# Another Calculus Question

1. Jan 22, 2006

### Jordan_

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.

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?

2. Jan 22, 2006

### durt

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.

3. Jan 22, 2006

### Jordan_

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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