




Minimize the surface area A of a tin can of fixed volume V.
Area of the tin can A = 2p(r^{2} + rh)
Volume of the tin can: V =pr^{2}h
We minimize fixed
Compute partial derivatives
The critical condition becomes
r^{2}(2r + h) = 2r^{2}h or 2r = h
G is minimized with F fixed when 2r = h.
(you must check that the endpoint behavior is consistent with the critical point being the minimum for G. If h goes to 0 then r must get large to keep V fixed, so that A goes to infinity. If r goes to 0 then h must behave as 1/r^{2} so that again A goes to infinity.)