# Ask Uncle Colin: Dimensions of a box

Dear Uncle Colin,

I’m told that a rectangular box has a surface area of 64cm2, and I have to find the maximum possible volume. How would I do that?

- Can Uncle Bring Obviousness Into Differentiation?

Hi, CUBOID, and thanks for your message - I certainly hope I can!

We have two relevant equations here: if the sides of the box are $x$, $y$, and $z$, then the volume is $V = xyz$ and the surface area is $2xy + 2yz + 2zx = 64$ - better still, $xy + yz + zx = 32$.

I’m going to try to do this without resorting to vector calculus (which would probably be simplest.)

Let’s start by assuming - for the moment - that $z$ is a fixed, known value. We can then say: $y(x+z) = 32 - xz$, so $y = \frac{32-xz}{x+z}$.

Now we can write the volume as $V = \frac{xz(32-xz)}{x+z}$.

You thought we were going to differentiate? Well, so did I, but I threw myself a symmetry dummy: we could just as easily have made $x$ the subject and got $V=\frac{yz(32-yz)}{y+z}$, which is identical: this means that for any fixed value of $z$, $x=y$ at the maximum.

That means, for any given value of $z$, we have $V=x^2z$ and a surface area of $x^2 + 2xz=32$.

But, we can rearrange that last relationship: $z = \frac{32-x^2}{2x}$. Putting that into the $V$ equation gives $V = \frac{x\br{32-x^2}}{2}$. Differentiating gives $\diff Vx = \frac{32 - 3x^2}{2}$, which has a maximum at $x= \sqrt{\frac{32}{3}}$.

That gives us $V = \frac{1}{2}\sqrt{\frac{32}{3}} \frac{64}{3}$, or $\frac{128}{9}\sqrt{6}$. ((It turns out, to the surprise of nobody who’s played with this before, that the maximum volume comes from a cube.))

Hope that helps!

- Uncle Colin