Answer to POW #3:

Given a fixed volume $V$, we need to minimize the surface area

\begin{displaymath}S=2 \pi r h + 2 \pi r^2 h\end{displaymath}

with the constraint that $V=\pi r^2 h$. Solving the constraint for $h$, we get $h=\frac{V}{\pi r^2}$, and thus

\begin{displaymath}S=2 \pi r \left(\frac{V}{\pi r^2}\right) + 2 \pi r^2 = 2Vr^{-1} + 2\pi r^2.\end{displaymath}

Differentiating with respect to $r$, we get

\begin{displaymath}S'=-2Vr^{-2}+2\pi r = \frac{-2V + 4\pi r^3}{r^2},\end{displaymath}

which does not exist when $r=0$, and is zero when $4\pi r^3 = 2V$, i.e. when

\begin{displaymath}r=\left(\frac{V}{2\pi}\right)^{\frac{1}{3}}.\end{displaymath}

For this $r$ we have

\begin{displaymath}h = \frac{V}{\pi r^2} = \frac{V}{\pi\left(\frac{V}{2\pi}\right)^{\frac{2}{3}}} = 2\left(\frac{V}{2\pi}\right)^{\frac{1}{3}}.\end{displaymath}

Therefore, the ratio of height to radius must be $\frac{h}{r} = 2$, regardless of the desired volume $V$ of the can.