Calculus - maximum problem

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An isosceles trapezoid is inscribed under the graph oof y =cos x. what are the dimensions which will give the greatest area?

  • Calculus - maximum problem -

    I will assume you want the base to be on the x-axis.
    Let's make the base as long as possible, that is, from (-pi/2,0) to (pi/2,0)
    Let the other point of contact be (x,y)

    Area of trap = (1/2)(pi + 2x)(y)
    = (1/2)(pi+2x)cosx = (1/2)picosx + xcosx
    d(Area)/dx = (-1/2)pi(sinx) + x(-sinx) + cosx
    = 0 for a max/min of Area
    -pi(sinx) - 2xsinx + 2cosx = 0
    2cosx = sinx(pi + 2x)
    2/(pi + 2x) = tanx

    I ran this through a primitive Newton's Method program and got
    x = .45797

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