Lagrange and his conclusions
Given the ellipse 9 ( x − 1 9 ) 2 + 1 6 ( y − 2 0 0 0 ) 2 = 2 5 Find a rectangle inscribed in this elipse which area is maximum.
Submit its area
(Topic: it's a mixture of calculus and geometry)
The answer is 600.
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1 solution
Thank you for your solution,Otto (+1) ↑ . You'll never cease to amaze(surprise) me."I" solved this problem using Lagrange multipliers and it didn't take me very little time... I always learn something with you, really thanks... I have always said that I have only three words for you: You are thundering...
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I was working through this problem with lagrange multipliers until I started running into numbers with 10 or more digits... that's when I started running out of room on my scratch pad and decided to call it quits. It probably wouldn't have been too tedious if the numbers in the expression for the ellipse were not as large.
The solution is incomplete as this area calculation is valid only for rectangles whose sides are parallel to the coordinate axes. What about the general case ?
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It is easy to see geometrically that the square is the largest quadrilateral that can be inscribed into a circle. Now we can the stretch the circle out into an ellipse.
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I don't think your second statement is rigorous enough. Yes, we can stretch the circle to an ellipse and the square gets stretched to a rectangle but what it has to do with the rectangle of largest area ?
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@Abhishek Sinha – If we pick a square in the circle and stretch parallel to two of its sides, we will get the largest quadrilateral in the ellipse, a rectangle with its sides parallel to the coordinate axes as claimed.
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@Otto Bretscher – Why that particular rectangle will have the maximum area ? Why not a tilted one ? Which mathematical theorem are you using ?
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@Abhishek Sinha – I think I have explained enough ;)
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@Otto Bretscher – We agree to disagree then.
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@Abhishek Sinha – Well, what more can I say? I take the largest quadrilateral in the circle, a square, and transform it into the largest quadrilateral in the ellipse, a rectangle with its sides parallel to the axes as claimed.
After almost three years, I took another look at this exchange as somebody posted made a post in this problem. I still don't quite understand your objection. Let me see which of the following points you agree with: (a) The largest quadrilaterals in a unit circle are squares, with area 2, and (b) The largest quadrilaterals in an ellipse with semiaxes a and b have area 2ab; among them is a rectangle.
Parameterize the ellipse as x = 1 9 + 1 5 cos t and y = 2 0 0 0 + 2 0 sin t . The area of an inscribed rectangle will be A = 4 × 1 5 cos t × 2 0 sin t = 6 0 0 × sin ( 2 t ) , with a maximum of 6 0 0 when sin ( 2 t ) = 1 .