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 $.