you start with a square, you will always end up removing more 1x1 squares from the top than you add on the right, and therefore the area will always decrease overall. (Try it for a few different size squares to see why.)
This process even works if you start with a rectangle, as long as you always remove squares from a long edge and put them back on a short edge. For example, you can start with a 13x13 square and go to 14x12 (removing some area) and then to 15x11 (removing some more area). This lets you start with a square and change the edges by more than 1, and still be sure that some area will be lost.
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
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add a comment |
$begingroup$
Robert, to keep it simple think of perimeter as 'Exposed Surface'.
Perhaps a visual illustration of 'why' will be of more use to you than an algebraic one?
Start with this:
Then with that in mind, this:
Notice how 'Area' and 'Total Surface' stay constant throughout the shape change (folding) process. Modifying the shape of an object directly influences the ratio of 'Exposed' & 'Concealed' surfaces (measured in terms of discrete units). Therfore the 'perimiter' of an object can be modified significantly without the need to change its area.
Also notice how the more compact we make the object (by reducing the amount of exposed surface, and therfore increasing the concealed surface) the more square like it becomes, this is because the square is the most compact form of rectangle, it has the least amount of exposed surface possible for a 'rectangular shaped object'.
As for the units used, they don't really matter, we could be referring to (m), (cm), (mm), etc. down to atomic distances if you like:
Either way the concept is the same; you break the object down to discrete units, and from there you can analyse the relationship between shape, area, and perimeter.
answered Dec 28 '14 at 9:54
nobodynobody
112
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add a comment |
$begingroup$
Constant perimeter does not imply constant area, because the later depends on $A+B$ while the former depends on $AB$. And recaling the AM-GM inequality:
$$ sqrt{A B } le frac{A+B}{2} implies S=A B le frac{(A+B)^2}{4}=frac{P^2}{16}$$
The equality happens when (and only when) $A=B$
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
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add a comment |
$begingroup$
Consider a rectangle with sides of length $a$ and $b$, and area $ab$. The average of those lengths is $c=frac{a+b}{2}$, and we can see that each of $a$ and $b$ differ from that average by the same amount, let's call it $d$. So assuming WLOG that $a>b$, $a=c+d$ and $b=c-d$, and the area of the rectangle $ab=(c+d)(c-d)=c^2-d^2$. For a constant perimeter $P$, the value of $a+b=frac{P}{2}$ does not change, so $c$ does not change. The way to obtain the largest possible area ($c^2-d^2$) given a constant perimeter is to make $d^2$ as small as possible. This is done by making $d=0$. In that case, $a=c$ and $b=c$ so $a=b$ and the rectangle is a square. Thus for a given perimeter, every (non-square) rectangle is smaller than the corresponding square.
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
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add a comment |
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The following proves that a square always has a greater area than a non-square rectangle of the same perimeter.
There are many ways to prove this. I have designed this particular proof so that almost anyone, even without much mathematical experience, can understand it.
(Click the image to make it bigger.)
answered Jan 5 at 5:08
MagnusMagnus
1113
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add a comment |
protected by Pedro Tamaroff♦ Dec 28 '14 at 11:05
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From the comments:
I can take that rectangle, cut the length off, add it to the width, and get the same area as the square.
Well, not quite.
While I wait to properly understand the question, here are some more "ridiculous algebraic formulas". Suppose the size of the square is $a$. Then the dimensions of the rectangle must be $a+b$ and $a-b$ for some $b$. Its area is $(a+b)times(a-b)=a^2-b^2$, which is less than the area of the square, $a^2$.
From the later comments:
Mathematics is a man made concept. Everything has a reason for being the way it is. . . There was a reason the inventors of mathematics created it the way they did. . . A human created the formula to determine the area and the perimeter, etc of something.
I see! You're asking the question "Why does the square have larger area than the rectangle?" in the sense of something like "Why is the speed limit only 20 miles per hour in this podunk town?" Someone was responsible for that speed limit. They could have chosen otherwise, but they chose this. Why?
But here I don't think people had that much choice in the matter. Area is just a name for the number of little squares that can fit inside a given shape. Perimeter is a name for the number of little line segments that can go along the boundary of the shape. Those would still be there even if we hadn't given those concepts a name, just as a lemon would have a colour and a taste even if we didn't have words for yellow and sour. I mean, you can draw a $4times2$ rectangle and a $3times3$ square on a piece of graph paper and count the number of little squares and the number of little lengths on the boundary by hand without using any formulas. Go ahead, try it right now! What part of that feels invented, or man-made, or arbitrary? Did someone else choose those numbers, when you counted them yourself?
It's kind of implicit in your original framing of the question, too, isn't it? You've built a $14times12$ foot room and a $13times13$ foot room, and you find that the $13times13$ room has more space in it. And you want to know why someone defined the formulas so that it works out this way. The idea that some mathematicians in antiquity are responsible, that if only they had defined area and perimeter differently it would somehow change the amount of stuff you can fit into your rooms, well, it doesn't make any sense to me.
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
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6
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I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
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– André 3000
Dec 27 '14 at 20:28
4
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@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
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– JMoravitz
Dec 27 '14 at 21:22
6
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What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
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– JMoravitz
Dec 27 '14 at 21:41
5
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
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– JMoravitz
Dec 28 '14 at 0:21
6
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
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– Rahul
Dec 28 '14 at 3:19
|
show 7 more comments
$begingroup$
From the comments:
I can take that rectangle, cut the length off, add it to the width, and get the same area as the square.
Well, not quite.
While I wait to properly understand the question, here are some more "ridiculous algebraic formulas". Suppose the size of the square is $a$. Then the dimensions of the rectangle must be $a+b$ and $a-b$ for some $b$. Its area is $(a+b)times(a-b)=a^2-b^2$, which is less than the area of the square, $a^2$.
From the later comments:
Mathematics is a man made concept. Everything has a reason for being the way it is. . . There was a reason the inventors of mathematics created it the way they did. . . A human created the formula to determine the area and the perimeter, etc of something.
I see! You're asking the question "Why does the square have larger area than the rectangle?" in the sense of something like "Why is the speed limit only 20 miles per hour in this podunk town?" Someone was responsible for that speed limit. They could have chosen otherwise, but they chose this. Why?
But here I don't think people had that much choice in the matter. Area is just a name for the number of little squares that can fit inside a given shape. Perimeter is a name for the number of little line segments that can go along the boundary of the shape. Those would still be there even if we hadn't given those concepts a name, just as a lemon would have a colour and a taste even if we didn't have words for yellow and sour. I mean, you can draw a $4times2$ rectangle and a $3times3$ square on a piece of graph paper and count the number of little squares and the number of little lengths on the boundary by hand without using any formulas. Go ahead, try it right now! What part of that feels invented, or man-made, or arbitrary? Did someone else choose those numbers, when you counted them yourself?
It's kind of implicit in your original framing of the question, too, isn't it? You've built a $14times12$ foot room and a $13times13$ foot room, and you find that the $13times13$ room has more space in it. And you want to know why someone defined the formulas so that it works out this way. The idea that some mathematicians in antiquity are responsible, that if only they had defined area and perimeter differently it would somehow change the amount of stuff you can fit into your rooms, well, it doesn't make any sense to me.
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
$endgroup$
6
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I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
$endgroup$
– André 3000
Dec 27 '14 at 20:28
4
$begingroup$
@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:22
6
$begingroup$
What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:41
5
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
$endgroup$
– JMoravitz
Dec 28 '14 at 0:21
6
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
$endgroup$
– Rahul
Dec 28 '14 at 3:19
|
show 7 more comments
$begingroup$
From the comments:
I can take that rectangle, cut the length off, add it to the width, and get the same area as the square.
Well, not quite.
While I wait to properly understand the question, here are some more "ridiculous algebraic formulas". Suppose the size of the square is $a$. Then the dimensions of the rectangle must be $a+b$ and $a-b$ for some $b$. Its area is $(a+b)times(a-b)=a^2-b^2$, which is less than the area of the square, $a^2$.
From the later comments:
Mathematics is a man made concept. Everything has a reason for being the way it is. . . There was a reason the inventors of mathematics created it the way they did. . . A human created the formula to determine the area and the perimeter, etc of something.
I see! You're asking the question "Why does the square have larger area than the rectangle?" in the sense of something like "Why is the speed limit only 20 miles per hour in this podunk town?" Someone was responsible for that speed limit. They could have chosen otherwise, but they chose this. Why?
But here I don't think people had that much choice in the matter. Area is just a name for the number of little squares that can fit inside a given shape. Perimeter is a name for the number of little line segments that can go along the boundary of the shape. Those would still be there even if we hadn't given those concepts a name, just as a lemon would have a colour and a taste even if we didn't have words for yellow and sour. I mean, you can draw a $4times2$ rectangle and a $3times3$ square on a piece of graph paper and count the number of little squares and the number of little lengths on the boundary by hand without using any formulas. Go ahead, try it right now! What part of that feels invented, or man-made, or arbitrary? Did someone else choose those numbers, when you counted them yourself?
It's kind of implicit in your original framing of the question, too, isn't it? You've built a $14times12$ foot room and a $13times13$ foot room, and you find that the $13times13$ room has more space in it. And you want to know why someone defined the formulas so that it works out this way. The idea that some mathematicians in antiquity are responsible, that if only they had defined area and perimeter differently it would somehow change the amount of stuff you can fit into your rooms, well, it doesn't make any sense to me.
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
$endgroup$
From the comments:
I can take that rectangle, cut the length off, add it to the width, and get the same area as the square.
Well, not quite.
While I wait to properly understand the question, here are some more "ridiculous algebraic formulas". Suppose the size of the square is $a$. Then the dimensions of the rectangle must be $a+b$ and $a-b$ for some $b$. Its area is $(a+b)times(a-b)=a^2-b^2$, which is less than the area of the square, $a^2$.
From the later comments:
Mathematics is a man made concept. Everything has a reason for being the way it is. . . There was a reason the inventors of mathematics created it the way they did. . . A human created the formula to determine the area and the perimeter, etc of something.
I see! You're asking the question "Why does the square have larger area than the rectangle?" in the sense of something like "Why is the speed limit only 20 miles per hour in this podunk town?" Someone was responsible for that speed limit. They could have chosen otherwise, but they chose this. Why?
But here I don't think people had that much choice in the matter. Area is just a name for the number of little squares that can fit inside a given shape. Perimeter is a name for the number of little line segments that can go along the boundary of the shape. Those would still be there even if we hadn't given those concepts a name, just as a lemon would have a colour and a taste even if we didn't have words for yellow and sour. I mean, you can draw a $4times2$ rectangle and a $3times3$ square on a piece of graph paper and count the number of little squares and the number of little lengths on the boundary by hand without using any formulas. Go ahead, try it right now! What part of that feels invented, or man-made, or arbitrary? Did someone else choose those numbers, when you counted them yourself?
It's kind of implicit in your original framing of the question, too, isn't it? You've built a $14times12$ foot room and a $13times13$ foot room, and you find that the $13times13$ room has more space in it. And you want to know why someone defined the formulas so that it works out this way. The idea that some mathematicians in antiquity are responsible, that if only they had defined area and perimeter differently it would somehow change the amount of stuff you can fit into your rooms, well, it doesn't make any sense to me.
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
edited Dec 29 '14 at 4:08
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
answered Dec 27 '14 at 15:55
RahulRahul
33k568166
33k568166
6
$begingroup$
I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
$endgroup$
– André 3000
Dec 27 '14 at 20:28
4
$begingroup$
@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:22
6
$begingroup$
What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:41
5
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
$endgroup$
– JMoravitz
Dec 28 '14 at 0:21
6
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
$endgroup$
– Rahul
Dec 28 '14 at 3:19
|
show 7 more comments
6
$begingroup$
I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
$endgroup$
– André 3000
Dec 27 '14 at 20:28
4
$begingroup$
@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:22
6
$begingroup$
What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:41
5
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
$endgroup$
– JMoravitz
Dec 28 '14 at 0:21
6
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
$endgroup$
– Rahul
Dec 28 '14 at 3:19
6
6
$begingroup$
I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
$endgroup$
– André 3000
Dec 27 '14 at 20:28
$begingroup$
I disagree. This answer shows that a claim you made in the comments is false, which seems to be the main misconception.
$endgroup$
– André 3000
Dec 27 '14 at 20:28
4
4
$begingroup$
@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:22
$begingroup$
@Robert Perhaps its not explicitly answering the question and is more aimed at attempting to fix a misconception you had, however you are unable to post images in a comment. If you are so bothered by it though, then perhaps Rahul should have began with "This is in response to your comment above, yada yada, it cant fit in a comment reply." Even so, by fixing the aforementioned misconception that should be answer enough to your original question.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:22
6
6
$begingroup$
What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:41
$begingroup$
What about the image is confusing? This clearly shows an example of two rectangular objects with same perimeter and different area. The red rectangle on the far left and the blue square on the right both have the same perimeter, yet the blue square has more area.
$endgroup$
– JMoravitz
Dec 27 '14 at 21:41
5
5
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
$endgroup$
– JMoravitz
Dec 28 '14 at 0:21
$begingroup$
@Robert Apparently we don't understand what you mean by the difference between "how" and "why". To mathematicians, "why something is true" is given by mathematical proof which is a connected series of logical if-then statements which start from basic principles and reach what it is we want to prove. This is exactly what has been done in each of the posts in various forms.
$endgroup$
– JMoravitz
Dec 28 '14 at 0:21
6
6
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
$endgroup$
– Rahul
Dec 28 '14 at 3:19
$begingroup$
@Robert: That said, I'm still not clear on what sort of explanation you're looking for. I generally don't know how to explain why two different things are not the same. But maybe you can help me understand. Suppose someone asks (just as an example), "Why is $1times 2=2$ even though $1+2=3$? Shouldn't $1times 2$ also be $3$?" what sort of explanation would you want to give them?
$endgroup$
– Rahul
Dec 28 '14 at 3:19
|
show 7 more comments
$begingroup$
If I may rephrase the problem, what is the maximum area $A$ of a rectangle of given perimeter $P$ ?
Let us call $a$ and $b$ the sizes of the rectangle. We then have $$A=a times b$$ $$P=2times(a+b)$$ Extract $b$ from $P$; this gives $b=frac P2 -a$; then $$A=a timesBig(frac P2 -aBig)$$ Now, compute the derivative with respect to $a$; so $$frac{dA}{da}=frac{P}{2}-2 a$$ So, an extremum is obtained for $a=frac P4=b$ (this means a square). The second derivative test confirms that this is a maximum.
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
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1
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This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
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– lcv
Dec 27 '14 at 16:01
2
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
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– Claude Leibovici
Dec 27 '14 at 16:06
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I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
add a comment |
$begingroup$
If I may rephrase the problem, what is the maximum area $A$ of a rectangle of given perimeter $P$ ?
Let us call $a$ and $b$ the sizes of the rectangle. We then have $$A=a times b$$ $$P=2times(a+b)$$ Extract $b$ from $P$; this gives $b=frac P2 -a$; then $$A=a timesBig(frac P2 -aBig)$$ Now, compute the derivative with respect to $a$; so $$frac{dA}{da}=frac{P}{2}-2 a$$ So, an extremum is obtained for $a=frac P4=b$ (this means a square). The second derivative test confirms that this is a maximum.
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
1
$begingroup$
This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
$endgroup$
– lcv
Dec 27 '14 at 16:01
2
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
$endgroup$
– Claude Leibovici
Dec 27 '14 at 16:06
$begingroup$
I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
add a comment |
$begingroup$
If I may rephrase the problem, what is the maximum area $A$ of a rectangle of given perimeter $P$ ?
Let us call $a$ and $b$ the sizes of the rectangle. We then have $$A=a times b$$ $$P=2times(a+b)$$ Extract $b$ from $P$; this gives $b=frac P2 -a$; then $$A=a timesBig(frac P2 -aBig)$$ Now, compute the derivative with respect to $a$; so $$frac{dA}{da}=frac{P}{2}-2 a$$ So, an extremum is obtained for $a=frac P4=b$ (this means a square). The second derivative test confirms that this is a maximum.
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
If I may rephrase the problem, what is the maximum area $A$ of a rectangle of given perimeter $P$ ?
Let us call $a$ and $b$ the sizes of the rectangle. We then have $$A=a times b$$ $$P=2times(a+b)$$ Extract $b$ from $P$; this gives $b=frac P2 -a$; then $$A=a timesBig(frac P2 -aBig)$$ Now, compute the derivative with respect to $a$; so $$frac{dA}{da}=frac{P}{2}-2 a$$ So, an extremum is obtained for $a=frac P4=b$ (this means a square). The second derivative test confirms that this is a maximum.
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
edited Dec 28 '14 at 15:04
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
answered Dec 27 '14 at 15:19
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
1
$begingroup$
This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
$endgroup$
– lcv
Dec 27 '14 at 16:01
2
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
$endgroup$
– Claude Leibovici
Dec 27 '14 at 16:06
$begingroup$
I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
add a comment |
1
$begingroup$
This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
$endgroup$
– lcv
Dec 27 '14 at 16:01
2
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
$endgroup$
– Claude Leibovici
Dec 27 '14 at 16:06
$begingroup$
I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
1
1
$begingroup$
This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
$endgroup$
– lcv
Dec 27 '14 at 16:01
$begingroup$
This is clearly correct but does it really answer the question? Somehow I find the real question - why? - more interesting. From the answer it's not clear why the maximum configuration has both sides of the same length (i.e. it's a square). I mean symmetry considerations are not taken into account (in an explicit way).
$endgroup$
– lcv
Dec 27 '14 at 16:01
2
2
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
$endgroup$
– Claude Leibovici
Dec 27 '14 at 16:06
$begingroup$
Interesting debate ! Beside showing that the maximum of $atimes b$ for $a+b=k$ is obtained when $a=b$, I really don't know. I would enjoy reading your opinion. Cheers :-)
$endgroup$
– Claude Leibovici
Dec 27 '14 at 16:06
$begingroup$
I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
$begingroup$
I suppose the "true" why is a little subtle as it depends on the geometry (which is implicitly assumed euclidean). Let me try this way. Fix a reference cartesian axis (horizontal and vertical) and measure sides $a, b$ to be the horizontal/vertical. Now you do the constrained maximization procedure and obtain $a^ast, b^ast$. We are trying to assume that $a^ast > b^ast$ for instance. Now flip the horizontal and vertical axis. The result of the constrained maximization now must be the flipped rectangle. But clearly the result cannot depend on the reference frame.
$endgroup$
– lcv
Dec 27 '14 at 16:51
add a comment |
$begingroup$
The perimeter and the area of a shape have little or no relation to one another.
Cut yourself a length of string and tie it together to make a loop. You can lay this out on the ground so that it surrounds many different sized areas.
There will be a maximum possible area. This is a famous problem known as the Isoperimetric Problem. It turns out that the maximum enclosed area is given by making a circle.
But what about the smallest possible enclosed area? Well, if you use "mathematical string", i.e. a one dimensional object with no thickness and with perfect "bendability", then you can contrive a shape that encloses an area as small as you like. Think about a long, thin rectangle. If you want the enclosed area to be less then make the rectangle longer and thinner.
Mathematically: Let the length of string be $p$ (for perimeter), and make a rectangle with width $w$ and thickness $t$. The perimeter is, by definition, given by $p = 2w+2t$. The area will, by definition, be given by $a=w t$. The perimeter is a constant and so we can rearrange $p=2w+2t$ to give
$$w=tfrac{1}{2}p -t $$
It follows that the area of the rectangle is given by
$$begin{eqnarray*}
a &=& wt \ \
&=& left(tfrac{1}{2}p -tright)t \ \
&=& tfrac{1}{2}pt - t^2
end{eqnarray*}$$
The area $a=tfrac{1}{2}pt - t^2$ has no minimum. No matter what the perimeter $p$ is, as the thickness $t to 0$, the area $a to 0$. As the thickness $t to 0$, the width $w to tfrac{1}{2}p$.
What about the maximum possible area for a rectangle? (A circle encloses the maximum possible area.) Completing the square on $a=tfrac{1}{2}pt - t^2$ gives
$$a = tfrac{1}{16}p^2-left(t-tfrac{1}{4}pright)^{! 2}$$
This shows that, in the case of the rectangle, the maximum enclosed area is given by making by $t-tfrac{1}{4}p=0$, i.e. by making the thickness one quarter of the perimeter. This means we have made a square. In such a case, the maximum area is $tfrac{1}{16}p^2$.
TLDR: There is no general connection between the perimeter and the enclosed area. You can make an enclosed area as small as you like for any given perimeter. There is, however, a maximum possible area for any given perimeter; this is formed by a circle.
edited Jan 5 at 15:05
greedoid
39.2k114797
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
$endgroup$
add a comment |
$begingroup$
The perimeter and the area of a shape have little or no relation to one another.
Cut yourself a length of string and tie it together to make a loop. You can lay this out on the ground so that it surrounds many different sized areas.
There will be a maximum possible area. This is a famous problem known as the Isoperimetric Problem. It turns out that the maximum enclosed area is given by making a circle.
But what about the smallest possible enclosed area? Well, if you use "mathematical string", i.e. a one dimensional object with no thickness and with perfect "bendability", then you can contrive a shape that encloses an area as small as you like. Think about a long, thin rectangle. If you want the enclosed area to be less then make the rectangle longer and thinner.
Mathematically: Let the length of string be $p$ (for perimeter), and make a rectangle with width $w$ and thickness $t$. The perimeter is, by definition, given by $p = 2w+2t$. The area will, by definition, be given by $a=w t$. The perimeter is a constant and so we can rearrange $p=2w+2t$ to give
$$w=tfrac{1}{2}p -t $$
It follows that the area of the rectangle is given by
$$begin{eqnarray*}
a &=& wt \ \
&=& left(tfrac{1}{2}p -tright)t \ \
&=& tfrac{1}{2}pt - t^2
end{eqnarray*}$$
The area $a=tfrac{1}{2}pt - t^2$ has no minimum. No matter what the perimeter $p$ is, as the thickness $t to 0$, the area $a to 0$. As the thickness $t to 0$, the width $w to tfrac{1}{2}p$.
What about the maximum possible area for a rectangle? (A circle encloses the maximum possible area.) Completing the square on $a=tfrac{1}{2}pt - t^2$ gives
$$a = tfrac{1}{16}p^2-left(t-tfrac{1}{4}pright)^{! 2}$$
This shows that, in the case of the rectangle, the maximum enclosed area is given by making by $t-tfrac{1}{4}p=0$, i.e. by making the thickness one quarter of the perimeter. This means we have made a square. In such a case, the maximum area is $tfrac{1}{16}p^2$.
TLDR: There is no general connection between the perimeter and the enclosed area. You can make an enclosed area as small as you like for any given perimeter. There is, however, a maximum possible area for any given perimeter; this is formed by a circle.
edited Jan 5 at 15:05
greedoid
39.2k114797
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
$endgroup$
add a comment |
$begingroup$
The perimeter and the area of a shape have little or no relation to one another.
Cut yourself a length of string and tie it together to make a loop. You can lay this out on the ground so that it surrounds many different sized areas.
There will be a maximum possible area. This is a famous problem known as the Isoperimetric Problem. It turns out that the maximum enclosed area is given by making a circle.
But what about the smallest possible enclosed area? Well, if you use "mathematical string", i.e. a one dimensional object with no thickness and with perfect "bendability", then you can contrive a shape that encloses an area as small as you like. Think about a long, thin rectangle. If you want the enclosed area to be less then make the rectangle longer and thinner.
Mathematically: Let the length of string be $p$ (for perimeter), and make a rectangle with width $w$ and thickness $t$. The perimeter is, by definition, given by $p = 2w+2t$. The area will, by definition, be given by $a=w t$. The perimeter is a constant and so we can rearrange $p=2w+2t$ to give
$$w=tfrac{1}{2}p -t $$
It follows that the area of the rectangle is given by
$$begin{eqnarray*}
a &=& wt \ \
&=& left(tfrac{1}{2}p -tright)t \ \
&=& tfrac{1}{2}pt - t^2
end{eqnarray*}$$
The area $a=tfrac{1}{2}pt - t^2$ has no minimum. No matter what the perimeter $p$ is, as the thickness $t to 0$, the area $a to 0$. As the thickness $t to 0$, the width $w to tfrac{1}{2}p$.
What about the maximum possible area for a rectangle? (A circle encloses the maximum possible area.) Completing the square on $a=tfrac{1}{2}pt - t^2$ gives
$$a = tfrac{1}{16}p^2-left(t-tfrac{1}{4}pright)^{! 2}$$
This shows that, in the case of the rectangle, the maximum enclosed area is given by making by $t-tfrac{1}{4}p=0$, i.e. by making the thickness one quarter of the perimeter. This means we have made a square. In such a case, the maximum area is $tfrac{1}{16}p^2$.
TLDR: There is no general connection between the perimeter and the enclosed area. You can make an enclosed area as small as you like for any given perimeter. There is, however, a maximum possible area for any given perimeter; this is formed by a circle.
edited Jan 5 at 15:05
greedoid
39.2k114797
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
$endgroup$
The perimeter and the area of a shape have little or no relation to one another.
Cut yourself a length of string and tie it together to make a loop. You can lay this out on the ground so that it surrounds many different sized areas.
There will be a maximum possible area. This is a famous problem known as the Isoperimetric Problem. It turns out that the maximum enclosed area is given by making a circle.
But what about the smallest possible enclosed area? Well, if you use "mathematical string", i.e. a one dimensional object with no thickness and with perfect "bendability", then you can contrive a shape that encloses an area as small as you like. Think about a long, thin rectangle. If you want the enclosed area to be less then make the rectangle longer and thinner.
Mathematically: Let the length of string be $p$ (for perimeter), and make a rectangle with width $w$ and thickness $t$. The perimeter is, by definition, given by $p = 2w+2t$. The area will, by definition, be given by $a=w t$. The perimeter is a constant and so we can rearrange $p=2w+2t$ to give
$$w=tfrac{1}{2}p -t $$
It follows that the area of the rectangle is given by
$$begin{eqnarray*}
a &=& wt \ \
&=& left(tfrac{1}{2}p -tright)t \ \
&=& tfrac{1}{2}pt - t^2
end{eqnarray*}$$
The area $a=tfrac{1}{2}pt - t^2$ has no minimum. No matter what the perimeter $p$ is, as the thickness $t to 0$, the area $a to 0$. As the thickness $t to 0$, the width $w to tfrac{1}{2}p$.
What about the maximum possible area for a rectangle? (A circle encloses the maximum possible area.) Completing the square on $a=tfrac{1}{2}pt - t^2$ gives
$$a = tfrac{1}{16}p^2-left(t-tfrac{1}{4}pright)^{! 2}$$
This shows that, in the case of the rectangle, the maximum enclosed area is given by making by $t-tfrac{1}{4}p=0$, i.e. by making the thickness one quarter of the perimeter. This means we have made a square. In such a case, the maximum area is $tfrac{1}{16}p^2$.
TLDR: There is no general connection between the perimeter and the enclosed area. You can make an enclosed area as small as you like for any given perimeter. There is, however, a maximum possible area for any given perimeter; this is formed by a circle.
edited Jan 5 at 15:05
greedoid
39.2k114797
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
edited Jan 5 at 15:05
greedoid
39.2k114797
edited Jan 5 at 15:05
greedoid
39.2k114797
edited Jan 5 at 15:05
greedoid
39.2k114797
39.2k114797
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
answered Dec 28 '14 at 0:32
Fly by NightFly by Night
25.8k32978
25.8k32978
add a comment |
add a comment |
$begingroup$
I'll try to give an intuitive answer. For the rigorous mathematical argument you have Claude's answer.
Let's imagine we build our rectangles and squares from small $1times 1$ blocks.
To build a rectangle with area $4$, we need to join $4$ small squares consecutively. You'll have two internal squares, in which two sides are belong to the perimeter, and two external squares, in which three sides belong to the perimeter.
So you'll have a perimeter of: $2 times 2 + 2times 3 = 10$
Now, to build a square with area $4$, you put $2$ small squares on top of another two small squares. All of these will have $2$ sides that contribute to the perimeter.
The perimeter will be: $4 times 2 = 8$.
This same argument can be increased to any size square/rectangle. Any rectangle will always have more of these blocks exposed to the outside than a square of the same area.
This proves that a rectangle will always have a larger perimeter than a square with the same area. This implies that if a rectangle and a square have the same perimeter, the rectangle must have a smaller area.
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
$endgroup$
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
8
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
add a comment |
$begingroup$
I'll try to give an intuitive answer. For the rigorous mathematical argument you have Claude's answer.
Let's imagine we build our rectangles and squares from small $1times 1$ blocks.
To build a rectangle with area $4$, we need to join $4$ small squares consecutively. You'll have two internal squares, in which two sides are belong to the perimeter, and two external squares, in which three sides belong to the perimeter.
So you'll have a perimeter of: $2 times 2 + 2times 3 = 10$
Now, to build a square with area $4$, you put $2$ small squares on top of another two small squares. All of these will have $2$ sides that contribute to the perimeter.
The perimeter will be: $4 times 2 = 8$.
This same argument can be increased to any size square/rectangle. Any rectangle will always have more of these blocks exposed to the outside than a square of the same area.
This proves that a rectangle will always have a larger perimeter than a square with the same area. This implies that if a rectangle and a square have the same perimeter, the rectangle must have a smaller area.
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
$endgroup$
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
8
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
add a comment |
$begingroup$
I'll try to give an intuitive answer. For the rigorous mathematical argument you have Claude's answer.
Let's imagine we build our rectangles and squares from small $1times 1$ blocks.
To build a rectangle with area $4$, we need to join $4$ small squares consecutively. You'll have two internal squares, in which two sides are belong to the perimeter, and two external squares, in which three sides belong to the perimeter.
So you'll have a perimeter of: $2 times 2 + 2times 3 = 10$
Now, to build a square with area $4$, you put $2$ small squares on top of another two small squares. All of these will have $2$ sides that contribute to the perimeter.
The perimeter will be: $4 times 2 = 8$.
This same argument can be increased to any size square/rectangle. Any rectangle will always have more of these blocks exposed to the outside than a square of the same area.
This proves that a rectangle will always have a larger perimeter than a square with the same area. This implies that if a rectangle and a square have the same perimeter, the rectangle must have a smaller area.
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
$endgroup$
I'll try to give an intuitive answer. For the rigorous mathematical argument you have Claude's answer.
Let's imagine we build our rectangles and squares from small $1times 1$ blocks.
To build a rectangle with area $4$, we need to join $4$ small squares consecutively. You'll have two internal squares, in which two sides are belong to the perimeter, and two external squares, in which three sides belong to the perimeter.
So you'll have a perimeter of: $2 times 2 + 2times 3 = 10$
Now, to build a square with area $4$, you put $2$ small squares on top of another two small squares. All of these will have $2$ sides that contribute to the perimeter.
The perimeter will be: $4 times 2 = 8$.
This same argument can be increased to any size square/rectangle. Any rectangle will always have more of these blocks exposed to the outside than a square of the same area.
This proves that a rectangle will always have a larger perimeter than a square with the same area. This implies that if a rectangle and a square have the same perimeter, the rectangle must have a smaller area.
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
answered Dec 27 '14 at 21:15
SciencertobeSciencertobe
405314
405314
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
8
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
add a comment |
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
8
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
$begingroup$
But why? Why does a square with a longer length and shorter width have a greater area than a rectangle with a shorter length and longer width? Why is the square larger than the rectangle?
$endgroup$
– user203450
Dec 27 '14 at 21:25
8
8
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
For me, this is the why. When you divide both into equal little parts, the rectangle will always have more of them touching the outside. So the perimeter of a rectangle is larger than that of an square of the same area. This means that if both have the same perimeter, the square will be larger Of course,this is just rewriting my answer, if that isn't an enough why for you, I apologize.
$endgroup$
– Sciencertobe
Dec 27 '14 at 21:34
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
$begingroup$
I think this is the most informative answer. Perhaps a more intuitive question is, why is it that given a rectangle and a square of equal area, the rectangle always has greater than or equal perimeter? And it's because the square has fewer unit square edges touching! Very good comment.
$endgroup$
– GWilliams
Dec 28 '14 at 7:11
add a comment |
$begingroup$
Let's says the rectangle has sides of length $A$ and $B$ and the square has sides of length $C$. Because they have the same perimeter we can say
$$
A + B = 2C , .
$$
Now let's assume that in the rectangle, $A$ is the bigger side and $B$ is the smaller side.
Which means $A$ will be bigger than $C$ and $B$ will be smaller than $C$. That means for some non-negative number $x$
begin{align*}
A = C + x\
B = C - x
end{align*}
so the area of the rectangle will be
$$
AB = (C+x)(C-x) = C^2 - x^2
$$
so the area of the square will be greater than that of the rectangle by the square of difference of the sides length.
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
$endgroup$
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
1
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
add a comment |
$begingroup$
Let's says the rectangle has sides of length $A$ and $B$ and the square has sides of length $C$. Because they have the same perimeter we can say
$$
A + B = 2C , .
$$
Now let's assume that in the rectangle, $A$ is the bigger side and $B$ is the smaller side.
Which means $A$ will be bigger than $C$ and $B$ will be smaller than $C$. That means for some non-negative number $x$
begin{align*}
A = C + x\
B = C - x
end{align*}
so the area of the rectangle will be
$$
AB = (C+x)(C-x) = C^2 - x^2
$$
so the area of the square will be greater than that of the rectangle by the square of difference of the sides length.
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
$endgroup$
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
1
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
add a comment |
$begingroup$
Let's says the rectangle has sides of length $A$ and $B$ and the square has sides of length $C$. Because they have the same perimeter we can say
$$
A + B = 2C , .
$$
Now let's assume that in the rectangle, $A$ is the bigger side and $B$ is the smaller side.
Which means $A$ will be bigger than $C$ and $B$ will be smaller than $C$. That means for some non-negative number $x$
begin{align*}
A = C + x\
B = C - x
end{align*}
so the area of the rectangle will be
$$
AB = (C+x)(C-x) = C^2 - x^2
$$
so the area of the square will be greater than that of the rectangle by the square of difference of the sides length.
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
$endgroup$
Let's says the rectangle has sides of length $A$ and $B$ and the square has sides of length $C$. Because they have the same perimeter we can say
$$
A + B = 2C , .
$$
Now let's assume that in the rectangle, $A$ is the bigger side and $B$ is the smaller side.
Which means $A$ will be bigger than $C$ and $B$ will be smaller than $C$. That means for some non-negative number $x$
begin{align*}
A = C + x\
B = C - x
end{align*}
so the area of the rectangle will be
$$
AB = (C+x)(C-x) = C^2 - x^2
$$
so the area of the square will be greater than that of the rectangle by the square of difference of the sides length.
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
edited Dec 28 '14 at 18:45
André 3000
12.5k22143
12.5k22143
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
answered Dec 28 '14 at 9:41
Ritesh SrivastavaRitesh Srivastava
312
312
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
1
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
add a comment |
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
1
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
$begingroup$
Welcome to math.se! You can use Mathjax to typeset math on this site; see here for a tutorial. I've edited the formatting of your answer, but feel free to change it if you don't like my edit.
$endgroup$
– André 3000
Dec 28 '14 at 18:42
1
1
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
Also, I really like this answer. It goes well with the picture in Rahul's answer.
$endgroup$
– André 3000
Dec 28 '14 at 18:47
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
$begingroup$
@SpamIAm Thanks for the edit. Also I think OP is asking the question in the line of this question: Why Circle encloses largest Area?
$endgroup$
– Ritesh Srivastava
Dec 29 '14 at 12:09
add a comment |
$begingroup$
My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.
answered Dec 28 '14 at 6:35
hobbshobbs
21628
$endgroup$
1
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
add a comment |
$begingroup$
My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.
answered Dec 28 '14 at 6:35
hobbshobbs
21628
$endgroup$
1
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
add a comment |
$begingroup$
My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.
answered Dec 28 '14 at 6:35
hobbshobbs
21628
$endgroup$
My attempt at making it as simple as possible:
Suppose you have a rectangle that isn't a square. Then it's longer in one dimension than the other.
Now take a little bit (say 1 inch) off of the long sides and add it to the short sides.
Since you added 1 inch to two sides, and removed 1 inch from two sides, the perimeter stays the same.
By adding 1 inch to the short sides, you added (1 inch) * (long side) to the area. By subtracting 1 inch from the long sides, you subtracted (1 inch) * (short side) from the area. Since the long sides were longer than the short sides to begin with, you added more area than you subtracted.
Therefore you can always increase a rectangle's area, without changing its perimeter, by transferring length from the long sides to the short sides (as long as the amount of length that you transfer isn't more than half the difference between them).
The only time this isn't possible is when all of the sides are the same length — that is, when the rectangle is a square. Therefore, a square maximizes area for a given perimeter.
answered Dec 28 '14 at 6:35
hobbshobbs
21628
edited Mar 1 '16 at 4:15
answered Dec 28 '14 at 6:35
hobbshobbs
21628
answered Dec 28 '14 at 6:35
hobbshobbs
21628
answered Dec 28 '14 at 6:35
hobbshobbs
21628
21628
1
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
add a comment |
1
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
1
1
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
$begingroup$
@IwillnotexistIdonotexist fixed, although as long as it's strictly less than the difference, you would still gain area, you would just also interchange the long and short sides :)
$endgroup$
– hobbs
Dec 28 '14 at 6:48
add a comment |
$begingroup$
Let $;R;$ be a rectange with sides $;x,y;$ s.t $;x+y=k;,;;kinBbb N;$ fixed, let us find the maximum of $;f(x,y):=xy$:
$$x+y=kimplies y=k-ximplies g(x)=f(x,k-x)=x(k-x)$$
From here:
$$g'(x)=k-2x=0iff x=frac k2implies y=frac k2$$
and the rectangle is in fact a square. you now prove that in the above we really get a maximum point.
Another more basic way with the same notation:
$$g(x)=-x^2+kx , ;;text{and this is a parabola that open downwards}$$
and thus its vertex is a maximal point. But the vertex is given by
$$left(;-frac k{-2};,;;-fracDelta{-4};right)=left(;frac k2;,;;frac{k^2}4;right)$$
and again $;x=frac k2;$ and etc.
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
$endgroup$
1
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
add a comment |
$begingroup$
Let $;R;$ be a rectange with sides $;x,y;$ s.t $;x+y=k;,;;kinBbb N;$ fixed, let us find the maximum of $;f(x,y):=xy$:
$$x+y=kimplies y=k-ximplies g(x)=f(x,k-x)=x(k-x)$$
From here:
$$g'(x)=k-2x=0iff x=frac k2implies y=frac k2$$
and the rectangle is in fact a square. you now prove that in the above we really get a maximum point.
Another more basic way with the same notation:
$$g(x)=-x^2+kx , ;;text{and this is a parabola that open downwards}$$
and thus its vertex is a maximal point. But the vertex is given by
$$left(;-frac k{-2};,;;-fracDelta{-4};right)=left(;frac k2;,;;frac{k^2}4;right)$$
and again $;x=frac k2;$ and etc.
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
$endgroup$
1
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
add a comment |
$begingroup$
Let $;R;$ be a rectange with sides $;x,y;$ s.t $;x+y=k;,;;kinBbb N;$ fixed, let us find the maximum of $;f(x,y):=xy$:
$$x+y=kimplies y=k-ximplies g(x)=f(x,k-x)=x(k-x)$$
From here:
$$g'(x)=k-2x=0iff x=frac k2implies y=frac k2$$
and the rectangle is in fact a square. you now prove that in the above we really get a maximum point.
Another more basic way with the same notation:
$$g(x)=-x^2+kx , ;;text{and this is a parabola that open downwards}$$
and thus its vertex is a maximal point. But the vertex is given by
$$left(;-frac k{-2};,;;-fracDelta{-4};right)=left(;frac k2;,;;frac{k^2}4;right)$$
and again $;x=frac k2;$ and etc.
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
$endgroup$
Let $;R;$ be a rectange with sides $;x,y;$ s.t $;x+y=k;,;;kinBbb N;$ fixed, let us find the maximum of $;f(x,y):=xy$:
$$x+y=kimplies y=k-ximplies g(x)=f(x,k-x)=x(k-x)$$
From here:
$$g'(x)=k-2x=0iff x=frac k2implies y=frac k2$$
and the rectangle is in fact a square. you now prove that in the above we really get a maximum point.
Another more basic way with the same notation:
$$g(x)=-x^2+kx , ;;text{and this is a parabola that open downwards}$$
and thus its vertex is a maximal point. But the vertex is given by
$$left(;-frac k{-2};,;;-fracDelta{-4};right)=left(;frac k2;,;;frac{k^2}4;right)$$
and again $;x=frac k2;$ and etc.
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
answered Dec 27 '14 at 15:18
TimbucTimbuc
30.9k22145
30.9k22145
1
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
add a comment |
1
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
1
1
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
This is age !! Thanks for pointing it.
$endgroup$
– Claude Leibovici
Dec 27 '14 at 15:30
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
$begingroup$
@ClaudeLeibovici Hehe...any time, Claude.
$endgroup$
– Timbuc
Dec 27 '14 at 15:31
add a comment |
$begingroup$
This is both in response to the comments made in Rahul's answer, and also as an additional graphical proof.
In addition, let me emphasize that to mathematicians, "Why" something is true is explained in mathematical proof. Explaining things via formula and pictures and such is not a "How" but is indeed a "Why". Also let me emphasize that when beginning anything it is important to remember the definitions. Area of a shape is loosely defined as "how many unit squares can fit in a shape", and for rectangles in particular it is explicitly defined as $Atimes B$ where $A$ and $B$ are the lengths of the sides.
Claim: All rectangles of a given perimeter have less than or equal area to that of the square of the same perimeter. Equivalently, a square of a given perimeter is of greater area than all rectangles of equal perimeter where the rectangle is not a square.
Without loss of generality let the rectangle we are comparing to be longer than it is tall. Having started with a square of side length $B$, choose some positive value of $xleq B$ and relabel the side length as $B=A+x$ where $A = B-x$
Make a horizontal slice a distance of $x$ away from the top extending across the entire square and a vertical slice a distance of $x$ away from the right which extends from the top to the previous slice. By removing the upper-right piece you effectively reduce the area. You may then move the upper-left piece and rotate it to fit nicely with the lower piece. The resulting formed rectangle is of the same perimeter as the original square, however the area is smaller.
As such the area of the blue square is $A^2 + 2Ax + x^2$ which is strictly greater than the area of the red square which is $A^2 + 2Ax$ while the perimeter remain the same.
By changing the value of $x$, you are able to form any rectangle of given perimeter, and if $x=B$ then you get a line segment which has zero area. $square$
If you are looking for a simpler answer, the "why" is because area and perimeter measure two completely different things. If you are looking for a more philosophical answer, then I would direct you to read Plato and Pythagoras.
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
$endgroup$
add a comment |
$begingroup$
This is both in response to the comments made in Rahul's answer, and also as an additional graphical proof.
In addition, let me emphasize that to mathematicians, "Why" something is true is explained in mathematical proof. Explaining things via formula and pictures and such is not a "How" but is indeed a "Why". Also let me emphasize that when beginning anything it is important to remember the definitions. Area of a shape is loosely defined as "how many unit squares can fit in a shape", and for rectangles in particular it is explicitly defined as $Atimes B$ where $A$ and $B$ are the lengths of the sides.
Claim: All rectangles of a given perimeter have less than or equal area to that of the square of the same perimeter. Equivalently, a square of a given perimeter is of greater area than all rectangles of equal perimeter where the rectangle is not a square.
Without loss of generality let the rectangle we are comparing to be longer than it is tall. Having started with a square of side length $B$, choose some positive value of $xleq B$ and relabel the side length as $B=A+x$ where $A = B-x$
Make a horizontal slice a distance of $x$ away from the top extending across the entire square and a vertical slice a distance of $x$ away from the right which extends from the top to the previous slice. By removing the upper-right piece you effectively reduce the area. You may then move the upper-left piece and rotate it to fit nicely with the lower piece. The resulting formed rectangle is of the same perimeter as the original square, however the area is smaller.
As such the area of the blue square is $A^2 + 2Ax + x^2$ which is strictly greater than the area of the red square which is $A^2 + 2Ax$ while the perimeter remain the same.
By changing the value of $x$, you are able to form any rectangle of given perimeter, and if $x=B$ then you get a line segment which has zero area. $square$
If you are looking for a simpler answer, the "why" is because area and perimeter measure two completely different things. If you are looking for a more philosophical answer, then I would direct you to read Plato and Pythagoras.
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
$endgroup$
add a comment |
$begingroup$
This is both in response to the comments made in Rahul's answer, and also as an additional graphical proof.
In addition, let me emphasize that to mathematicians, "Why" something is true is explained in mathematical proof. Explaining things via formula and pictures and such is not a "How" but is indeed a "Why". Also let me emphasize that when beginning anything it is important to remember the definitions. Area of a shape is loosely defined as "how many unit squares can fit in a shape", and for rectangles in particular it is explicitly defined as $Atimes B$ where $A$ and $B$ are the lengths of the sides.
Claim: All rectangles of a given perimeter have less than or equal area to that of the square of the same perimeter. Equivalently, a square of a given perimeter is of greater area than all rectangles of equal perimeter where the rectangle is not a square.
Without loss of generality let the rectangle we are comparing to be longer than it is tall. Having started with a square of side length $B$, choose some positive value of $xleq B$ and relabel the side length as $B=A+x$ where $A = B-x$
Make a horizontal slice a distance of $x$ away from the top extending across the entire square and a vertical slice a distance of $x$ away from the right which extends from the top to the previous slice. By removing the upper-right piece you effectively reduce the area. You may then move the upper-left piece and rotate it to fit nicely with the lower piece. The resulting formed rectangle is of the same perimeter as the original square, however the area is smaller.
As such the area of the blue square is $A^2 + 2Ax + x^2$ which is strictly greater than the area of the red square which is $A^2 + 2Ax$ while the perimeter remain the same.
By changing the value of $x$, you are able to form any rectangle of given perimeter, and if $x=B$ then you get a line segment which has zero area. $square$
If you are looking for a simpler answer, the "why" is because area and perimeter measure two completely different things. If you are looking for a more philosophical answer, then I would direct you to read Plato and Pythagoras.
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
$endgroup$
This is both in response to the comments made in Rahul's answer, and also as an additional graphical proof.
In addition, let me emphasize that to mathematicians, "Why" something is true is explained in mathematical proof. Explaining things via formula and pictures and such is not a "How" but is indeed a "Why". Also let me emphasize that when beginning anything it is important to remember the definitions. Area of a shape is loosely defined as "how many unit squares can fit in a shape", and for rectangles in particular it is explicitly defined as $Atimes B$ where $A$ and $B$ are the lengths of the sides.
Claim: All rectangles of a given perimeter have less than or equal area to that of the square of the same perimeter. Equivalently, a square of a given perimeter is of greater area than all rectangles of equal perimeter where the rectangle is not a square.
Without loss of generality let the rectangle we are comparing to be longer than it is tall. Having started with a square of side length $B$, choose some positive value of $xleq B$ and relabel the side length as $B=A+x$ where $A = B-x$
Make a horizontal slice a distance of $x$ away from the top extending across the entire square and a vertical slice a distance of $x$ away from the right which extends from the top to the previous slice. By removing the upper-right piece you effectively reduce the area. You may then move the upper-left piece and rotate it to fit nicely with the lower piece. The resulting formed rectangle is of the same perimeter as the original square, however the area is smaller.
As such the area of the blue square is $A^2 + 2Ax + x^2$ which is strictly greater than the area of the red square which is $A^2 + 2Ax$ while the perimeter remain the same.
By changing the value of $x$, you are able to form any rectangle of given perimeter, and if $x=B$ then you get a line segment which has zero area. $square$
If you are looking for a simpler answer, the "why" is because area and perimeter measure two completely different things. If you are looking for a more philosophical answer, then I would direct you to read Plato and Pythagoras.
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
answered Dec 28 '14 at 0:18
JMoravitzJMoravitz
46.8k33886
46.8k33886
add a comment |
add a comment |
$begingroup$
Take a 13x13 square, and for convenience's sake, draw it on a 1x1 grid, so it takes up 13x13 squares.
Pick one corner of the square (let's say top left), and remove the 1x1 square in that corner from the big square. The shape you have now (big-square-minus-little-square) has the same perimeter as the big square that you started with, but obviously its area is 1 square less than the area of the big square that you started with.
Now remove another square, adjacent to the one that you first removed (let's say top almost-left). You now have a 13x13 square from which you removed a 2x1 rectangle. The perimeter is still the same, but the area has been diminished by two.
Keep removing 1x1 squares along the same edge (top edge), until there is only one 1x1 square left on that edge. You now have a 13x13 square from which you removed a 12x1 rectangle, without changing its perimeter. You can also think of it as a 13x12 rectangle with a single 1x1 square hanging off one corner (at the top of the top right corner).
Now take that one extra 1x1 square and move it from the edge it's on (top edge) to its adjacent edge (right edge), so that now you have a 13x12 rectangle with a single 1x1 square hanging off the same corner, but in the other direction (at the right of the top right corner). You can also think of it as a 14x12 rectangle with a 11x1 rectangle missing. In this step, you once again did not change the perimeter, and you also did not change the area.
Now, starting from the 1x1 square that's hanging off the corner, add one missing 1x1 square back next to it (just below it). You now have a 14x12 rectangle with a 10x1 square missing. In this step, you did not change the perimeter, but you increased the area by 1.
Keep adding 1x1 squares back until you have a full 14x12 rectangle. Every time you add a 1x1 square back, you are increasing the area by 1, without changing perimeter.
Now look at what you did over all: you removed 13-1=12 squares off the top, but you added 12-1=11 squares on the right. Therefore, the area decreased overall from start to finish.
If you think about how this process works, you will see that if you start with a square, you will always end up removing more 1x1 squares from the top than you add on the right, and therefore the area will always decrease overall. (Try it for a few different size squares to see why.)
This process even works if you start with a rectangle, as long as you always remove squares from a long edge and put them back on a short edge. For example, you can start with a 13x13 square and go to 14x12 (removing some area) and then to 15x11 (removing some more area). This lets you start with a square and change the edges by more than 1, and still be sure that some area will be lost.
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
$endgroup$
add a comment |
$begingroup$
Take a 13x13 square, and for convenience's sake, draw it on a 1x1 grid, so it takes up 13x13 squares.
Pick one corner of the square (let's say top left), and remove the 1x1 square in that corner from the big square. The shape you have now (big-square-minus-little-square) has the same perimeter as the big square that you started with, but obviously its area is 1 square less than the area of the big square that you started with.
Now remove another square, adjacent to the one that you first removed (let's say top almost-left). You now have a 13x13 square from which you removed a 2x1 rectangle. The perimeter is still the same, but the area has been diminished by two.
Keep removing 1x1 squares along the same edge (top edge), until there is only one 1x1 square left on that edge. You now have a 13x13 square from which you removed a 12x1 rectangle, without changing its perimeter. You can also think of it as a 13x12 rectangle with a single 1x1 square hanging off one corner (at the top of the top right corner).
Now take that one extra 1x1 square and move it from the edge it's on (top edge) to its adjacent edge (right edge), so that now you have a 13x12 rectangle with a single 1x1 square hanging off the same corner, but in the other direction (at the right of the top right corner). You can also think of it as a 14x12 rectangle with a 11x1 rectangle missing. In this step, you once again did not change the perimeter, and you also did not change the area.
Now, starting from the 1x1 square that's hanging off the corner, add one missing 1x1 square back next to it (just below it). You now have a 14x12 rectangle with a 10x1 square missing. In this step, you did not change the perimeter, but you increased the area by 1.
Keep adding 1x1 squares back until you have a full 14x12 rectangle. Every time you add a 1x1 square back, you are increasing the area by 1, without changing perimeter.
Now look at what you did over all: you removed 13-1=12 squares off the top, but you added 12-1=11 squares on the right. Therefore, the area decreased overall from start to finish.
If you think about how this process works, you will see that if you start with a square, you will always end up removing more 1x1 squares from the top than you add on the right, and therefore the area will always decrease overall. (Try it for a few different size squares to see why.)
This process even works if you start with a rectangle, as long as you always remove squares from a long edge and put them back on a short edge. For example, you can start with a 13x13 square and go to 14x12 (removing some area) and then to 15x11 (removing some more area). This lets you start with a square and change the edges by more than 1, and still be sure that some area will be lost.
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
$endgroup$
add a comment |
$begingroup$
Take a 13x13 square, and for convenience's sake, draw it on a 1x1 grid, so it takes up 13x13 squares.
Pick one corner of the square (let's say top left), and remove the 1x1 square in that corner from the big square. The shape you have now (big-square-minus-little-square) has the same perimeter as the big square that you started with, but obviously its area is 1 square less than the area of the big square that you started with.
Now remove another square, adjacent to the one that you first removed (let's say top almost-left). You now have a 13x13 square from which you removed a 2x1 rectangle. The perimeter is still the same, but the area has been diminished by two.
Keep removing 1x1 squares along the same edge (top edge), until there is only one 1x1 square left on that edge. You now have a 13x13 square from which you removed a 12x1 rectangle, without changing its perimeter. You can also think of it as a 13x12 rectangle with a single 1x1 square hanging off one corner (at the top of the top right corner).
Now take that one extra 1x1 square and move it from the edge it's on (top edge) to its adjacent edge (right edge), so that now you have a 13x12 rectangle with a single 1x1 square hanging off the same corner, but in the other direction (at the right of the top right corner). You can also think of it as a 14x12 rectangle with a 11x1 rectangle missing. In this step, you once again did not change the perimeter, and you also did not change the area.
Now, starting from the 1x1 square that's hanging off the corner, add one missing 1x1 square back next to it (just below it). You now have a 14x12 rectangle with a 10x1 square missing. In this step, you did not change the perimeter, but you increased the area by 1.
Keep adding 1x1 squares back until you have a full 14x12 rectangle. Every time you add a 1x1 square back, you are increasing the area by 1, without changing perimeter.
Now look at what you did over all: you removed 13-1=12 squares off the top, but you added 12-1=11 squares on the right. Therefore, the area decreased overall from start to finish.
If you think about how this process works, you will see that if you start with a square, you will always end up removing more 1x1 squares from the top than you add on the right, and therefore the area will always decrease overall. (Try it for a few different size squares to see why.)
This process even works if you start with a rectangle, as long as you always remove squares from a long edge and put them back on a short edge. For example, you can start with a 13x13 square and go to 14x12 (removing some area) and then to 15x11 (removing some more area). This lets you start with a square and change the edges by more than 1, and still be sure that some area will be lost.
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
$endgroup$
Take a 13x13 square, and for convenience's sake, draw it on a 1x1 grid, so it takes up 13x13 squares.
Pick one corner of the square (let's say top left), and remove the 1x1 square in that corner from the big square. The shape you have now (big-square-minus-little-square) has the same perimeter as the big square that you started with, but obviously its area is 1 square less than the area of the big square that you started with.
Now remove another square, adjacent to the one that you first removed (let's say top almost-left). You now have a 13x13 square from which you removed a 2x1 rectangle. The perimeter is still the same, but the area has been diminished by two.
Keep removing 1x1 squares along the same edge (top edge), until there is only one 1x1 square left on that edge. You now have a 13x13 square from which you removed a 12x1 rectangle, without changing its perimeter. You can also think of it as a 13x12 rectangle with a single 1x1 square hanging off one corner (at the top of the top right corner).
Now take that one extra 1x1 square and move it from the edge it's on (top edge) to its adjacent edge (right edge), so that now you have a 13x12 rectangle with a single 1x1 square hanging off the same corner, but in the other direction (at the right of the top right corner). You can also think of it as a 14x12 rectangle with a 11x1 rectangle missing. In this step, you once again did not change the perimeter, and you also did not change the area.
Now, starting from the 1x1 square that's hanging off the corner, add one missing 1x1 square back next to it (just below it). You now have a 14x12 rectangle with a 10x1 square missing. In this step, you did not change the perimeter, but you increased the area by 1.
Keep adding 1x1 squares back until you have a full 14x12 rectangle. Every time you add a 1x1 square back, you are increasing the area by 1, without changing perimeter.
Now look at what you did over all: you removed 13-1=12 squares off the top, but you added 12-1=11 squares on the right. Therefore, the area decreased overall from start to finish.
If you think about how this process works, you will see that if you start with a square, you will always end up removing more 1x1 squares from the top than you add on the right, and therefore the area will always decrease overall. (Try it for a few different size squares to see why.)
This process even works if you start with a rectangle, as long as you always remove squares from a long edge and put them back on a short edge. For example, you can start with a 13x13 square and go to 14x12 (removing some area) and then to 15x11 (removing some more area). This lets you start with a square and change the edges by more than 1, and still be sure that some area will be lost.
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
answered Dec 28 '14 at 7:07
Ben ArtinBen Artin
1211
1211
add a comment |
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$begingroup$
Robert, to keep it simple think of perimeter as 'Exposed Surface'.
Perhaps a visual illustration of 'why' will be of more use to you than an algebraic one?
Start with this:
Then with that in mind, this:
Notice how 'Area' and 'Total Surface' stay constant throughout the shape change (folding) process. Modifying the shape of an object directly influences the ratio of 'Exposed' & 'Concealed' surfaces (measured in terms of discrete units). Therfore the 'perimiter' of an object can be modified significantly without the need to change its area.
Also notice how the more compact we make the object (by reducing the amount of exposed surface, and therfore increasing the concealed surface) the more square like it becomes, this is because the square is the most compact form of rectangle, it has the least amount of exposed surface possible for a 'rectangular shaped object'.
As for the units used, they don't really matter, we could be referring to (m), (cm), (mm), etc. down to atomic distances if you like:
Either way the concept is the same; you break the object down to discrete units, and from there you can analyse the relationship between shape, area, and perimeter.
answered Dec 28 '14 at 9:54
nobodynobody
112
$endgroup$
add a comment |
$begingroup$
Robert, to keep it simple think of perimeter as 'Exposed Surface'.
Perhaps a visual illustration of 'why' will be of more use to you than an algebraic one?
Start with this:
Then with that in mind, this:
Notice how 'Area' and 'Total Surface' stay constant throughout the shape change (folding) process. Modifying the shape of an object directly influences the ratio of 'Exposed' & 'Concealed' surfaces (measured in terms of discrete units). Therfore the 'perimiter' of an object can be modified significantly without the need to change its area.
Also notice how the more compact we make the object (by reducing the amount of exposed surface, and therfore increasing the concealed surface) the more square like it becomes, this is because the square is the most compact form of rectangle, it has the least amount of exposed surface possible for a 'rectangular shaped object'.
As for the units used, they don't really matter, we could be referring to (m), (cm), (mm), etc. down to atomic distances if you like:
Either way the concept is the same; you break the object down to discrete units, and from there you can analyse the relationship between shape, area, and perimeter.
answered Dec 28 '14 at 9:54
nobodynobody
112
$endgroup$
add a comment |
$begingroup$
Robert, to keep it simple think of perimeter as 'Exposed Surface'.
Perhaps a visual illustration of 'why' will be of more use to you than an algebraic one?
Start with this:
Then with that in mind, this:
Notice how 'Area' and 'Total Surface' stay constant throughout the shape change (folding) process. Modifying the shape of an object directly influences the ratio of 'Exposed' & 'Concealed' surfaces (measured in terms of discrete units). Therfore the 'perimiter' of an object can be modified significantly without the need to change its area.
Also notice how the more compact we make the object (by reducing the amount of exposed surface, and therfore increasing the concealed surface) the more square like it becomes, this is because the square is the most compact form of rectangle, it has the least amount of exposed surface possible for a 'rectangular shaped object'.
As for the units used, they don't really matter, we could be referring to (m), (cm), (mm), etc. down to atomic distances if you like:
Either way the concept is the same; you break the object down to discrete units, and from there you can analyse the relationship between shape, area, and perimeter.
answered Dec 28 '14 at 9:54
nobodynobody
112
$endgroup$
Robert, to keep it simple think of perimeter as 'Exposed Surface'.
Perhaps a visual illustration of 'why' will be of more use to you than an algebraic one?
Start with this:
Then with that in mind, this:
Notice how 'Area' and 'Total Surface' stay constant throughout the shape change (folding) process. Modifying the shape of an object directly influences the ratio of 'Exposed' & 'Concealed' surfaces (measured in terms of discrete units). Therfore the 'perimiter' of an object can be modified significantly without the need to change its area.
Also notice how the more compact we make the object (by reducing the amount of exposed surface, and therfore increasing the concealed surface) the more square like it becomes, this is because the square is the most compact form of rectangle, it has the least amount of exposed surface possible for a 'rectangular shaped object'.
As for the units used, they don't really matter, we could be referring to (m), (cm), (mm), etc. down to atomic distances if you like:
Either way the concept is the same; you break the object down to discrete units, and from there you can analyse the relationship between shape, area, and perimeter.
answered Dec 28 '14 at 9:54
nobodynobody
112
answered Dec 28 '14 at 9:54
nobodynobody
112
answered Dec 28 '14 at 9:54
nobodynobody
112
answered Dec 28 '14 at 9:54
nobodynobody
112
112
add a comment |
add a comment |
$begingroup$
Constant perimeter does not imply constant area, because the later depends on $A+B$ while the former depends on $AB$. And recaling the AM-GM inequality:
$$ sqrt{A B } le frac{A+B}{2} implies S=A B le frac{(A+B)^2}{4}=frac{P^2}{16}$$
The equality happens when (and only when) $A=B$
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
$endgroup$
add a comment |
$begingroup$
Constant perimeter does not imply constant area, because the later depends on $A+B$ while the former depends on $AB$. And recaling the AM-GM inequality:
$$ sqrt{A B } le frac{A+B}{2} implies S=A B le frac{(A+B)^2}{4}=frac{P^2}{16}$$
The equality happens when (and only when) $A=B$
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
$endgroup$
add a comment |
$begingroup$
Constant perimeter does not imply constant area, because the later depends on $A+B$ while the former depends on $AB$. And recaling the AM-GM inequality:
$$ sqrt{A B } le frac{A+B}{2} implies S=A B le frac{(A+B)^2}{4}=frac{P^2}{16}$$
The equality happens when (and only when) $A=B$
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
$endgroup$
Constant perimeter does not imply constant area, because the later depends on $A+B$ while the former depends on $AB$. And recaling the AM-GM inequality:
$$ sqrt{A B } le frac{A+B}{2} implies S=A B le frac{(A+B)^2}{4}=frac{P^2}{16}$$
The equality happens when (and only when) $A=B$
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
answered Dec 27 '14 at 19:40
leonbloyleonbloy
40.6k645107
40.6k645107
add a comment |
add a comment |
$begingroup$
Consider a rectangle with sides of length $a$ and $b$, and area $ab$. The average of those lengths is $c=frac{a+b}{2}$, and we can see that each of $a$ and $b$ differ from that average by the same amount, let's call it $d$. So assuming WLOG that $a>b$, $a=c+d$ and $b=c-d$, and the area of the rectangle $ab=(c+d)(c-d)=c^2-d^2$. For a constant perimeter $P$, the value of $a+b=frac{P}{2}$ does not change, so $c$ does not change. The way to obtain the largest possible area ($c^2-d^2$) given a constant perimeter is to make $d^2$ as small as possible. This is done by making $d=0$. In that case, $a=c$ and $b=c$ so $a=b$ and the rectangle is a square. Thus for a given perimeter, every (non-square) rectangle is smaller than the corresponding square.
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
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add a comment |
$begingroup$
Consider a rectangle with sides of length $a$ and $b$, and area $ab$. The average of those lengths is $c=frac{a+b}{2}$, and we can see that each of $a$ and $b$ differ from that average by the same amount, let's call it $d$. So assuming WLOG that $a>b$, $a=c+d$ and $b=c-d$, and the area of the rectangle $ab=(c+d)(c-d)=c^2-d^2$. For a constant perimeter $P$, the value of $a+b=frac{P}{2}$ does not change, so $c$ does not change. The way to obtain the largest possible area ($c^2-d^2$) given a constant perimeter is to make $d^2$ as small as possible. This is done by making $d=0$. In that case, $a=c$ and $b=c$ so $a=b$ and the rectangle is a square. Thus for a given perimeter, every (non-square) rectangle is smaller than the corresponding square.
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
$endgroup$
add a comment |
$begingroup$
Consider a rectangle with sides of length $a$ and $b$, and area $ab$. The average of those lengths is $c=frac{a+b}{2}$, and we can see that each of $a$ and $b$ differ from that average by the same amount, let's call it $d$. So assuming WLOG that $a>b$, $a=c+d$ and $b=c-d$, and the area of the rectangle $ab=(c+d)(c-d)=c^2-d^2$. For a constant perimeter $P$, the value of $a+b=frac{P}{2}$ does not change, so $c$ does not change. The way to obtain the largest possible area ($c^2-d^2$) given a constant perimeter is to make $d^2$ as small as possible. This is done by making $d=0$. In that case, $a=c$ and $b=c$ so $a=b$ and the rectangle is a square. Thus for a given perimeter, every (non-square) rectangle is smaller than the corresponding square.
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
$endgroup$
Consider a rectangle with sides of length $a$ and $b$, and area $ab$. The average of those lengths is $c=frac{a+b}{2}$, and we can see that each of $a$ and $b$ differ from that average by the same amount, let's call it $d$. So assuming WLOG that $a>b$, $a=c+d$ and $b=c-d$, and the area of the rectangle $ab=(c+d)(c-d)=c^2-d^2$. For a constant perimeter $P$, the value of $a+b=frac{P}{2}$ does not change, so $c$ does not change. The way to obtain the largest possible area ($c^2-d^2$) given a constant perimeter is to make $d^2$ as small as possible. This is done by making $d=0$. In that case, $a=c$ and $b=c$ so $a=b$ and the rectangle is a square. Thus for a given perimeter, every (non-square) rectangle is smaller than the corresponding square.
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
answered Jan 6 at 3:50
Keith BackmanKeith Backman
1,1491712
1,1491712
add a comment |
add a comment |
$begingroup$
The following proves that a square always has a greater area than a non-square rectangle of the same perimeter.
There are many ways to prove this. I have designed this particular proof so that almost anyone, even without much mathematical experience, can understand it.
(Click the image to make it bigger.)
answered Jan 5 at 5:08
MagnusMagnus
1113
$endgroup$
add a comment |
$begingroup$
The following proves that a square always has a greater area than a non-square rectangle of the same perimeter.
There are many ways to prove this. I have designed this particular proof so that almost anyone, even without much mathematical experience, can understand it.
(Click the image to make it bigger.)
answered Jan 5 at 5:08
MagnusMagnus
1113
$endgroup$
add a comment |
$begingroup$
The following proves that a square always has a greater area than a non-square rectangle of the same perimeter.
There are many ways to prove this. I have designed this particular proof so that almost anyone, even without much mathematical experience, can understand it.
(Click the image to make it bigger.)
answered Jan 5 at 5:08
MagnusMagnus
1113
$endgroup$
The following proves that a square always has a greater area than a non-square rectangle of the same perimeter.
There are many ways to prove this. I have designed this particular proof so that almost anyone, even without much mathematical experience, can understand it.
(Click the image to make it bigger.)
answered Jan 5 at 5:08
MagnusMagnus
1113
edited Jan 7 at 14:36
answered Jan 5 at 5:08
MagnusMagnus
1113
answered Jan 5 at 5:08
MagnusMagnus
1113
answered Jan 5 at 5:08
MagnusMagnus
1113
1113
add a comment |
add a comment |
protected by Pedro Tamaroff♦ Dec 28 '14 at 11:05
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
4
$begingroup$
The reason is that given a fixed perimeter, a convex 4-gon's area is always less than or equal that of a square (with the same perimeter), i.e.: the square maximizes the area.
$endgroup$
– Timbuc
Dec 27 '14 at 15:12
9
$begingroup$
@Robert That is exactly the answer. Its proof is in the answer below. And no: you cannot get what you say you can, and again: the proof is below.
$endgroup$
– Timbuc
Dec 27 '14 at 15:18
5
$begingroup$
It would be helpful if the OP posted a sample answer to their 'why' question; either one that answers another 'why' question to which he knows the answer, or a mock-answer to this question that he does not accept but at least has the form that he desires (preferably the first option); this way we would at least know what it is the OP expects as an answer in spirit, which is more than can be said at the moment.
$endgroup$
– guest
Dec 28 '14 at 3:58
5
$begingroup$
@Robert Let's keep things civil. People are trying to help you here. If you find their answers unsatisfactory, you can always say so, but mind your manners, please.
$endgroup$
– Pedro Tamaroff♦
Dec 28 '14 at 4:39
7
$begingroup$
@Robert asking "why is the area different even though the perimeter is the same" is like asking "why do bananas and lemons taste different even though the color is the same?" There's no reason why two things that are the same color should taste the same, and there's no reason why two shapes that have the same perimeter should have the same area. As it turns out, they usually don't!
$endgroup$
– hobbs
Dec 28 '14 at 6:42