differentiation abt maximize

12 Let the coordinate of upper-right corner is (x,y)

Then the area of the rectangle is A = 4xy

As y = (1/4) √(144 - 9x^2)

A = x√(144 - 9x^2)

Let dA/dx = 0

√(144 - 9x^2) - 9x^2/√(144 - 9x^2) = 0

144 - 9x^2 = 9x^2

x = √8

The maximum calue of the rectangle is √8 * √72 = 24

Let the vertex of the rectangle (in first quadrant) be (4 sin θ, 3 cos θ), therefore, the area of the rectangle A(θ) is
2(4 sin θ)*2(3 cos θ)
= 48 sin θ cos θ
= 24 sin 2θ
A'(θ) = 48 cos 2θ
A''(θ) = - 96 sin 2θ
when A'(θ) = 0, θ = π/4, A''(π/4) = -96. So θ = π/4 is the max.
ie. Maximum area of the rectangle is 24 sin (π/2) = 24 sq. units.