A puzzle that came to me by way of Barney Maunder-Taylor:

A long sheet of plastic is turned into a gutter by folding up two sides. What fraction of the width should be bent up and through what angle to maximise the amount of water it can hold?

(Unstated, but implied: both sides are the same length and folded up by the same angle.)

Have a crack at it, and I’ll take you through a plodding method and a lovely way.

Spoilers below the line.

Ploddingly

What he have here is a trapezium. If the original sheet has width 1, and we fold up sides of length $x$ by an angle of $\theta$, the trapezium has a base of length $1-2x$, a height of $x \sin(\theta)$, and an upper side of $(1-2x) + 2x\cos(\theta)$.

The resulting area is $\br{x\sin(\theta)}{(1-2x+x\cos(\theta))}$. Which is a bit of a mess, so let’s tidy it up very slightly:

  • $A = \br{x\sin(\theta)}\br{1 + x\br{\cos(\theta)-2}}$

… and we need to maximise that.

It’s a ready-factorised quadratic! The turning point must be midway between the roots.

One root is at 0, the other at $\frac{1}{2-\cos(\theta)}$, so the maximum is at $x=\frac{1}{4-2\cos(\theta)}$ for any $0 < \theta < \piby 2$.

Now we have $A = \frac{3}{2}\br{ \frac{\sin(\theta)}{4-2\cos(\theta)}}$. What value of $\theta$ maximises this? It’s time for some ugly quotient rule:

$\diff{A}{\theta} = \frac{(4-2\cos(\theta))\cos(\theta) - 2\sin^2(\theta)}{(4-2\cos(\theta))^2}$, which has to be zero.

So:

  • $4\cos(\theta) - 2\cos^2(\theta) - 2\sin^2(\theta) = 0$
  • $4\cos(\theta) = 2$
  • $\cos(\theta) = \frac{1}{2}$ and $\theta = \piby3$.

Going back, this makes $x = \frac{1}{3}$. Horrible calculation, nice answer.

Quickly

Put another sheet upside down on top, connected along the long sides, so that it mirrors the first. The gutter is now a tube in the shape of a hexagon with a fixed perimeter.

The fixed-perimeter hexagon with the largest cross-sectional area is regular – so the solution is to fold the sheet in thirds, with an angle of $\piby 3$.

Lovely puzzle! Did you tackle it differently?