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 $\frac{128}{9}\sqrt{6}$.1

Hope that helps!

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. It turns out, to the surprise of nobody who's played with this before, that the maximum volume comes from a cube. []

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