Math CentralQuandaries & Queries


Question from Margot, a student:

I need to do a PA for maths and I'm a bit stuck.
The PA is about folding a box with a volume that is as big as possible. The first few questions where really easy but then this one came up.

These are the questions that I have to answer:
1. Calculate the exact measurements in mm of A4 paper.
2. Same size squares are cut from each corner of an A4 paper. After that a basket is folded. Find out what the size of the squares must be for the volume to be maximum.
3. Find out the maximum using a graph.
4. Find out the maximum by differentiating.
5. Same task as at 2. Start with a square piece of paper of 15 by 15 cm.
6. Choose at least another two square sizes of paper and determine the maximum volume of those sizes.
7. Start with a square paper with sides of z cm by z cm. Make another basket with a maximum volume. Find a formula to express the maximum content in z.
8. Prove by differentiating that the formula at 7 does indeed give you the maximum volume for each value of z.

So I already have the answers for questions 1 to 7 but I thought that it would maybe be useful to have an overview of the PA. My question is about number 8. For that you need question 7 so here are my calculations for 7:

V = l x h x w
l = w = z - 2h

V = h(z - 2h)^2
V(h) = h(z^2 - 4hz + 4h^2)
V(h) = 4h^3 - 4zh^2 + (z^2)h

So with the last formula you can calculate the volume and by using a graph you can read off the maximum volume. Or you can use the derived function: V'(h) = 12h^2 - 8zh + z^2 = 0

That is my answer to question 7. But how do you prove (by differentiating) that it's true for all values of z?
I tried to do something with the fact that z/h = 6 (I can send the calculations if anyone wants to). But I don't really see the use of it.
Also I thought that you could maybe use the second derivative?

Hi Margot,

You are almost done. You have the derivative of $V(h)$ and you have set it equal to zero.

\[12h^2 - 8zh + z^2 = 0.\]

This quadratic factors to yield

\[(6h - z)(2h - z) = 0\]

and thus $h = \frac{z}{6}$ or $h = \frac{z}{2}.$ The first gives you what you discovered in part 7. What about the second?


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