Mixing
How can I maximize the area of a rectangular base and semicircular top when perimeter is not given
$begingroup$
I know that the perimeter would be $(2x+y)+pi r=P$
I am confused in these kinds of optimization problems, which variable to solve for.
I know that area is $xy+frac{pi r^2}{2}$
If I solve for r, I get $r = frac{P-(2x+y)}{pi}$
Then $A = xy+ frac{pi (frac{P-(2x+y)}{pi})^{2}}{2}$
Am I supposed to take the derivative now? I am very confused any help would be appreciated. Thanks!
EDIT: $P = 2xr+pi r = 2x+r(1+pi)$
$A = xr +frac{pi r^2}{2}$
$x = frac{P-r(1+pi)}{x}$
$A = frac{P-r(1+pi)}{x} r + frac{pi r^2}{2}$
$A = frac{P-r^2+2pi r^2}{2}$
calculus
asked Jan 24 at 17:09
user8358234user8358234
342110
$endgroup$
add a comment |
$begingroup$
I know that the perimeter would be $(2x+y)+pi r=P$
I am confused in these kinds of optimization problems, which variable to solve for.
I know that area is $xy+frac{pi r^2}{2}$
If I solve for r, I get $r = frac{P-(2x+y)}{pi}$
Then $A = xy+ frac{pi (frac{P-(2x+y)}{pi})^{2}}{2}$
Am I supposed to take the derivative now? I am very confused any help would be appreciated. Thanks!
EDIT: $P = 2xr+pi r = 2x+r(1+pi)$
$A = xr +frac{pi r^2}{2}$
$x = frac{P-r(1+pi)}{x}$
$A = frac{P-r(1+pi)}{x} r + frac{pi r^2}{2}$
$A = frac{P-r^2+2pi r^2}{2}$
calculus
asked Jan 24 at 17:09
user8358234user8358234
342110
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22
add a comment |
$begingroup$
I know that the perimeter would be $(2x+y)+pi r=P$
I am confused in these kinds of optimization problems, which variable to solve for.
I know that area is $xy+frac{pi r^2}{2}$
If I solve for r, I get $r = frac{P-(2x+y)}{pi}$
Then $A = xy+ frac{pi (frac{P-(2x+y)}{pi})^{2}}{2}$
Am I supposed to take the derivative now? I am very confused any help would be appreciated. Thanks!
EDIT: $P = 2xr+pi r = 2x+r(1+pi)$
$A = xr +frac{pi r^2}{2}$
$x = frac{P-r(1+pi)}{x}$
$A = frac{P-r(1+pi)}{x} r + frac{pi r^2}{2}$
$A = frac{P-r^2+2pi r^2}{2}$
calculus
asked Jan 24 at 17:09
user8358234user8358234
342110
$endgroup$
I know that the perimeter would be $(2x+y)+pi r=P$
I am confused in these kinds of optimization problems, which variable to solve for.
I know that area is $xy+frac{pi r^2}{2}$
If I solve for r, I get $r = frac{P-(2x+y)}{pi}$
Then $A = xy+ frac{pi (frac{P-(2x+y)}{pi})^{2}}{2}$
Am I supposed to take the derivative now? I am very confused any help would be appreciated. Thanks!
EDIT: $P = 2xr+pi r = 2x+r(1+pi)$
$A = xr +frac{pi r^2}{2}$
$x = frac{P-r(1+pi)}{x}$
$A = frac{P-r(1+pi)}{x} r + frac{pi r^2}{2}$
$A = frac{P-r^2+2pi r^2}{2}$
calculus
calculus
asked Jan 24 at 17:09
user8358234user8358234
342110
asked Jan 24 at 17:09
user8358234user8358234
342110
edited Jan 24 at 17:21
user8358234
asked Jan 24 at 17:09
user8358234user8358234
342110
asked Jan 24 at 17:09
user8358234user8358234
342110
asked Jan 24 at 17:09
user8358234user8358234
342110
342110
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$P=2x+y+pi r=2x+2r+pi r$ because the horizontal dimension of the rectangle is equal in length to the diameter of the semi-circle.$$A=xy+pi r^2/2=2rx+pi r^2/2$$
Substitute $2x$ from the first equation: $2x=P-2r-pi r$, to get$$A(r)=r[P-(2+pi)r]+pi r^2/2$$Differentiate with respect to $r$ and equate to $0$, giving$$A'(r)=P-2(2+pi)r+pi r=0implies r=frac P{4+pi}$$Ensure that this is the value of $r$ that gives the maximum area, then use the first equation to get the corresponding $x$. You could have expressed the area as a function of $x$ by substituting for $r$ too, and then differentiated with respect to $x$ to land at the same answer.
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
$endgroup$
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
$P=2x+y+pi r=2x+2r+pi r$ because the horizontal dimension of the rectangle is equal in length to the diameter of the semi-circle.$$A=xy+pi r^2/2=2rx+pi r^2/2$$
Substitute $2x$ from the first equation: $2x=P-2r-pi r$, to get$$A(r)=r[P-(2+pi)r]+pi r^2/2$$Differentiate with respect to $r$ and equate to $0$, giving$$A'(r)=P-2(2+pi)r+pi r=0implies r=frac P{4+pi}$$Ensure that this is the value of $r$ that gives the maximum area, then use the first equation to get the corresponding $x$. You could have expressed the area as a function of $x$ by substituting for $r$ too, and then differentiated with respect to $x$ to land at the same answer.
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
$endgroup$
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
add a comment |
$begingroup$
$P=2x+y+pi r=2x+2r+pi r$ because the horizontal dimension of the rectangle is equal in length to the diameter of the semi-circle.$$A=xy+pi r^2/2=2rx+pi r^2/2$$
Substitute $2x$ from the first equation: $2x=P-2r-pi r$, to get$$A(r)=r[P-(2+pi)r]+pi r^2/2$$Differentiate with respect to $r$ and equate to $0$, giving$$A'(r)=P-2(2+pi)r+pi r=0implies r=frac P{4+pi}$$Ensure that this is the value of $r$ that gives the maximum area, then use the first equation to get the corresponding $x$. You could have expressed the area as a function of $x$ by substituting for $r$ too, and then differentiated with respect to $x$ to land at the same answer.
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
$endgroup$
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
add a comment |
$begingroup$
$P=2x+y+pi r=2x+2r+pi r$ because the horizontal dimension of the rectangle is equal in length to the diameter of the semi-circle.$$A=xy+pi r^2/2=2rx+pi r^2/2$$
Substitute $2x$ from the first equation: $2x=P-2r-pi r$, to get$$A(r)=r[P-(2+pi)r]+pi r^2/2$$Differentiate with respect to $r$ and equate to $0$, giving$$A'(r)=P-2(2+pi)r+pi r=0implies r=frac P{4+pi}$$Ensure that this is the value of $r$ that gives the maximum area, then use the first equation to get the corresponding $x$. You could have expressed the area as a function of $x$ by substituting for $r$ too, and then differentiated with respect to $x$ to land at the same answer.
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
$endgroup$
$P=2x+y+pi r=2x+2r+pi r$ because the horizontal dimension of the rectangle is equal in length to the diameter of the semi-circle.$$A=xy+pi r^2/2=2rx+pi r^2/2$$
Substitute $2x$ from the first equation: $2x=P-2r-pi r$, to get$$A(r)=r[P-(2+pi)r]+pi r^2/2$$Differentiate with respect to $r$ and equate to $0$, giving$$A'(r)=P-2(2+pi)r+pi r=0implies r=frac P{4+pi}$$Ensure that this is the value of $r$ that gives the maximum area, then use the first equation to get the corresponding $x$. You could have expressed the area as a function of $x$ by substituting for $r$ too, and then differentiated with respect to $x$ to land at the same answer.
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
answered Jan 24 at 17:46
Shubham JohriShubham Johri
5,262718
5,262718
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
add a comment |
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
thank you! when i figured out what you were saying with y = 2r i got the same answer. so now i plug in $P/4+pi$ into the original equation and i got $x = 4+3 pi + 1/2 pi^2 $ is this correct? this value is not in terms of P though
$endgroup$
– user8358234
Jan 24 at 17:55
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
@user8358234 Check again. $x$ should be in terms of $P$. Using $2x=P-(2+pi)r$, I'm getting $r=x=dfrac P{4+pi}$
$endgroup$
– Shubham Johri
Jan 24 at 17:57
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
ah sorry i made another mistake. I should get $frac{1}{P} frac{4+pi}{2+pi}$
$endgroup$
– user8358234
Jan 24 at 17:58
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
$begingroup$
@user8358234 You can't get $P$ in the denominator. It is dimensionally incorrect
$endgroup$
– Shubham Johri
Jan 24 at 18:03
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 26 at 10:22