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Math Help - Optimization problem

  1. #1
    VkL
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    Optimization problem

    You need to manufacture a cylindrical pot, without a top,
    with a volume of 1 ft3. The cylindrical part of the pot is to be made of aluminum, the bottom of copper. Copper is five times as expensive as aluminum. What dimensions would minimize the total cost of the pot?


    I Know how to do optimization. Find the derivative, set it equal to zero, find the absolute max and min, and compare it to the intervals given. But this problem is really bothering me. Can anyone help? what are the intervals here? and how do you set up the geometry/sketch of this?
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  2. #2
    MHF Contributor
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    Hi

    The cylinder is characterized by its height h and its radius r

    Volume is V_0 = \pi r^2 h

    V_0 is fixed equal to 1ft3 therefore h = \frac{V_0}{\pi r^2}

    Let a be the price of aluminum per ft2 ; price of copper is 5a per ft2

    Price of the cylinder is P = 5a \pi r^2 + a 2 \pi r h

    P(r) = 5a \pi r^2 + a 2 \pi r \frac{V_0}{\pi r^2}

    P(r) = a(5 \pi r^2 + 2 \frac{V_0}{r})

    \frac{dP}{dr}(r) = a(10 \pi r - 2 \frac{V_0}{r^2})

    \frac{dP}{dr}(r) = 0 for r = \sqrt[3]{\frac{V_0}{5\pi}} = 0.4 ft

    h = \frac{V_0}{\pi r^2} = 2 ft
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  3. #3
    VkL
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    Thank you For the explanation, Although I do not see how you determined the price of the cylinder running-gag. 5A(pie)r^2 + a2(pie)rh ?

    Everything else seems fine.
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  4. #4
    MHF Contributor
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    Well I assumed that the price is proportional to the area

    The area of the bottom is \pi r^2

    The area of the side is 2 \pi r h
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