The curve in this problem is symmetric with respect to the y axis.

The rectangle will be composed of 2 rectangles so it suffices to maximize the area of the

rectangle in the first quadrant.

A = x[3/(4+x^{2})]

The numerator of dA/dx when set = 0= 4-x^{2}.

Therefore, maximum area occurs when x=2 and the vertices required are (2,3/8) and (-2,3/8)

and (-2,0) and (2,0).