We start with the area of a rectangle A=BH=(2x)(y)=2x(1-x^2)=2x-2x^3
We must find the critical point of this function with x>0.
A'=2-6x^2=0
1-3x^2=0: 1/3=x^2: sqrt(3)/3=x
So plugging in this critical value we obtain:
A=2qrt(3)/3-2sqrt(3)/9=4sqrt(3)/9