Rectangle's base inscribed on x-axis, upper corners on the parabola y=4-x^2. Find width height of maximum area

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=4-x^2. What are the dimensions (length and width) of such a rectangle with the greatest possible area?

Answer 1

Let
2x = width rectangle
4 - x² = height rectangle
A = area rectangle

A = 2x(4 - x²) = 8x - 2x³

Take the derivative and set it equal to zero to find the critical points.

dA/dx = 8 - 6x² = 0
8 = 6x²
x² = 8/6 = 4/3
x = 2/√3

Width = 2x = 2(2/√3) = 4/√3
Height = 4 - x² = 4 - 4/3 = 8/3

Area = (4/√3)(8/3) = 32/(3√3) = 32√3 / 9

Answer 2

A = 2x(4 - x)^2 = 8x - 2x^2
dA/dx = 8 - 4x^2 = 0 for max area
x = √2
l = 2√2
w = 2

Answer 3

width is 2x, height is 4-x², so area is
A(x) = 2x(4-x²) = 8x - 2x^3
max area when dA/dx = 0, so
d[8x - 2x^3]/dx = 8 - 6x² = 0
6x² = 8
x² = 4/3
x = 2/√3
width = 4/√3
height = 4 - 4/3 = 8/3