South African Mathematics Olympiad

Suppose that the rectangular sheet of paper has dimensions $a$ and $b$, with $b$ being the longer side. We calculate the volume of the two cylinders formed by joining the long sides and short sides, respectively.

Suppose the short sides are glued together. Then the height of the cylinder is $a$ and the circumference of the cylinder is $b$. If $r$ is the radius of the cylinder, it means that $2\pi r=b$, or $r=\frac{b}{2\pi }$. Hence the cylinder's volume is $\pi r^{2}h=\pi (\frac{b}{2\pi }{)}^2\cdot a=\frac{a{b}^2}{4\pi }$.

Suppose the long sides are glued together. Then the height of the cylinder is $b$ and the circumference of the cylinder is $a$. If $r$ is the radius of the cylinder, it means that $2\pi r=a$, or $r=\frac{a}{2\pi }$. Hence the cylinder's volume is $\pi r^{2}h=\pi (\frac{a}{2\pi }{)}^2\cdot b=\frac{a^{2}b}{4\pi }$.

Comparing these two numbers, we see that the first volume can be written as $b(\frac{ab}{4\pi })$, while the second volume is $a(\frac{ab}{4\pi })$. Since the number in brackets is the same for both and $b>a$, it means that the first volume $b(\frac{ab}{4\pi })$ is the largest. Hence the largest volume is obtained when the short sides of the sheet of paper are glued together.