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Math Help - Optimization

  1. #1
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    Optimization

    1) A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 40 ft, find the value of x so that the greatest possible amount of light is admitted. (Give your answer correct to two decimal places.)

    A_r = x(40-x)
    A_s = \frac{\pi (\frac{x}{2})^2}{2} =  \frac{\pi x^2}{8}
    A_T = A_r + A_s = x(40-x) + \frac{\pi x^2}{8}

    Derive and set to 0:

    40-2x + \frac{\pi x}{4} = 0

    x = 32.93

    ?

    2) A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)

    A_t = x^2 * \frac{\sqrt{3}}{4}
    A_r = (10-x)^2

    A_{TOT} = A_r + A_r = (10-x)^2 + x^2 * \frac{\sqrt{3}}{4}

    Derive and set to 0:

    x\sqrt{3} - 20 + 2x => x = 5.36

    ?
    Last edited by Open that Hampster!; October 27th 2009 at 06:40 PM.
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  2. #2
    Senior Member Pinkk's Avatar
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    If we let the diameter of the semicircle (which is the bottom, horizontal edge of the rectangle) be x, then the perimeter of the semicircle is \frac{\pi x}{2}. Then if we let each vertical edge of the rectangle be y, then 2y + \frac{\pi x}{2} + x = 40, so y = 20 - \frac{x}{2} - \frac{\pi x}{4}. The area of the rectangle, A_{r} = xy = 20x - \frac{x^{2}}{2} -\frac{\pi x^{2}}{4}. The area of the semicircle, A_{sc}=\frac{\pi x^{2}}{8}. So area total is:

    A = A_{r} + A_{sc} = 20x - \frac{x^{2}}{2} -\frac{\pi x^{2}}{4} + \frac{\pi x^{2}}{8}.
    Set A'=0 to find the x value that gives maximum area.
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