Rectangle incribed in an ellipse

1. Jul 28, 2008

brizer

Problem: What is the area of the largest rectangle that can be inscribed in the ellipse 9x2+4y2=36

Relevant Equations:
equation of an ellipse: (x - h)2/a2 + (y - k)2/b2 = 1

I only got as far as
x2/4 + y2/9 = 1
(x - 0)2/22 + (y - 0)2/32 = 1

I have no idea where to go from here

2. Jul 28, 2008

HallsofIvy

Staff Emeritus
I think you can assume that the rectangle has sides parallel to the x and y axes. Okay, if one vertex of the rectangle is at (x,y), the other four vertices are at (x, -y), (-x, y), and (-x,-y). The lengths of the two sides are 2x and 2y and the area is xy.
Knowing that x2/4+ y2/9= 1, you could solve for y as a function of x, write A= xy as a function of x only and differentiate.

Or, if you know the method, it would be simpler to use the "Lagrange multiplier method" to directly minimize xy with the condition x2/4+ y2/9= 1.

The problem with not showing any work is that we don't know what methods you are familiar with.

3. Jul 28, 2008

brizer

Sorry, it's been two years since geometry so I was glad I could even remember the formula for an ellipse. My work was pretty much an ellipse with a triangle in it and my book is no help since my math teacher made up this summer homework on a whim. At any rate, I'm not familiar with the Lagrange multiplier method, so I wrote A=xy as a function of x and differentiated.

(x2/4) + (y2/9) = 1
(x/2) +(y/9) = 1
3x + 2y = 6
2y = 6-3x

A=(2x)(6-3x)
A=(12x-6x2)
A'=12-12x

So, from here do I just plug in variable a for x and solve? That would give me a negative area. Basically, I get that the points where the rectangle and ellipse touch are parallel, but I don't understand how you find them.

4. Jul 28, 2008

Defennder

Why is the area xy? Shouldn't it be A = 2x(2y) instead?

5. Jul 28, 2008

Defennder

How does this follow from the previous step?

6. Jul 29, 2008

HallsofIvy

Staff Emeritus
Yes, you are right. Sorry about that.