Arectangle is inscribed with its base on the x -axis and its upper corners on the parabola
y=2-x^2
what are the dimensions of such a rectangle with the greatest possible area?
width=
height=

The area of a rectangle is given by A = length * width.
width of the given rectangle is x
height is y = 2 - x^2
Area = x(2 - x^2) = 2x - x^3
For area to be maximum, dA/dx = 0
dA/dx = 2 - 3x^2 = 0
3x^2 = 2
x^2 = 2/3
x = √(2/3)
y = 2 - (√(2/3))² = 2 - 2/3 = 4/3

Therefore, the required rectangle has a width of 2√(2/3) ≈ 1.63 and a height of 4/3 ≈ 1.33