Rectangle incribed in an ellipse

[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

 

[HallsofIvy]

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.

 

[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.

 

[Defennder]

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

 

[Defennder]

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

How does this follow from the previous step?

 

[HallsofIvy]

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

Yes, you are right. Sorry about that.