Calculus

A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y= 12 - x^2. What are the dimensions of such a rectangle with the greatest possible area?

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  1. The vertices of the rectangle are points:
    (-x,0),(x,0),(x,12-x^2),(-x,12-x^2).
    The area A(x)=2x(12-x^2)

    A(x)=24x-2x^3
    A'(x)=24-6x^2=0 => x=+-2
    max A(x)=A(2)=48-16=32

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