We have a special type of cylinder whose height, h, is always twice its radius r. What is the rate of change of its volume V with respect to r in terms of the total surface area, A?

I figured V= (pi)(r^2)(h) and since h=2r , V = (pi)(r^2)(2r)

Surface Area = 2(pi)(r^2)+(pi)(r)(h) = 2(pi)(r^2)+(pi)(2r^2)

I'm really not sure what to do after that though, thanks.

- Adrian