= Solution
For a closed <circular cylinder>, $S=2\pi r^2+2\pi rh$ and the volume constraint is $\pi r^2h=V$. Use a <Lagrange multiplier>:
$$
\mathcal L=2\pi r^2+2\pi rh-\lambda(\pi r^2h-V).
$$
The stationary equations are
$$
4\pi r+2\pi h-2\lambda\pi rh=0,\qquad 2\pi r-\lambda\pi r^2=0.
$$
For $r>0$, the second gives $\lambda=2/r$; the first then gives $h=2r$. The constraint yields
$$
\boxed{r=\left(\frac{V}{2\pi}\right)^{1/3},\qquad h=2\left(\frac{V}{2\pi}\right)^{1/3}},
$$
and
$$
\boxed{S_{\min}=6\pi\left(\frac{V}{2\pi}\right)^{2/3}}.
$$
To verify a global <minimum>, eliminate $h$ to obtain $S(r)=2\pi r^2+2V/r$. It tends to infinity at both ends of $r>0$ and has $S''(r)=4\pi+4V/r^3>0$, so its unique <stationary point> is the global <minimum>.
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