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Math Help - Maximum

  1. #1
    Junior Member ihmth's Avatar
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    Maximum

    Need help, I don't know what this trapezoid gutter really look like

    1.) A trapezoid gutter is to be made from a strip of metal 22 in. wide, by bending up the edges. If the bases is 14 in. wide, what width across the top gives the greatest carrying capacity? (use trig function)
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  2. #2
    Super Member

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    Hello, ihmth!

    1) A trapezoid gutter is to be made from a strip of metal 22 in. wide, by bending up the edges.
    If the base is 14 in. wide, what width across the top gives the greatest carrying capacity?
    (use trig function)
    Code:
                F               E
        A * - - * - - - - - - - * - - * D
           \    :               :    /
            \   :               :   /
           4 \  :              h:  / 4
              \ :               :θ/
               \:               :/
                *---------------*
                B      14       C

    The isosceles trapezoid is ABCD, where BC = 14, and AB = CD = 4.
    Let \theta = \angle ECD

    Area Formula: . A \;=\;\frac{h}{2}(b_1 + b_2)

    In right triangle DEC\!:\;\;DE = 4\sin\theta,\;EC = 4\cos\theta

    So we have: . h \:=\:4\cos\theta,\;b_1 \:=\: 14,\;b_2 \:=\:14 + 8\sin\theta

    Then: . A \;=\;\frac{4\cos\theta}{2}\bigg[14 + (14 + 8\sin\theta)\bigg] \quad\Rightarrow\quad A \;=\;8\cos\theta(7 + 2\sin\theta)

    . . and that is the function to be maximized . . .

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