# Maximizing volume of a box

Say you have a sheet of gold with dimensions $$5$$ units by $$8$$ units, and you want to cut out a square with side-length $$z$$ from each corner of the box so that you can subsequently fold the sides up in order to construct an open-topped box. What is the optimal value of $$z$$ if we want to maximize the volume of the resulting box?

Honestly I think the problem I am having is visualizing this. I've tried drawing pictures, and I still don't get it. I think this is like a calculus optimization question, so I tried setting up equations.

Volume = $$40 \cdot z$$.

A picture of what is left when you cut a square of size $$z$$ from each corner of the sheet is below with $$z=1$$. The height will be $$z$$, but the bottom will be smaller than $$5 \times 8$$. Once you figure out the area of the bottom, the volume will be a cubic in $$z$$. Differentiate and set to zero....  $$V = (8 - 2 z) (5 - 2 z) z$$ $$\frac{\partial V}{\partial z} = 12 z^2 -52 z + 40$$

Set to $$0$$ to find $$z = 1$$ (or $$10/3$$ can be rejected) Hint, maybe this drawing helps: $$V=Z*(W-2Z)(H-2Z)$$

Given $$W=5,H=8$$

$$V=Z*(5-2Z)(8-2Z)$$

So now we need to maximize: $$V=Z*(5-2Z)(8-2Z)$$ Subject to: $$(5-2Z)(8-2Z)=40$$

V has 1 or more critical point(s) when: $$\frac{dV}{dZ}=0$$

That is:

$$\frac{d}{dZ}\left(Z\left(5-2Z\right)\left(8-2Z\right)\right)=12Z^2-52Z+40=0$$ $$Z=0, Z=\frac{10}{3}, Z=1$$

When $$Z=0$$, we don't get a box. We get a square. Reject this value.

When $$Z=10/3$$, we get $$W=5-2Z$$ negative, so its no good.

When $$Z=1$$ we use the 2nd derivative test to identify the whether $$Z=1$$ is a min. or max. for $$V$$:

$$\frac{d^2}{dZ^2}\left(Z\left(5-2Z\right)\left(8-2Z\right)\right)=24Z-52$$

At $$Z=1$$ the above expression is negative. Hence $$V$$ has a local maximum at $$Z=1$$.

The volume becomes: $$V=5 * 8 * 1 \hspace{.4cm} unit^{3}$$

It is convenient to work with symbols.

Let the length and width be denoted by $$L,W.$$ The Volume $$V= x(L-2x)(W-2 x)$$

For convenience let $$2x = u \rightarrow 2V= u (u-L)(u-W)$$

Differentiate w.r.t. $$u$$ and simplify $$3u^2-2u(L+W)+LW=0$$ Solve the quadratic equation $$u=2x= \frac{ (L+W)\mp \sqrt{L^2+W^2-LW}}{2}$$ Now insert the given values $$L=8, W=5$$ $$2x= (10,3), \,x = 1.5$$

First solution discarded when finding remaining length/width because we cannot subtract 10 from 8 to result in a positive solution.

Remaining length and width are $$(8- 2*1.5, 5-2*1.5)= ( 5, 2)$$ The corners are cut or folded and the 3D box has dimensions $$(L,W,H)$$ $$(5,2,1.5)$$ and open topped box Volume $$LWH= 15$$ cubic units. It remains to verify by second differentiation in calculus procedure maxima/minima that this is in fact a maximum.

Think of it like the red cross symbol. It fits snuggly inside of a square, but the missing squares in the four corners are the ones with side length $$z$$. Then if you fold the "arms" of the symbol up, it creates a rectangular prism, with height $$z$$, and base lengths $$s-2z$$, where $$s$$ is the sidelength of the original square we cut up. This has a volume of

$$V = z(s-2z)^2$$

Could you come up for a formula where the original shape we cut from was a rectangle instead of a square?

• The base is not square, it is rectangular Sep 17 '19 at 5:18
• I did. You assume the base is $(s-2z)$ square in your final formula. In the text you say it is $s-2z$, but the units are not correct for that. Sep 17 '19 at 5:19
• @RossMillikan which it is, if we start from a square, which I did. I wanted to give an example to OP to let them work out the rectangular case for themselves. Sep 17 '19 at 5:20
• @RossMillikan Please, your edit is only to save face, I am very sure that is not what you meant to point out initially. However, thank you for catching that typo, I appreciate it regardless. Sep 17 '19 at 5:24