Originally Posted by

**galactus**

We have $\displaystyle V={\pi}r^{2}h=\frac{\pi}{8}$...[1]

The surface area, assuming the bin has only a bottom:

$\displaystyle S=2{\pi}rh+1{\pi}r^{2}$...[2]

This is what must be minimized.

So, we solve [1] for h and sub into [2]:

Solving [1] for h, we get $\displaystyle h=\frac{1}{8r^{2}}$

That goes into the surface formula:

$\displaystyle 2{\pi}r(\frac{1}{8r^{2}})+1{\pi}r^{2}$

Differentiate:

$\displaystyle S'(r)=2{\pi}r - \frac{\pi}{4r^{2}}$

Set to 0 and solve for r and we get

$\displaystyle 2 \pi r=\frac{\pi}{4 r^2}$

$\displaystyle r^3=\frac{1}{8}$

$\displaystyle r=\frac{1}{2}$

Plug that into h.

$\displaystyle h=\frac{1}{8(0.25)}=\frac{1}{2}$