Rectangle inscribed in an ellipse.

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To find the area of the largest rectangle inscribed in the ellipse defined by the equation x^2/a^2 + y^2/b^2 = 1, the rectangle's corners can be positioned at (0,0), (p,0), (p,q), and (0,q). The area A of the rectangle is given by A = pq, which needs to be maximized. To do this, solve for y in terms of x using the ellipse equation, substitute it into the area equation, and then take the derivative to find critical points. It's important to multiply the area by 4 to account for all quadrants since the initial coordinates only represent the first quadrant. This method will yield the maximum area of the rectangle inscribed in the ellipse.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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