find the dimensions of the rectangle of maximum area that can be inscribed inside the ellipse (x/4)^2+(y/3)^2 = 1

Answer 1

The area of an enclosed rectangle will be 4xy, where +/- x and +/- y are corner coordinates.

A will have a maximum when A^2 has a maximum. This will make the math easier.

A^2 = 16 x^2*y^2 = 16 x^2*9[1 - (x/4)^2]

dA^2/dx = 288x - 9*4x^3 = 0
x^2 = 8
x = sqrt8 = 2.828
(y/3)^2 = 1 - 1/2 = 1/2
y/3 = 1/sqrt2
y = 3/sqrt2 = 2.121

Side lengths are 5.657 and 4.242 for maximum area. (2x and 2y)

Check my thinking; my algebra tends to be sloppy.