calculus

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Consider a window the shape of which is a rectangle of height h surmounted by a triangle having a height T that is 1.1 times the width w of the rectangle

If the cross-sectional area is A, determine the dimensions of the window which minimize the perimeter.

h =
w =

  • calculus -

    p = w+2h+2s
    where (w/2)^2 + (1.1w)^2 = s^2
    s = 1.46w

    A = wh + 1/2 w * 1.1w
    = wh + 1.6w
    so, h = (A-1.6w)/w = A/w = 1.6

    p = w+2(A/w+1.6) + 2(1.46w)
    = w + 2A/w + 3.2 + 2.92w
    = 3.92w + 3.2 + A/w

    dp/dw = 3.92 - A/w^2
    dp/dw = 0 when 3.92w^2 = A, or
    w = A/1.98
    h = A/1.80

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