max question

Let A be the area of rectangle with (x,y) as a vertex.

A = 2xy = 2x(4 - x^2) = 8x - 2x^3
A' = 8 - 6x^2
If A' = 0, the x = √(4/3) = 2/3 √3
A'' = -12x
A attains local max. at x = 2/3 √3

The area of the largest rectangle is 2(2/3 √3)(4 - 4/3) = 32/9 √3

Consider the area of the rectangle is xy=x(4-x^2)=-x^3+4x
Take derivative -3x^2+4
Set -3x^2+4 = 0 = > x^2=4/3
And xy=-x^3+4x=-1.5396+4.6188=3.0792
The largest rectangle that can be inscribed in the region bounded by the parabola with equation y=4-X^2 and the x-axis is 3.0792