find the the area of the largest rectangle that can be under the parabola y=9-x^2 above the x axis?

Answer 1

Let the base of the rectangle have length 2x.
Then the area of the rectangle is
a = 2xy = 2x(9-x^2)
da/dx = 6(3-x^2)
so da/dx = 0 when x = ±√3
we want x>0, so a has a max of 12√3