If A is the area of the required rectangle between the 2 parabola, then

A/2 = (-x^2 + 1 - x^2)×x

A/2 = -2x^3 + x ---- 1

Now differentiating wrt x and equating it to 0

-6x^2 + 1 = 0

x = +/-(1/(6^0.5)) --- 2

Sub 2 in 1

A/2 = 2/(3×(6^0 5))

**A = 4/(3×(6^0.5))** is the max area possible for a rectangle between the 2 parabolas with dimensions **(2/6^0.5) & (2/3).**