# Simple optimization problem - Box [on hold]

"You're asked to design a box of volume 1m^3 with height 1/2m. What dimensions (length and width) will use the least material?

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Does the box have a top? –  André Nicolas Feb 28 '13 at 1:21
The problem doesn't specify, I would work it out as if it did. –  John McKevitt Feb 28 '13 at 1:22

We assume that the box has a top, and is of the usual "box" shape. If there is no top, the same general procedure applies. For a discussion of cases where the bottom is not a rectangle, please see the remark at the end.

Let the base of the box be $x\times y$. Then the volume is $\frac{1}{2}xy$. This should be $1$, so $xy=2$

The amount of material used is $2xy + \frac{1}{2}(2x+2y)$. Since $xy=2$, we conclude that the amount of material is $4+x+y$.

We want to minimize $4+x+y$, given that $xy=2$.

There are many ways to do this, without calculus or with. For example, to minimize $x+y$ is, for positive quantities, equivalent to minimizing $(x+y)^2$.

Note that $$(x+y)^2=(x-y)^2+4xy=(x-y)^2+8.$$ This is smallest when $(x-y)^2=0$. Thus we need to set $x=y=\sqrt{2}$.

Another way: We want to minimize $4+x+y$ given that $xy=2$. Thus $y=\frac{2}{x}$. Then the amount of material used, as a function $A(x)$ of $x$, is given by $$A(x)=4+x+\frac{2}{x}.$$ Calculate $A'(x)$. We get $A'(x)=1-\frac{2}{x^2}$. That gives critical point $x=\sqrt{2}$. It is not hard to argue that at this critical point $A(x)$ reaches a minimum.

Remark: Suppose that we make the bottom of the box circular. So the box is a cylinder.

Let the radius of the base be $x$. Then the area of the base is $\pi x^2$. Since the box has height $\frac{1}{2}$ and volume $1$, we find that $x^2=\frac{2}{\pi}$.

For the top and bottom, we use a total amount of material $\pi x^2+\pi x^2$, that is, $4$. For the material on the "side," we need $\frac{1}{2}(2\pi x)$. This turns out to be $\sqrt{2\pi}$. The total amount of material is therefore $4+\sqrt{2\pi}$. This easily beats the box-shaped box we discussed in the answer.

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thank you andre –  John McKevitt Feb 28 '13 at 1:36
You are welcome. I gave two solutions to show that calculus is not the only tool in max/min problems. –  André Nicolas Feb 28 '13 at 1:41

Let's have 2 variables l for length and w for width

Since we know that $V = \dfrac{1}{2}lw = 1$, thus $w = \dfrac{2}{l}$

So now we have $$Length = l\ \ ;Width = \dfrac{2}{l}$$

Since a box has 6 sides(I assume this is a closed box), the total area of material we are going to use is: $$A = 2\cdot (l\cdot \dfrac{2}{l}+\dfrac{1}{2}\cdot l+\dfrac{1}{2}\cdot \dfrac{2}{l})$$

(Think why this is true.)

So we have $$A = 2(2+\dfrac{l}{2}+\dfrac{1}{l}) = 4+l+\dfrac{2}{l}$$

So $$\dfrac{dA}{d\ l} = 1-\dfrac{2}{l^2}$$

Set this to 0, we get $l = \sqrt{2}$. Hence $w = \dfrac{2}{\sqrt{2}} = \sqrt{2}$

Now you can verify that answer by taking the second derivative and plug in the value of l to verify that it is indeed a minimum.

Another question for you to think about is why this turns out to be a square? What's the geometric implication of that in terms of using minimal material?

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thank you very much Enzo –  John McKevitt Feb 28 '13 at 1:36
No problem :) You are very welcome, also Andre's solution is really neat since it doesn't use calculus. –  Enzo Feb 28 '13 at 1:49