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Math Help - Optimization

  1. #1
    Senior Member topher0805's Avatar
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    Optimization

    A rectangular storage container with an open top is to have a volume of 10 meters cubed. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

    My work:

    The area of the base is given by:

    A_b = 2w^2

    The area of the sides is given by:

    A_s = 4wh

    I know the cost per square meter so:

    C_b = 20w^2

    C_s = 24wh

    So the total cost is represented by:

    C = 20w^2 + 24wh

    And I know that the volume is equal to 10 cubic meters so:

    V = l\cdot w\cdot h

    l\cdot w\cdot h = 10

    2w^2\cdot h = 10

    w^2\cdot h = 5

    h = \frac {5}{w^2}

    I plug that into my cost formula:

    C(w) = 20w^2 + \frac {120}{w}

    Take the derivative:

    C'(w) = 40w - \frac {120}{w^2}

    I find the critical points to be 0, 1.44.

    Then I find that the absolute minimum is 1.44 but when I use this number I get the incorrect answer of 124 (ish).

    Any suggestions?
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  2. #2
    Senior Member topher0805's Avatar
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    Area of the sides should be 6wh, you idiot.
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  3. #3
    Senior Member topher0805's Avatar
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    Even when I use 6wh as the area of the sides, I get an incorrect answer of 81.18

    GRRRR, this question is frustrating me.
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  4. #4
    Senior Member topher0805's Avatar
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    Hmmm, if I use the formula:

    <br />
C = 20w^2 + \frac {180}{w}

    And I plug in the cubed root of 4.5 for w, then I get an answer of 163.54 but if I do it in two steps and use the knowledge that h = 5/w^2 to solve it I get an answer of 81.18....oh well, 163.54 appears to be correct.
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