Rectangle inscribed 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 an ellipse defined by the equation x^2/a^2 + y^2/b^2 = 1, with the rectangle's sides parallel to the axes. The original poster attempts to visualize the problem by positioning the rectangle within the ellipse and identifying the coordinates of its corners.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss methods for maximizing the area of the rectangle, including substituting for y in the area equation and taking derivatives. There are questions about the implications of the ellipse's center being at the origin and whether to multiply the area by 2 or 4 based on quadrant considerations.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to derive the maximum area. Some guidance has been offered regarding the use of derivatives and area calculations, but there is no explicit consensus on the best method yet.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to consider the rectangle's position relative to the ellipse and the implications of working within specific quadrants. There is also uncertainty regarding the correct application of calculus in this context.

tysonk
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Find the area of the largest rectangle that can be inscribed (with sides parallel to the axes in the ellipse).
x^2/a^2 +y^2/b^2 = 1

I came across the above problem and am not sure how to proceed with it. I drew the ellipse with the inscribed rectangle and tried repositioning the ellipse so that the corner of the rectangle is placed at the origin.

Then the corners have coordinates.
(0,0) , (p, 0), (p, q), (0, q)
A = pq so we want the maximum A however I'm not sure where to go from here.
 
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Solve for y, plug that value into your area equation... then take the derivative and set it to zero. Solve for x and then plug that value for x back into the area equation(the one in which you've substituted for the y value) and then simplify.

Hopefully someone can back me up, but I believe that's the best way to do it.
 
When finding the derivative of that then I'll have a dy/dx term? Also would I have to times that area by 2 since the ellipse equation is such that the center of the ellipse is at the origin...
 
tysonk said:
When finding the derivative of that then I'll have a dy/dx term? Also would I have to times that area by 2 since the ellipse equation is such that the center of the ellipse is at the origin...
If you are referring to your "(0,0) , (p, 0), (p, q), (0, q)" then you would multiply by 4 to get the area of the entire rectangle since that is only in the first quadrant.
 

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