Proving smallest surface area of cylinder is when h=2r

[sanddrag]

Originally Posted by Kevin Sevcik

sanddrag: V is a constant in this case since you said yourself for a given volume treat it as another number. Thus:
A' = 4*pi*r - 2*V/r^2

I can follow you here. The key IS to think of V as something like some random number (has a derivative of 0). Then when I set this equal to zero and solve I get 2V=4*pi*r^3. Then I substitue in pi*r^2*h for V and voila! it reduces down to 2h=r.



Thanks so much everyone!