# Homework Help: Rectangle inscribed in generic ellipse

1. Oct 25, 2008

### v0id19

1. The problem statement, all variables and given/known data
Largest possible area of a rectangle inscribed in the ellipse (x2/a2)+(y2/b2)=1

2. Relevant equations
Area of the rectangle = length*height

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 25, 2008

### HallsofIvy

Looks to me like "Laplace multipliers" would make this much simpler. You have an ellipse of the form b2x2+ a2y2= 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.
$\nabla A= y\vec{i}+ x\vec{j}$ and $\nabla F= 2b^2x\vec{i}+ 2a^2y\vec{j}$.

One must be a multiple of the other.

3. Oct 25, 2008

### v0id19

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