Maximum Cross-Sectional Area Implies Maximum Volume?

What ratio gives us the maximum volume? Call the horizontal dimension $x$ and the vertical dimension $y$ . The volume is (when revolved around the vertical axis):

$V = \pi x^2 y$

The perimeter constraint is:

$P = 2 x + 2 y \\ y = \frac{P}{2} - x$

Substituting in for $y$ :

$V = \pi x^2 \Big(\frac{P}{2} - x\Big) = \frac{P \pi}{2} x^2 - \pi x^3$

Find the $x$ value corresponding to the max volume (the second derivative test, etc. are left as exercises for the reader):

$P \pi x - 3 \pi x^2 = 0 \\ x = \frac{P}{3}$

In the square case, $x = \frac{P}{4}$ . It makes sense that the volume should be maximized when the horizontal component takes up a larger share of the perimeter, since the volume depends on the square of the horizontal component, and only on the first power of the vertical component.