A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola f(x)=8-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

Let the base of the rectangle go from -x to x. The area of the rectangle is thus

a = 2xy = 2x(8-x^2)
da/dx = 16 - 6x^2
da/dx = 0 when x = 4/√6

So, the rectangle is 8/√6 by 16/3