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Math Help - Optimization

  1. #1
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    Optimization

    A piece of wire long L is cut in two pieces. One piece is bent into a semicircle and the other is bent into a equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximun and (b) a minimun?

    I'm stuck in this exercise. I dont know what to do
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  2. #2
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    Quote Originally Posted by Jabés
    A piece of wire long L is cut in two pieces. One piece is bent into a semicircle and the other is bent into a equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximun and (b) a minimun?

    I'm stuck in this exercise. I dont know what to do
    Here is one way.

    Cut the L long wire into x and (L-x) pieces, where x is in terms of L.

    Bend the x into a semi-cirle. The (L-x) into an equilateral triangle.
    (Let us say you can do that.)

    Area enclosed by x:
    Circumference = x = [2pi(r)]/2 +2r = r(pi +2)
    r = x/(2+pi) = (0.1945)x
    Area1 = [pi(r^2)]/2 = (1/2)*pi*[(0.1945)x]^2 = (0.0594)(x^2) ----***

    Area enclosed by (L-x):
    Each side of the equilateral triangle is (L-x)/3
    Area = (1/2)(a)(b)sinC
    Area2 = (1/2)[(L-x)/3][(L-x)/3]sin(60deg)
    Area2 = (sqrt(3) /36)[(L-x)^2] = (0.0481)*(L-x)^2 -------------***

    Total area enclosed, A, is:
    A = (0.0594)(x^2) +(0.0481)[(L-x)^2]

    For max or min A, set dA/dx to zero.
    dA/dx = (0.1188)x +(0.0962)(L-x)(-1)
    dA/dx = (0.1188)x +(0.0962)x -(0.0962)L

    (I accidentally pressed the send button)

    To continue...
    dA/dx = (0.215)x -(0.0962)L
    Set that to zero,
    x = (0.0962)L /(0.215) = (0.4474)L

    Therefore, for maximum enclosed areas, cut the L long wire into (0.4474)L and (0.5526)L pieces. -------------answer.

    For minimum enclosed areas?
    I guess cut the L long wire into the same pieces as above, but make semi-circle out of the (0.5526)L piece and equilateral triangle out of the (0.4474)L piece.
    Last edited by ticbol; June 19th 2006 at 11:55 AM.
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  3. #3
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    Thanks for your help!
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