Rectangle inscribed in generic ellipse

[v0id19]

Homework Statement

Largest possible area of a rectangle inscribed in the ellipse (x²/a²)+(y²/b²)=1

Homework Equations

Area of the rectangle = length*height

The Attempt at a Solution

I have it set up so that the four corners of the rectangle are at (x,y) (-x,y) (-x,-y) (x,-y) and that area therefore is A=(2x)(2y).
In order to find the max area, I know I need to differentiate the equation, so I need to eliminate either x or y.
Using the ellipse equation, I found x to be equal to sqrt(a²-(y²a²)/b²).
Substituting that into the area equation I get:
A=2(sqrt(a²-(y²a²)/b²))*(2y).

And differentiating that has been a nightmare and I haven't gotten it right yet. I know a and b are constants, and become one in the derivative.

But is there any easier way of solving this problem?

 

[HallsofIvy]

Looks to me like "Laplace multipliers" would make this much simpler. You have an ellipse of the form b²x²+ a²y<sup>2= a2b2. I notice you are assuming that the maximal area rectangle has its sides parallel to the axes of the ellipse. I'm not sure that follows easily but I can't see any more general way of doing this. You want to maximize A(xy)= xy with the constraint F(x,y)= b2x2+ a2y2= a2.
[itex]\nabla A= y\vec{i}+ x\vec{j}[/itex] and [itex]\nabla F= 2b^2x\vec{i}+ 2a^2y\vec{j}[/itex].

One must be a multiple of the other.</sup>

 

[v0id19]

I'm just in calculus BC and am not familiar with 'laplace multipliers' how does that work?