A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20 per m^2, while that for the side costs $10 per m^2. Determine the radius of the drum that minimizes the cost of the material used.

Question

A cylindrical drum is made to hold exactly 1m^3 in its interior. Assume that the material for the top and the bottom costs $20 per m^2, while that for the side costs $10 per m^2. Determine the radius of the drum that minimizes the cost of the material used. 
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/ \ 
|\____/| <--------Drum 
| | h 
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I posted this question before but I wasn't sure about one part. 
Eqn's: 
(1) Area = 2PI*r^2 + 2PI*rh 
(2) Cost = 20*2PI*r^2 + 2PI*rh

Solving for "h" for eqn (1) 
h = -Area(2PI*r^2/(2PI*r)

did I do this step right cuz after i solve for h I am able to sub that into eqn 2, and find DCost/dt but I feel Im missing a step. Any help will be appreciated! :)

Use the volume equation to solve for h.

Volume= PI r^2 h 
h= volume/PIr^2 
Then put that into cost. 
Cost = 20*2PI*r^2 + 2PI*r(volume/PIr^2)

Now simplify, and take the derivative of cost with respect to r.