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Math Help - 2 variable optimization problem

  1. #1
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    2 variable optimization problem[SOLVED]

    Find the minimum cost of a box (with dimensions xyz) that holds 16 m^3. The top and bottom cost $20 and the sides cost $10.

    So the formula for the cost is C=40xy + \frac{320}{x} + \frac{320}{y}

    the partial derivatives are 40y-\frac{320}{x^2} and 40x-\frac{320}{y} and I set them equal to 0 to get
    y=\frac{8}{x^2} (1) and x=\frac{8}{y^2} (2)

    Then I do something like (1)+(2) to get x^3=8 but I get lost along the way.
    Last edited by superdude; October 18th 2009 at 10:20 AM. Reason: solved
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  2. #2
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    Quote Originally Posted by superdude View Post
    Find the minimum cost of a box (with dimensions xyz) that holds 16 m^3. The top and bottom cost $20 and the sides cost $10.

    So the formula for the cost is C=40xy + \frac{320}{x} + \frac{320}{y}

    the partial derivatives are 40y-\frac{320}{x^2} and 40x-\frac{320}{y} and I set them equal to 0 to get
    y=\frac{8}{x^2} (1) and x=\frac{8}{y^2} (2)

    Then I do something like (1)+(2) to get x^3=8 but I get lost along the way.
    Yes, you're right!

     y = \frac{8}{x^2}  \rightarrow x = \pm \sqrt{\frac{8}{y}}

    And you also know that  x = \frac{8}{y^2}

    Therefore  \frac{8}{y^2} =  \pm \sqrt{\frac{8}{y}}

    Square both sides:

     \frac{64}{y^4} =  \frac{8}{y}

     \therefore y^3 = 8 \rightarrow y = 2

    Now you can also find x by substituting this value into either equation you had at the start.
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  3. #3
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    thanks
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