What are the optimal dimensions for a soup can to minimize material costs?

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Alright, here is another problem I'm having trouble with. I used a technique I got off of someone on my other thread, and am just looking to see what should be done next.

A soup manufacturer wants to sell it's soup in 500mL cans. The metal for the top and bottom costs $1.20/m^2. The metal for the side costs $0.40/m^2. After the circles for the top and bottom are cut out of rectangle, the remaining metal will be scrapped. Find the dimensions of the can that will minimize the cost of materal.

ANS: 2.54m, 24.68m

Alright so first I got my equation for the surface area of a cylinder:
SA = 2(pi)rh + 2(pi)r^2

and converted 500mL to 0.0005m^3

Then I put in my cost numbers:
SA = 1.2[2(pi)r^2] + 0.4[2(pi)rh]

Now at this stage I used what I learned in the other thread and got the 'h' out of there by switching the volume formula around (V = (pi)r^2h) and substituting it for 'h'.

SA = (1.2)2(pi)r^2 + (0.4)2(pi)r[.0005/(pi)r^2)

Now I think I have to simplify and then take the derivative, but when I do that I get dr/dx's and then I get confused and don't know where to go. Anyone give me some tips?
 
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Before I even attempt this problem, I can tell you that there are no soup cans out there the size of a six-story building... 24 meters?

Anyway, you'll probably get confused by naming the cost function "SA," so rename it to "C." Then take the derivative of C with respect to r (dC/dr). There are no x's in there that you have to deal with.
 
durt said:
Before I even attempt this problem, I can tell you that there are no soup cans out there the size of a six-story building... 24 meters?
Anyway, you'll probably get confused by naming the cost function "SA," so rename it to "C." Then take the derivative of C with respect to r (dC/dr). There are no x's in there that you have to deal with.

Yes I know that number looks retardedly big. I copied it from the board but I'm sure she just made a mistake writing it. Just thought I'd post it incase you guys needed it.
 
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