Thread: Absolute Maximum Value

A can (cylinder) must have a volume of 24π m^3. Material for the bottom costs $0.15/m^2 and all other material costs $0.05/m^2. Which dimensions minimize the cost of material?

So far I know that A= πr^2 + 2πrh

v=π(r^2)h

and therefore

h=(24π)/(πr^2)

Rearranging the formula for area, I got

A= πr^2 +48π/r

I thought the next step would be to find the first derivative of the area with respect to r, then solve for r (my result was about 2.5). Then I figured I would find the second derivative and sub in r to make sure it was greater than 0. (because the answer must be 0).

Then I realized it didn't account for the difference in cost between the bottom and side of the can. How can I include this? Thanks in advance!

cost in cents ...