Minimize the cost of a box of fixed volume if the sides are twice as expensive as the base and top
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Suppose, to build a box (a rectangular solid) of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. Choose dimensions for the box that minimize the cost of building it, if the volume is (say) $3000$ cubic inches.
Here is what I have:
Let $x$ be the length (in.) of a side of the base. Let $h$ be the height (in.). Volume is given by $$V = x^2h = 3000;.$$
Furthermore let $m$ be the cost (per square in.) of the base and top, and $n$ the cost (per square in.) of the sides. Then total cost, $C$, is given by $$C = m(2x^2) + n(4xh);.$$
We also know $$m = 2n;.$$
To do: Express $C$ in terms of one variable, then use standard elementary calculus techniques to minimize $C$. I am stuck at this point: How do I get $C$ in terms of a single variable? I can substitute $h = 2000/x^2$, and either $m = 2n$ or $n = m/2$, but this leaves $C$ in terms of $x$ and $n$ or $m$.
calculus optimization volume
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up vote
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Suppose, to build a box (a rectangular solid) of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. Choose dimensions for the box that minimize the cost of building it, if the volume is (say) $3000$ cubic inches.
Here is what I have:
Let $x$ be the length (in.) of a side of the base. Let $h$ be the height (in.). Volume is given by $$V = x^2h = 3000;.$$
Furthermore let $m$ be the cost (per square in.) of the base and top, and $n$ the cost (per square in.) of the sides. Then total cost, $C$, is given by $$C = m(2x^2) + n(4xh);.$$
We also know $$m = 2n;.$$
To do: Express $C$ in terms of one variable, then use standard elementary calculus techniques to minimize $C$. I am stuck at this point: How do I get $C$ in terms of a single variable? I can substitute $h = 2000/x^2$, and either $m = 2n$ or $n = m/2$, but this leaves $C$ in terms of $x$ and $n$ or $m$.
calculus optimization volume
The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose, to build a box (a rectangular solid) of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. Choose dimensions for the box that minimize the cost of building it, if the volume is (say) $3000$ cubic inches.
Here is what I have:
Let $x$ be the length (in.) of a side of the base. Let $h$ be the height (in.). Volume is given by $$V = x^2h = 3000;.$$
Furthermore let $m$ be the cost (per square in.) of the base and top, and $n$ the cost (per square in.) of the sides. Then total cost, $C$, is given by $$C = m(2x^2) + n(4xh);.$$
We also know $$m = 2n;.$$
To do: Express $C$ in terms of one variable, then use standard elementary calculus techniques to minimize $C$. I am stuck at this point: How do I get $C$ in terms of a single variable? I can substitute $h = 2000/x^2$, and either $m = 2n$ or $n = m/2$, but this leaves $C$ in terms of $x$ and $n$ or $m$.
calculus optimization volume
Suppose, to build a box (a rectangular solid) of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. Choose dimensions for the box that minimize the cost of building it, if the volume is (say) $3000$ cubic inches.
Here is what I have:
Let $x$ be the length (in.) of a side of the base. Let $h$ be the height (in.). Volume is given by $$V = x^2h = 3000;.$$
Furthermore let $m$ be the cost (per square in.) of the base and top, and $n$ the cost (per square in.) of the sides. Then total cost, $C$, is given by $$C = m(2x^2) + n(4xh);.$$
We also know $$m = 2n;.$$
To do: Express $C$ in terms of one variable, then use standard elementary calculus techniques to minimize $C$. I am stuck at this point: How do I get $C$ in terms of a single variable? I can substitute $h = 2000/x^2$, and either $m = 2n$ or $n = m/2$, but this leaves $C$ in terms of $x$ and $n$ or $m$.
calculus optimization volume
edited Aug 17 at 4:33
Math Lover
12.6k21232
12.6k21232
asked Aug 17 at 3:32
A. Mack
112
112
The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34
add a comment |Â
The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34
The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34
The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34
add a comment |Â
1 Answer
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You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correctâÂÂI did not check them)
$$ C(x)
= m(2x^2) + n(4xh)
= 4nx^2 + frac8000 nx
= 4n left(x^2 + frac2000x right). $$
This tends to infinity both as $xto 0^+$ and as $xto +infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve:
beginalign
&0 = C'(x)
= fracmathrmdmathrmdx 4n left(x^2 + frac2000x right)
= 4n left( 2x - frac2000x^2 right) \
&qquadimplies 0 = 2x - frac2000x^2 \
&qquadimplies 0 = 2x^3 - 2000 \
&qquadimplies x^3 = 1000 \
&qquadimplies x = sqrt[3]1000 = 10.
endalign
As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correctâÂÂI did not check them)
$$ C(x)
= m(2x^2) + n(4xh)
= 4nx^2 + frac8000 nx
= 4n left(x^2 + frac2000x right). $$
This tends to infinity both as $xto 0^+$ and as $xto +infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve:
beginalign
&0 = C'(x)
= fracmathrmdmathrmdx 4n left(x^2 + frac2000x right)
= 4n left( 2x - frac2000x^2 right) \
&qquadimplies 0 = 2x - frac2000x^2 \
&qquadimplies 0 = 2x^3 - 2000 \
&qquadimplies x^3 = 1000 \
&qquadimplies x = sqrt[3]1000 = 10.
endalign
As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.
add a comment |Â
up vote
1
down vote
You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correctâÂÂI did not check them)
$$ C(x)
= m(2x^2) + n(4xh)
= 4nx^2 + frac8000 nx
= 4n left(x^2 + frac2000x right). $$
This tends to infinity both as $xto 0^+$ and as $xto +infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve:
beginalign
&0 = C'(x)
= fracmathrmdmathrmdx 4n left(x^2 + frac2000x right)
= 4n left( 2x - frac2000x^2 right) \
&qquadimplies 0 = 2x - frac2000x^2 \
&qquadimplies 0 = 2x^3 - 2000 \
&qquadimplies x^3 = 1000 \
&qquadimplies x = sqrt[3]1000 = 10.
endalign
As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correctâÂÂI did not check them)
$$ C(x)
= m(2x^2) + n(4xh)
= 4nx^2 + frac8000 nx
= 4n left(x^2 + frac2000x right). $$
This tends to infinity both as $xto 0^+$ and as $xto +infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve:
beginalign
&0 = C'(x)
= fracmathrmdmathrmdx 4n left(x^2 + frac2000x right)
= 4n left( 2x - frac2000x^2 right) \
&qquadimplies 0 = 2x - frac2000x^2 \
&qquadimplies 0 = 2x^3 - 2000 \
&qquadimplies x^3 = 1000 \
&qquadimplies x = sqrt[3]1000 = 10.
endalign
As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.
You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correctâÂÂI did not check them)
$$ C(x)
= m(2x^2) + n(4xh)
= 4nx^2 + frac8000 nx
= 4n left(x^2 + frac2000x right). $$
This tends to infinity both as $xto 0^+$ and as $xto +infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve:
beginalign
&0 = C'(x)
= fracmathrmdmathrmdx 4n left(x^2 + frac2000x right)
= 4n left( 2x - frac2000x^2 right) \
&qquadimplies 0 = 2x - frac2000x^2 \
&qquadimplies 0 = 2x^3 - 2000 \
&qquadimplies x^3 = 1000 \
&qquadimplies x = sqrt[3]1000 = 10.
endalign
As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.
answered Aug 17 at 3:43
Xander Henderson
13.2k83150
13.2k83150
add a comment |Â
add a comment |Â
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The costs are just constants---you don't know what they are, but they don't change. The only variable involved is $x$, so push forward with your optimization techniques there.
â Xander Henderson
Aug 17 at 3:34