Find the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4=1.

Question

Thanks.

Answer 1

If the base of the rectangle has length 2x, and height y, the area is

a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)

da/dx = 4(8-x^2)/√(16-x^2)

so, max area occurs at x=√8, and the rectangle is 2√8 by √2