Mixing
The sum of two numbers is $28.$ Find the numbers if the sum of their squares is a minimum.
$begingroup$
I have no clue where to start.
The sum of two numbers is $28$. Find the numbers assuming that the sum of their squares is a minimum
I am an eleventh-grader. I only learned how to find the minimum for functions like $y=ax^2+bx+c$ or $y=a(x-h)^2+K.$ I don't believe I know how to do calculus and don't know what a derivative is.
Thanks!
algebra-precalculus quadratics maxima-minima
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
$endgroup$
|
show 1 more comment
$begingroup$
I have no clue where to start.
The sum of two numbers is $28$. Find the numbers assuming that the sum of their squares is a minimum
I am an eleventh-grader. I only learned how to find the minimum for functions like $y=ax^2+bx+c$ or $y=a(x-h)^2+K.$ I don't believe I know how to do calculus and don't know what a derivative is.
Thanks!
algebra-precalculus quadratics maxima-minima
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
$endgroup$
3
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
2
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
1
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
1
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43
|
show 1 more comment
$begingroup$
I have no clue where to start.
The sum of two numbers is $28$. Find the numbers assuming that the sum of their squares is a minimum
I am an eleventh-grader. I only learned how to find the minimum for functions like $y=ax^2+bx+c$ or $y=a(x-h)^2+K.$ I don't believe I know how to do calculus and don't know what a derivative is.
Thanks!
algebra-precalculus quadratics maxima-minima
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
$endgroup$
I have no clue where to start.
The sum of two numbers is $28$. Find the numbers assuming that the sum of their squares is a minimum
I am an eleventh-grader. I only learned how to find the minimum for functions like $y=ax^2+bx+c$ or $y=a(x-h)^2+K.$ I don't believe I know how to do calculus and don't know what a derivative is.
Thanks!
algebra-precalculus quadratics maxima-minima
algebra-precalculus quadratics maxima-minima
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
edited Jan 21 at 12:19
José Carlos Santos
165k22132235
165k22132235
asked Sep 25 '18 at 16:34
ZebertZebert
364
asked Sep 25 '18 at 16:34
ZebertZebert
364
asked Sep 25 '18 at 16:34
ZebertZebert
364
364
3
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
2
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
1
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
1
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43
|
show 1 more comment
3
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
2
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
1
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
1
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43
3
3
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
2
2
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
1
1
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
1
1
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43
|
show 1 more comment
8 Answers
8
active
oldest
votes
$begingroup$
Let $x, y$ be such that $x+y = 28 iff y= 28-x$.
Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2tag{1}$$
Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$
answered Sep 25 '18 at 16:46
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
Begin by writing down what the problem is asking you to do.
The sum of two numbers is 28
So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.
the sum of their squares is a minimum
Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$begin{align}S &= x^2 + (28 - x)^2 \ &= 2x^2-56x + 784.end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{bover 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$begin{align}x &= -{-56over 2(2)} \ &= 14.end{align}$$
Here we have found $x$. I will leave it to you to find $y$.
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
$endgroup$
add a comment |
$begingroup$
By the RMS-AM inequality (root-mean square vs. arithmetic mean):
$$sqrt{frac{a^2+b^2}{2}} ge frac{|a|+|b|}{2} ge frac{a+b}{2} = frac{28}{2}=14$$
Equalities hold iff $,a=b ge 0,$, therefore the minimum of $,a^2+b^2,$ is attained for $,a=b=14,$.
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
$endgroup$
add a comment |
$begingroup$
Let's name the numbers $a$ and $b$.
$$a + b = 28$$
Furthermore, let's define the sum of the squares (so we can minimize that):
$$f(a,b) = a^2 + b^2$$
Let's use the first formula to reduce the amount of variables:
$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 cdot28cdot a +28^2$$
now let's find the minimum of $f(x)$:
$$dfrac{df(a)}{da} = 4a-2cdot28$$
Finally, we will set this to zero and solve for a:
$$4a-2cdot28=0$$
$$4a=2cdot28$$
$$a = 14$$
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
$endgroup$
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
add a comment |
$begingroup$
$$min_{x+y=28}x^2+y^2$$
is the same as
$$min_xx^2+(28-x)^2$$ which you find by differentiation.
$$x-(28-x)=0implies x=14.$$
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
Note that $$(a-b)^2 ge 0 implies a^2+b^2 ge 2ab implies 2(a^2+b^2) ge a^2+b^2 +2ab.$$
That is,
$$a^2+b^2 gefrac{(a+b)^2}{2},$$
where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
$begingroup$
"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "
Then experiment.
$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.
Can you make a hypothesis? Can you argue why.
Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $sqrt ab le frac {a+b}2$ Can you see how that would be relevent.)
I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K
Well, that's more than I knew in the 11th grade!
Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?
but... experiment and muck around and see what you find. No one is expectiong you to get it right away.
====
If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.
Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.
If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.
In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.
I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.
If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - frac d 2$ or $a = 14 - frac d2$ or $a , 14 - frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$
Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $frac {a+b}2 = frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.
So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.
How do we minimize that? Well $e^2 ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.
The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.
Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2930450%2fthe-sum-of-two-numbers-is-28-find-the-numbers-if-the-sum-of-their-squares-is%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $x, y$ be such that $x+y = 28 iff y= 28-x$.
Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2tag{1}$$
Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$
answered Sep 25 '18 at 16:46
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
Let $x, y$ be such that $x+y = 28 iff y= 28-x$.
Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2tag{1}$$
Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$
answered Sep 25 '18 at 16:46
NamasteNamaste
1
$endgroup$
add a comment |
$begingroup$
Let $x, y$ be such that $x+y = 28 iff y= 28-x$.
Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2tag{1}$$
Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$
answered Sep 25 '18 at 16:46
NamasteNamaste
1
$endgroup$
Let $x, y$ be such that $x+y = 28 iff y= 28-x$.
Then, the function we want to minimize looks something like this: $$x^2 + y^2 = x^2 +(28-x)^2 = x^2+ 784-56 x + x^2tag{1}$$
Simplify $(1)$ to get $$f(x) = 2x^2 -56x + 784tag{2}$$ And given you know how to find the minimum of a function of the form $$f(x) = ax^2 +bx+c$$ use your knowledge to minimize the function $(2)$
answered Sep 25 '18 at 16:46
NamasteNamaste
1
edited Sep 25 '18 at 16:59
answered Sep 25 '18 at 16:46
NamasteNamaste
1
answered Sep 25 '18 at 16:46
NamasteNamaste
1
answered Sep 25 '18 at 16:46
NamasteNamaste
1
1
add a comment |
add a comment |
$begingroup$
Begin by writing down what the problem is asking you to do.
The sum of two numbers is 28
So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.
the sum of their squares is a minimum
Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$begin{align}S &= x^2 + (28 - x)^2 \ &= 2x^2-56x + 784.end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{bover 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$begin{align}x &= -{-56over 2(2)} \ &= 14.end{align}$$
Here we have found $x$. I will leave it to you to find $y$.
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
$endgroup$
add a comment |
$begingroup$
Begin by writing down what the problem is asking you to do.
The sum of two numbers is 28
So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.
the sum of their squares is a minimum
Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$begin{align}S &= x^2 + (28 - x)^2 \ &= 2x^2-56x + 784.end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{bover 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$begin{align}x &= -{-56over 2(2)} \ &= 14.end{align}$$
Here we have found $x$. I will leave it to you to find $y$.
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
$endgroup$
add a comment |
$begingroup$
Begin by writing down what the problem is asking you to do.
The sum of two numbers is 28
So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.
the sum of their squares is a minimum
Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$begin{align}S &= x^2 + (28 - x)^2 \ &= 2x^2-56x + 784.end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{bover 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$begin{align}x &= -{-56over 2(2)} \ &= 14.end{align}$$
Here we have found $x$. I will leave it to you to find $y$.
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
$endgroup$
Begin by writing down what the problem is asking you to do.
The sum of two numbers is 28
So we take two numbers, $x$ and $y$. From this, we know that $x + y = 28$.
the sum of their squares is a minimum
Here, we want to minimize $S = x^2 + y^2$. From the above, we can see that $y = 28 - x$, which by substituting this into $S$, we get $$begin{align}S &= x^2 + (28 - x)^2 \ &= 2x^2-56x + 784.end{align}$$ This is a quadratic equation, which the minimum or maximum of the parabola (for any $y = ax^2 + bx + c$) occurs at $x = -{bover 2a}$. Here, $a = 2$ and $b = -56$. Hence, $$begin{align}x &= -{-56over 2(2)} \ &= 14.end{align}$$
Here we have found $x$. I will leave it to you to find $y$.
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
answered Sep 25 '18 at 16:43
Decaf-MathDecaf-Math
3,422825
3,422825
add a comment |
add a comment |
$begingroup$
By the RMS-AM inequality (root-mean square vs. arithmetic mean):
$$sqrt{frac{a^2+b^2}{2}} ge frac{|a|+|b|}{2} ge frac{a+b}{2} = frac{28}{2}=14$$
Equalities hold iff $,a=b ge 0,$, therefore the minimum of $,a^2+b^2,$ is attained for $,a=b=14,$.
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
$endgroup$
add a comment |
$begingroup$
By the RMS-AM inequality (root-mean square vs. arithmetic mean):
$$sqrt{frac{a^2+b^2}{2}} ge frac{|a|+|b|}{2} ge frac{a+b}{2} = frac{28}{2}=14$$
Equalities hold iff $,a=b ge 0,$, therefore the minimum of $,a^2+b^2,$ is attained for $,a=b=14,$.
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
$endgroup$
add a comment |
$begingroup$
By the RMS-AM inequality (root-mean square vs. arithmetic mean):
$$sqrt{frac{a^2+b^2}{2}} ge frac{|a|+|b|}{2} ge frac{a+b}{2} = frac{28}{2}=14$$
Equalities hold iff $,a=b ge 0,$, therefore the minimum of $,a^2+b^2,$ is attained for $,a=b=14,$.
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
$endgroup$
By the RMS-AM inequality (root-mean square vs. arithmetic mean):
$$sqrt{frac{a^2+b^2}{2}} ge frac{|a|+|b|}{2} ge frac{a+b}{2} = frac{28}{2}=14$$
Equalities hold iff $,a=b ge 0,$, therefore the minimum of $,a^2+b^2,$ is attained for $,a=b=14,$.
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
edited Sep 25 '18 at 18:50
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
answered Sep 25 '18 at 18:40
dxivdxiv
57.9k648101
57.9k648101
add a comment |
add a comment |
$begingroup$
Let's name the numbers $a$ and $b$.
$$a + b = 28$$
Furthermore, let's define the sum of the squares (so we can minimize that):
$$f(a,b) = a^2 + b^2$$
Let's use the first formula to reduce the amount of variables:
$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 cdot28cdot a +28^2$$
now let's find the minimum of $f(x)$:
$$dfrac{df(a)}{da} = 4a-2cdot28$$
Finally, we will set this to zero and solve for a:
$$4a-2cdot28=0$$
$$4a=2cdot28$$
$$a = 14$$
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
$endgroup$
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
add a comment |
$begingroup$
Let's name the numbers $a$ and $b$.
$$a + b = 28$$
Furthermore, let's define the sum of the squares (so we can minimize that):
$$f(a,b) = a^2 + b^2$$
Let's use the first formula to reduce the amount of variables:
$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 cdot28cdot a +28^2$$
now let's find the minimum of $f(x)$:
$$dfrac{df(a)}{da} = 4a-2cdot28$$
Finally, we will set this to zero and solve for a:
$$4a-2cdot28=0$$
$$4a=2cdot28$$
$$a = 14$$
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
$endgroup$
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
add a comment |
$begingroup$
Let's name the numbers $a$ and $b$.
$$a + b = 28$$
Furthermore, let's define the sum of the squares (so we can minimize that):
$$f(a,b) = a^2 + b^2$$
Let's use the first formula to reduce the amount of variables:
$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 cdot28cdot a +28^2$$
now let's find the minimum of $f(x)$:
$$dfrac{df(a)}{da} = 4a-2cdot28$$
Finally, we will set this to zero and solve for a:
$$4a-2cdot28=0$$
$$4a=2cdot28$$
$$a = 14$$
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
$endgroup$
Let's name the numbers $a$ and $b$.
$$a + b = 28$$
Furthermore, let's define the sum of the squares (so we can minimize that):
$$f(a,b) = a^2 + b^2$$
Let's use the first formula to reduce the amount of variables:
$$f(a)=a^2 + (28 - a)^2 = 2 a^2 - 2 cdot28cdot a +28^2$$
now let's find the minimum of $f(x)$:
$$dfrac{df(a)}{da} = 4a-2cdot28$$
Finally, we will set this to zero and solve for a:
$$4a-2cdot28=0$$
$$4a=2cdot28$$
$$a = 14$$
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
answered Sep 25 '18 at 16:45
Finn EggersFinn Eggers
364213
364213
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
add a comment |
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
$begingroup$
btw, this is also why a square has compared to all the other rectangles the best area to circumference ratio.
$endgroup$
– Finn Eggers
Sep 25 '18 at 16:46
add a comment |
$begingroup$
$$min_{x+y=28}x^2+y^2$$
is the same as
$$min_xx^2+(28-x)^2$$ which you find by differentiation.
$$x-(28-x)=0implies x=14.$$
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
$$min_{x+y=28}x^2+y^2$$
is the same as
$$min_xx^2+(28-x)^2$$ which you find by differentiation.
$$x-(28-x)=0implies x=14.$$
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
$$min_{x+y=28}x^2+y^2$$
is the same as
$$min_xx^2+(28-x)^2$$ which you find by differentiation.
$$x-(28-x)=0implies x=14.$$
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
$endgroup$
$$min_{x+y=28}x^2+y^2$$
is the same as
$$min_xx^2+(28-x)^2$$ which you find by differentiation.
$$x-(28-x)=0implies x=14.$$
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
answered Sep 25 '18 at 16:50
Yves DaoustYves Daoust
129k676227
129k676227
add a comment |
add a comment |
$begingroup$
Note that $$(a-b)^2 ge 0 implies a^2+b^2 ge 2ab implies 2(a^2+b^2) ge a^2+b^2 +2ab.$$
That is,
$$a^2+b^2 gefrac{(a+b)^2}{2},$$
where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
$begingroup$
Note that $$(a-b)^2 ge 0 implies a^2+b^2 ge 2ab implies 2(a^2+b^2) ge a^2+b^2 +2ab.$$
That is,
$$a^2+b^2 gefrac{(a+b)^2}{2},$$
where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
$begingroup$
Note that $$(a-b)^2 ge 0 implies a^2+b^2 ge 2ab implies 2(a^2+b^2) ge a^2+b^2 +2ab.$$
That is,
$$a^2+b^2 gefrac{(a+b)^2}{2},$$
where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
$endgroup$
Note that $$(a-b)^2 ge 0 implies a^2+b^2 ge 2ab implies 2(a^2+b^2) ge a^2+b^2 +2ab.$$
That is,
$$a^2+b^2 gefrac{(a+b)^2}{2},$$
where the equality holds when $a-b=0$, or equivalently $a = b$. In your case, $a+b=28$. Therefore, $a^2+b^2$ will be minimum when $a = b = 28/2 = 14$.
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
edited Sep 25 '18 at 16:55
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
answered Sep 25 '18 at 16:50
Math LoverMath Lover
14.1k31437
14.1k31437
add a comment |
add a comment |
$begingroup$
"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "
Then experiment.
$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.
Can you make a hypothesis? Can you argue why.
Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $sqrt ab le frac {a+b}2$ Can you see how that would be relevent.)
I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K
Well, that's more than I knew in the 11th grade!
Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?
but... experiment and muck around and see what you find. No one is expectiong you to get it right away.
====
If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.
Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.
If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.
In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.
I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.
If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - frac d 2$ or $a = 14 - frac d2$ or $a , 14 - frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$
Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $frac {a+b}2 = frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.
So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.
How do we minimize that? Well $e^2 ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "
Then experiment.
$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.
Can you make a hypothesis? Can you argue why.
Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $sqrt ab le frac {a+b}2$ Can you see how that would be relevent.)
I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K
Well, that's more than I knew in the 11th grade!
Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?
but... experiment and muck around and see what you find. No one is expectiong you to get it right away.
====
If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.
Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.
If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.
In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.
I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.
If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - frac d 2$ or $a = 14 - frac d2$ or $a , 14 - frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$
Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $frac {a+b}2 = frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.
So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.
How do we minimize that? Well $e^2 ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
$endgroup$
add a comment |
$begingroup$
"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "
Then experiment.
$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.
Can you make a hypothesis? Can you argue why.
Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $sqrt ab le frac {a+b}2$ Can you see how that would be relevent.)
I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K
Well, that's more than I knew in the 11th grade!
Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?
but... experiment and muck around and see what you find. No one is expectiong you to get it right away.
====
If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.
Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.
If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.
In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.
I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.
If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - frac d 2$ or $a = 14 - frac d2$ or $a , 14 - frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$
Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $frac {a+b}2 = frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.
So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.
How do we minimize that? Well $e^2 ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
$endgroup$
"I have no clue where to start" "I don't believe I know how to do calculus and don't know what a derivative is. "
Then experiment.
$a + b =28$. So you could have $14 + 14 = 28$ and $14^2 + 14^2 = 196 + 196 = 392$. Or you could have $1 + 27 = 28$ and $1^2 + 27^2 = 1 + 729 = 730 > 392$. Or you can have $13 + 15 = 28$ and $13^2 + 15^2 = 169 +225 = 394 > 392$.
Can you make a hypothesis? Can you argue why.
Hint: Have you ever hear of the A.M/G.M inequality? (If $a> 0; b> 0$ then $sqrt ab le frac {a+b}2$ Can you see how that would be relevent.)
I only learned how to find the minimum for functions like y=ax2+bx+c or y=a(x−h)2+K
Well, that's more than I knew in the 11th grade!
Consider $a + b = 28$ so $b = 28 -a$ and you want to find the minimum of $a^2 + b^2 = a^2 + (28 -a)^2= 28^3 - 2*28a + 2a^2$. Can you use what you know to find the minimum?
but... experiment and muck around and see what you find. No one is expectiong you to get it right away.
====
If it were up to me and I only had the knowledge I had in the $11$th grade I would notice that if $a$ and $b$ are close together it seems that the sums of the squares are smaller than than if the are far apart. I would then figure if $a = b = 14$ would be the smallest $a^2 + b^2$.
Then I'd try to see if I can justify it. I'd figure $a + b = 28$ so $b = 28-a$ so I am trying to minimalize $a^2 + b^2 = a^2 + (28 - a)^2 = 2a^2 - 56a + 28^2$. That is smallest when $2a^2 - 56a$ or when $a^2 - 28a = a(a-28)$ is smallest.
If $a$ goes up by, say $1$ then $a'^2 - 28a' = (a + 1)^2 - 28(a+1) = a^2 - 28 + (2a + 1)-28 $. That's an increase if $(2a + 1) -28$ is positive and a decrease if $(2a+1) - 28$ is negative.
In other words if $2a + 1 -28 > 0$ or $2a > 27$ or $a > 13.5$ then be increasing by $a$ by one will increase $a^2 + b^2$ but it if $a < 13.5$ then increasing $a$ by one will be decreasing the value of $a^2 + b^2$. So if $a = 13.5$ is when adding $1$ to $a$ will stop decreasing $a^2 + b^2$ and start increasing it. So it's reasonable that somewhere near $a = 13.5$ and $b = 14.5$ that $a^2 + b^2$ reaches a minimum and has nowhere to go but up.
I suppose I'd then think that $1$ was a jerky to big of a jump and I'd consider a jump of $d$ where $d$ can be something small like $frac 1{1000}$. Then jumping from $a$ to $a + d$ would change $a^2 -28a$ to $(a + d)^2 - 28(a+d) = a^2 - 28a + 2ad + d^2 - 28d$ or a differencd of $2ad + d^2 - 28d$. So to find the point where that "stops" being a negative decrease and starts being a poisitive increase I want to find when $2ad + d^2 - 28d > 0$ and when $2ad + d^2 - 28d =0$ and when $2ad + d^2 - 38d < 0$.
If we divide by $d$ (which is a tiny positive amount) in other words when $2a + d > 28$ and $2a + d = 28$ and if $2a + d < 28$. Or with $a > 14 - frac d 2$ or $a = 14 - frac d2$ or $a , 14 - frac d2$. As $d$ can be very small this shift when we go from this being a decrease to this being an increase is at $a = 14$
Now maybe it would have occured to me (I honestly don't know if it would have or not) that if we can replace $b$ with $28 -a$ we could "center" by averaging $frac {a+b}2 = frac {28}2 = 14$ and replacing $a = 14 -e$ and $b = 14 + e$ so $a + b = (14 -e) + (14 + e) = 28$.
So now we want to minimize $(14 - e)^2 + (14 + e)^2 = 14^2 - 28 e + e^2 + 14^2 + 2e + e^2 = 2*14^2 + 2e^2$.
How do we minimize that? Well $e^2 ge 0$ so to minimize it we simply take $e = 0$. So it is minimum at $e =0$ or $a = 14 - 0 = 14 = b = 14 + 0$.
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
edited Sep 25 '18 at 18:43
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
answered Sep 25 '18 at 17:12
fleabloodfleablood
71.9k22687
71.9k22687
add a comment |
add a comment |
$begingroup$
Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.
The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.
Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
$endgroup$
add a comment |
$begingroup$
Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.
The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.
Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
$endgroup$
add a comment |
$begingroup$
Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.
The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.
Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
$endgroup$
Suppose you take a line segment of length 28, and divide it into two pieces, $left$ and $right$, with the $left$ part smaller than the $right$ part. Now draw the squares with sides $left$ and $right$. Now suppose you move the dividing point over by $h$ to the right. The $left$ square will increase by height $h$, while the $right$ square will lose $h$ in height. While the amount of height that $left$ gains is equal to the height that $right$ loses, $right$ loses that height over a greater distance, so the area that $right$ loses is more than the area $left$ gains.
The above hold until $left$ is equal to $right$. So if you want to decrease the total area, you should increase the smaller length until the lengths are equal. Hence, you want $left=right$, which happens when they're both equal to $14$.
Another way of looking at it is that if you have a right triangle, the length squared of the hypotenuse is the sum of the squares of the legs. So this is equivalent to "You have a right triangle whose legs add up to 28. What leg lengths minimize the hypotenuse?"
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
answered Sep 25 '18 at 18:49
AcccumulationAcccumulation
7,0952619
7,0952619
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2930450%2fthe-sum-of-two-numbers-is-28-find-the-numbers-if-the-sum-of-their-squares-is%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
We have $x+y =28$. We are requested to find $x,y$ s.t. $x^2+y^2$ attains minimum. Now $y = 28-x$, so…… could you see how to proceed now?
$endgroup$
– xbh
Sep 25 '18 at 16:37
2
$begingroup$
Do you have calculus as an available tool? Specifically, do you know how to find the derivative of a function, or not?
$endgroup$
– Namaste
Sep 25 '18 at 16:39
1
$begingroup$
Based on your previous hint, we want to minimize $f(x) = x^2 +(28-x)^2 = ... $
$endgroup$
– Namaste
Sep 25 '18 at 16:40
1
$begingroup$
No, can you continue? I only learned how to find min in y=ax^2+bx+c or y=a(x-h)^2+K. I don't believe I know how to do calculus and don't know what a dervative is
$endgroup$
– Zebert
Sep 25 '18 at 16:42
$begingroup$
Would factoring it help?
$endgroup$
– Zebert
Sep 25 '18 at 16:43