A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=12−x2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

let the points of contact be (x,y) and (-x,y)
let the area of the rectangle be A

A = 2xy
= 2x(12-x^2) = 24x - 2x^3
dA/dx = 24 -6x^2
= 0 for a max of A
6x^2 = 24
x^2 = 4
x = ± 2
y = 12 - 4 = 8

the rectangle will be 4 units for the base and the height will be 8
(touching at the points (2,8) and (-2,8)