Shaun of Nottingham High School and also Ben
and Thomas sent in good solutions to this problem. Here is Shaun's
solution.
There are three possibilities we must consider, in each one the 'middle' rod (vertical in the diagram ) is one of $p$, $q$ or $r$. Interchanging the horizontal rods either side makes no difference to the area of the shape, it is just being flipped vertically. Remember the area of a trapezium can be expressed as ${1\over 2}(a+b)h$, where $a$ and $b$ are parallel sides, and $h$ the distance between them. |

This means we can express the areas of the three possible
shapes as:

$$\frac {1}{2}(p+q)r, \frac {1}{2}(q+r)p, \frac
{1}{2}(r+p)q.$$

Let's compare the first two. As $qr> qp$ we have $qr + pr
> qp + pr$ and so

$${1\over 2}(p+q)r > {1\over 2}(q+r)p.$$

I hope you can see an identical argument can be used to
compare the other pairings of expressions for area, and we end up
with:

$${1\over 2}(q+p)r > {1\over 2}(p+r)q > {1\over
2}(r+q)p.$$

So the rods should be arranged with $r$ being the middle one
in order to enclose the maximum area.