Find the area of the largest rectangle with lower base on the x axis and upper vertices on the parabola y=12-x^2

 Draw a picture of the downward-facing parabola, and of a rectangle of the type described. Let (x,y) be the upper right-hand corner of the rectangle. Then by symmetry, the base of the rectangle has length 2x, and the height is y, that is, 12−x2. So the area A(x) of the rectangle is given by A(x)=2x(12−x2).

The answer is all about the number 4