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Math Help - Optimization

  1. #1
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    Optimization

    A box is to be made out of a 10 by 14 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that W is less than or requal to L).


    Find two numbers A and B(with A is less than or equal to B) whose difference is 32 and whose product is minimized.


    Find the minimum distance from the parabola
    to the point (0,-3).
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  2. #2
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    Find the minimum distance from the parabola
    to the point (0,-3).
    y=\pm\sqrt{x}

    L=\sqrt{(x-0)^{2}+(y+3)^{2}}

    L=\sqrt{x^{2}+(-\sqrt{x}+3)^{2}}

    There is a trick that is helpful in minimizing or maximizing a distance. It is

    based on the observation that the distance and the square of the distance

    have their max or min at the same point. Therefore, we can do it sans radical


    S=L^{2}=x^{2}+(-\sqrt{x}+3)^{2}

    \frac{dS}{dx}=2x-\frac{3}{\sqrt{x}}+1

    2x-\frac{3}{\sqrt{x}}+1=0

    Now, solve for x.

    If you want to determine the nature of this critical point, use the second derivative test.
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
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    I got 1 for x, but it isn't right. Any ideas?
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  4. #4
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    I figured it out, y=-1 and x=1, so 2.2361 is the answer out of the equation. Any help on the other two?
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  5. #5
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    A box is to be made out of a 10 by 14 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that W is less than or equal to L).
    These problems are all very cliche. If you google them, you will probably find them or, at least, something very similar.

    Let x be the length and width of the square cut out of each corner.

    Then, when we fold them up the volume would be:

    V=x(10-2x)(14-2x)

    When you differentiate, set to 0 and solve for x, you will get two solutions. Only one will be in the domain of the problem.

    #2 is very straighforward. Try that one yourself. OK?.
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