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Math Help - Absolute Maximum Value

  1. #1
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    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!
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  2. #2
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    Re: Absolute Maximum Value

    cost in cents ...

    C = 15(\pi r^2) + 5(\pi r^2 + 2\pi rh)
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