Consider the following problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.

For this I started by drawing some pictures and writing some things down as shown:

enter image description here

I determined that the formula for volume in this case is $V=l^2 \cdot h$

The problem asked me to also draw some examples and determine the volumes for these and guess the optimum. I was able to do this with the absolute maximum volume is $2 ft^3$.

However, I need to solve this using Calculus and the next steps I would take would to find another formula to relate $l$ and $h$ together. So after thinking about this I came up with:

$A = (l + 2h)^2 = 9$ where the formula uses the "cut" box geometry to determine the original area of the pre-cut cardboard box, which is $9ft^2$ in this case. I simplified this again by just relating $l$ and $h$ to be just one side of the pre-cut box. So I ended up with just $l + 2h = 3$.

So now I can use this to substitute the $l$ in the volume formula in terms of $h$. So:

$l + 2h = 3$

$l = {3 \over 2h}$


$V=({3 \over 2h})^2 \cdot h$

When I simplify this I get:

$V={9h \over 4h^2}$

and the derivative:

$V'={9h^2 \over 4h^4}$

The derivative doesn't solve for 0, so this implies not global maximum. So either I made a mistake somewhere, or I used the wrong formula to relate $l$ and $h$. I've spent and hour on this problem and am not sure what else to do.


You have:


From this does not follow that $l=\frac 3{2h}$.

Instead, you should have:


  • 2
    $\begingroup$ Despite looking at this for an hour, I didn't catch this mistake. Well that certainly can make this a problem. $\endgroup$ – Matt Jul 5 '14 at 22:08

One side of the box is $l-2x$ where x is the length of one of the four corners. The area of the bottom is $A=(l-2x)^2$. And the volume is $V=(l-2x)^2\cdot x$. This mathematical term can be optimized with respect to x.


But you can solve this without using calculus... (or at least without any hard calculation)

Consider an infinitesimal change $\delta$ in the position of the fold between walls and floor. At the maximum, this change must add/subtract the same volume from the change in height as from the moving in/out of the walls, and this change is $\delta$ multiplied by the relevant area. So this maximum is when the area of the base and the area of the walls are the same, i.e.:

$l^2 = 4lh$


$l = 4h$ i.e. the box shape is 4:4:1

In your example it follows immediately that the base is 2 ft square, and the volume is 2 cu. ft.

(Crumbs, it must be several decades since I last used those quaint old units.)


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