A solid is formed by adjoining two hemispheres to the ends of a (right) cylinder.

Given that the volume of the solid is 12cm^3, find the radius of the cylinder that produces the minimum surface area for the solid.

Note that the radius of the cylinder and the radius of the hemispheres will be the same. The surface area of the solid:

SA = 2πrh + 2πr^2 + 2πr^2
SA = 2πrh + 4πr^2

The volume of the solid:

V = πr^2h + (2/3)πr^3 + (2/3)πr^3
V = πr^2h + (4/3)πr^3
12 = πr^2h + (4/3)πr^3
h = [12 - (4/3)πr^3] / [πr^2]

Sub that into the surface area formula:

SA = 2πr[12 - (4/3)πr^3] / [πr^2] + 4πr^2
SA = [24 - (8/3)πr^3] / [r] + 4πr^2

Take the derivative; set it equal to zero:

SA = [-8πr^3 - 24 + (8/3)πr^3] / [r^2] + 8πr
0 = [-8πr^3 - 24 + (8/3)πr^3] / [r^2] + 8πr
0 = -8πr^3 - 24 + (8/3)πr^3 + 8πr^3
0 = (8/3)πr^3 - 24
9 / π = r^3
r = [9 / π]^(1/3)

Verify that this is a minimum with the second derivative, or just take my word for it.

Done!