Rectangle incribed in an ellipse

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Homework Help Overview

The problem involves finding the area of the largest rectangle that can be inscribed in the ellipse defined by the equation 9x² + 4y² = 36. The discussion centers around the geometric properties of the ellipse and the rectangle, as well as the mathematical methods to approach the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss assuming the rectangle has sides parallel to the axes and explore the relationship between the rectangle's vertices and the ellipse. There are mentions of differentiating the area function and using the Lagrange multiplier method as potential approaches. Some participants express uncertainty about the methods and the reasoning behind the area calculation.

Discussion Status

The discussion is active, with participants offering various approaches and questioning the assumptions made in the problem. There is a recognition of the need for clarity on the methods being used, and some guidance has been provided regarding the differentiation of the area function.

Contextual Notes

Some participants express a lack of familiarity with certain mathematical methods, such as the Lagrange multiplier method, and there are indications of confusion regarding the area formula for the rectangle. The original poster also notes challenges due to the nature of the homework assignment.

brizer
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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
 
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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.
 
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.
 
Why is the area xy? Shouldn't it be A = 2x(2y) instead?
 
brizer said:
(x2/4) + (y2/9) = 1
(x/2) +(y/9) = 1
3x + 2y = 6
2y = 6-3x
How does this follow from the previous step?
 
Defennder said:
Why is the area xy? Shouldn't it be A = 2x(2y) instead?

Yes, you are right. Sorry about that.
 

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