Calculus

Find the area of the largest rectangle with sides parallel to the coordinate axes which can be inscribed in the area bounded by the two parabolas y=26-x^2 and y=x^2+2

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  1. Make a sketch and draw in the rectangle.
    Let the top-right point be P(x,26-x^2)
    and the bottom-right point be Q(x, x^2 + 2)

    So the width of the rectangle is 2x
    and the height is 26-x^2 - x^2 - 2
    = 24 - 2x^2

    area = 2x(24-2x^2)
    = 48x - 4x^3
    d(area)/dx = 48 - 12x^2
    = 0 for a max of area
    12x^2 = 48
    x^2 = 4
    x = 2

    the largest area = 48(2) - 4(2^3)
    = 96 - 32
    = 64 units^2

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  2. thank you so much!

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