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The cross-section of a drain is a trapezoid. The sides and the bottom of the trapezoid each have a length of 10 feet. Determine the angle θ such that the drain will have maximal cross-sectional area.

  • calculus -

    For lack of specificity, I will let θ be the angle between the upper base and a side. I expect it will end up 90°, making a square cross-section, but let's wee what happens.

    The height of the trapezoid will be 10sinθ, so the area is

    a = 10sinθ (10+10+2*10cosθ)
    = 10sinθ (20+20cosθ)
    = 200sinθ (1+cosθ)

    da/dθ = 200cosθ (1+cosθ) + 200sinθ(-sinθ)
    = 200(cos^2θ + cosθ - sin^2θ)
    = 200(2cos^2θ+cosθ-1)

    da/dθ = 0 when cosθ = 1/2 or -1.
    Looks like the max area occurs at θ=π/3

    Looks like my gut feeling was wrong. Good thing, as it would have been a boring trapezoid.

  • calculus - correction -

    Oops. I forgot to divide by 2 when getting the average of the bases, but that does not affect the answer in this case.

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