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find the perimeter of the rectangle with maximum area that can be inscribed in a semicircle of radius 2 ft. if two of its vertices are on the diameter of the semicircle and the other two are on the semicircle.

  • calculas - ,

    Did you make a sketch?
    draw a line from the centre to a vertex of the rectangle lying on the semicircle.
    Let the length of the rectangle be 2x and its height y
    I see a right-angled triangle with sides x and y and hypotenuse 2
    x^2 + y^2 = 4

    let the area be A
    A = 2xy
    A^2 = 4x^2y^2 = 4x^2(4-x^2)
    = 16x^2 - 4x^4
    2A dA/dx = 32x - 16x^3 = 0 for max of A
    16x(2 - x^2) = 0
    x^2=2
    x=√2
    sub back into equation above
    y = √2

    so length = 2√2 and width = √2
    Perimeter = 6√2

  • calculas - ,

    thank you so muchh :)

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