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Math Help - Problem with Optimization

  1. #1
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    Question Problem with Optimization

    This question really threw me for a loop:
    A closed rectangular container with a square base is to have a volume of 2250 cubic inches. The material for the top and bottom of the container will cost $2 per square inch and the material for the sides will cost $3 per square inch. Find the dimensions of the container of least cost.

    What I have so far:
    V=x^2h = 2250 so h = \frac{2250}{x^2}

    S = 2x^2 + 4xh so S = 2x^2 + 4x * \frac{2250}{x^2} = 2x + \frac{9000}{x}

    But when I thought about trying to find the feasible domain, I got stumped. The question doesn't specify a minimum or maximum value for either a side of the base or the height of the container. So, theoretically, x would get infinitely large as h got infinitely small and vice versa. Is there a generally accepted thing to do in this situation or am I doing something wrong here?
    Thanks for the help


    Oh, and i had a few more equations relative to the problem:
    I separated the equation for surface area into S = s_{1} + s_{2} where S is the total surface area, s_{1} is the area of the base( 2x^2), and s_{2} is the area of a side( 4xh).
    I then made the equation P = 2s_{1} + 3s_{2} where P is the total price. But I didn't know what to do with all this...


    Edit: I suppose I'll just attempt to finish the problem with a feasible domain of 0 < x < \infty, although I won't be able to test for the endpoints... oh well.
    Last edited by wombat; November 27th 2010 at 06:45 PM.
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  2. #2
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    You were close, but the question asks for price minimisation, not surface area minimisation.

    x^2h = 2250

    \text{Price} = 2x^2 \cdot \$2 + 4xh \cdot \$ 3

    Make the substitution:

    \text{Price} = 2x^2 \cdot \$2 + 4x(\frac{2250}{x^2}) \cdot \$ 3

    \text{Price} = 4x^2 + \frac{27000}{x}

    Now differentiate:

    P(x) = 4x^2 + \frac{27000}{x}

    P'(x) = 8x - \frac{27000}{x^2}


    Set P'(x) to 0 and solve for x.
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  3. #3
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    Ah, yes. It makes so much more sense now. Thanks for the help.
    In the end, i got x = 15 and h = 10. So a box with dimensions of 15" x 15" x 10" would do the trick.
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