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Rectangle incribed in an ellipse

  1. Jul 28, 2008 #1
    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
  2. jcsd
  3. Jul 28, 2008 #2


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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.
  4. Jul 28, 2008 #3
    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


    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.
  5. Jul 28, 2008 #4


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    Why is the area xy? Shouldn't it be A = 2x(2y) instead?
  6. Jul 28, 2008 #5


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    How does this follow from the previous step?
  7. Jul 29, 2008 #6


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    Yes, you are right. Sorry about that.
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