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?

Answer 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