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8 March, 22:56

Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs of f (x) = 18-x^2 and

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  1. 9 March, 00:19
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    F (x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.

    g (x) = 2x^2-9 is a parabola having vertex at (0, - 9) and opening upwards.

    By symmetry, let the x-coordinates of the vertices of rectangle be x and - x = > its width is 2x.

    Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.

    => Area, A

    = 2x (y1 - y2)

    = 2x (18 - x^2 - 2x^2 + 9)

    = 2x (27 - 3x^2)

    = 54x - 6x^3

    For area to be maximum, dA/dx = 0 and d²A/dx² < 0

    => 54 - 18x^2 = 0

    => x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)

    d²A/dx² = - 36x < 0 for x = √3

    => maximum area

    = 54 (√3) - 6 (√3) ^3

    = 54√3 - 18√3

    = 36√3.
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