Solid

Problem:



Solution:

Minimize $4\pi r^2 + 2\pi r h$ subject to $\frac{4}{3}\pi r^3 + \pi r^2 h = 6$.

The Lagrangian is $4\pi r^2 + 2\pi r h - \lambda(\frac{4}{3}\pi r^3 + \pi r^2 h - 6)$.

The partial derivatives must be 0, so we have

$8\pi r + 2\pi h - \lambda (4 \pi r^2 + 2\pi r h) = 0$
$2 \pi r - \lambda(\pi r^2) = 0$

The second equation implies $\lambda = \frac{2}{r}$.

Substitute this back to the first equation, solving get $h = 0$.

This seems reasonable because we know sphere maximize volume and minimize surface area, it only make sense if the cylindrical part vanish.

So it is simple now, $\frac{4}{3}\pi r^3 = 6$, $r = \sqrt[3]{\frac{9}{2\pi}} = 1.127251652$.