Most efficient cost for a cylinder

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Homework Statement


An open-topped cylinder is to have a volume of 250 cm3. The material for the bottom of the pot costs 4 cents per cm2, and the material for the side of the pot costs 2 cents per cm2. What dimensions will minimize the total cost of this pot?

The Attempt at a Solution


$$
A_{bottom}=πr^2
\\
C_{bottom}=4(πr^2)
$$

$$
A_{side}=2πrh
\\
C_{side}=2(2πrh)
$$

$$
V=πr^2h
\\
250=πr^2h
\\
h=\frac {250}{πr^2}
\\
∴C_{side}=2(2πr\frac {250}{πr^2})
$$

$$
C_{total}=4(πr^2+2(2πr\frac {250}{πr^2})
\\
\frac {d(C_{total})}{d(r)}=8πr-\frac{1000}{πr^3}
$$

Then I tried to use the first derivative test. I am stuck.
 
Last edited:
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Uhh I messed something up there with itex.
 
I really messed up the formatting in that first post so it kind of looks like a mess. Until I figure that out, perhaps someone could point me in the right direction to solving the question?
 
Okay I think I fixed all the formatting. Silly me.
 
NEVERMIND. I figured out my silly error. It's all good now. Can I delete this thread?
 
EricPowell said:
NEVERMIND. I figured out my silly error. It's all good now. Can I delete this thread?
We don't delete threads as a matter of course. Even though it's of no use to you now, others might find it helpful.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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