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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 = $40 \cdot z$.

But I don't know how to come up with the constraint function. Can someone please help?

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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....

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$V = (8 - 2 z) (5 - 2 z) z$

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$\frac{\partial V}{\partial z} = 12 z^2 -52 z + 40$

Set to $0$ to find $z = 1$ (or $10/3$ can be rejected)

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Hint, maybe this drawing helps:

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$$V=Z*(W-2Z)(H-2Z)$$

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

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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.

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It remains to verify by second differentiation in calculus procedure maxima/minima that this is in fact a maximum.

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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?

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  • $\begingroup$ The base is not square, it is rectangular $\endgroup$ Sep 17 '19 at 5:18
  • $\begingroup$ 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. $\endgroup$ Sep 17 '19 at 5:19
  • $\begingroup$ @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. $\endgroup$ Sep 17 '19 at 5:20
  • $\begingroup$ @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. $\endgroup$ Sep 17 '19 at 5:24

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