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

[Joshua May]

Originally Posted by sanddrag

I need to prove that surface area A of a cylinder (of volume V) is smallest when the height, h equals twice the radius, r.

We have two equations

1. V=(pi)(r^2)(h)
2. A=2(pi)(r^2) + 2(pi)(r)(h)

Here's what I think I need to do in order to prove the smallest A occurs when h=2r. I think I need to take equation 1 and solve for h which comes out to h=V/(pi)(r^2). Then I need to substitue that in for h in equation 2. This would make equation 2 become:

A=2(pi)(r^2) + 2(pi)(r)[V/(pi)(r^2)]

Now, I know this can be simplified, but I'm not sure if I'm doing it right. I think it would go to:

A=2(pi)(r^2) + 2V/r

Is this right? If so, what do I need to do now? I know I have to take the derivative to find the absolute minium. This is where I'm havign trouble. Can someone please help. Thanks.

I think you've got it so far.

I believe the next step to go is to find the derivative of A=2(pi)(r^2) + 2V/r, set that to zero, solve it, and do a sign test to show that this is a minimum point.

Sorry, but I haven't done optimization problems in a few months, so I might be a little rusty, but this should set you on the right track.