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Math Help - Maxima/minima: Still confused

  1. #1
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    Unhappy Maxima/minima: Still confused

    Suppose that a weight is to be held 10ft below a horizontal line AB by a wire in the shape of a Y. If the points A & B are 8ft apart, a.) Estimate on your graphics calculator to the nearest foot the shortest length of wire that can be used. b.) Confirm your estimates in part a analytically... ( Ill graph it)
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  2. #2
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    Hello, ^_^Engineer_Adam^_^!

    Suppose that a weight is to be held 10ft below a horizontal line AB by a wire in the shape of a Y.
    If the points A & B are 8ft apart, find the shortest length of wire that can be used.
    Code:
            A    4    C    4    B
          - * - - - - + - - - - *
          :   *       |     θ *
          :     *     |     *
          :       *   |   *
          :         * | *
         10           *D
          :           *
          :           *
          :           *
          -           *W

    The weight is at W. .Let /CBD = θ

    The length of the wire is: .L .= .AD + BD + DW

    In right triangle BCD, we have: .BD = 4Ěsecθ, CD = 4Ětanθ
    . . So: .DW .= .10 - 4Ětanθ

    And we have: .L .= .2(4Ěsecθ) + (10 - 4Ětanθ) .= .8Ěsecθ - 4Ětanθ + 10

    Now you can minimize L . . .

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  3. #3
    Grand Panjandrum
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    See attachment for notation

    x/4 = tan(theta)

    4/l1 = cos(theta)

    Total length of the wire is:

    l = 2*l1+(10-x) = (8-4 sin(theta))/cos(theta) +10.

    So if l is a minimum we have dl/dtheta=0, and:

    dl/dtheta = [8 sin(theta)-4 sin^2(theta)]/cos^2(theta) -4,

    so for a mininmum we have:

    [8 sin(theta)-4 sin^2(theta)]/cos^2(theta) = 4

    or:

    8 sin(theta)= 4 (cos^2(theta)+sin^2(theta)) = 4.

    Therefore:

    Sin(theta) = Ż, of theat=30 degrees.

    Putting this back into the expression for l gives:

    l = 10 + 4 sqrt(3).

    RonL
    Attached Thumbnails Attached Thumbnails Maxima/minima: Still confused-gash.jpg  
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