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Re: Proving smallest surface area of cylinder is when h=2r
Josh has the rest of the process right. It looks like you're going to get a 3rd order polynominial when you get dA/dr. So you should get 3 points where it equals zero. After you find the minimum, you'll have an answer for R in terms of V. You'll need to plug this back into your equation for h in terms of V and r to show that h = 2r.
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