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Math Help - Rectangle and Ellipse

  1. #1
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    Question Rectangle and Ellipse

    We have a Rectangle with sides parallel to axes (x and y). The rectangle is inside of the Ellipse x^2/a^2 + y^2/b^2  = 1
    a and b > 1.

    Calculate the dimensions of the rectangle with maximum area.

    Thanks
    Last edited by osodud; June 16th 2009 at 11:34 AM.
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  2. #2
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    Quote Originally Posted by osodud View Post
    We have a Rectangle with sides parallel to axes (x and y). The rectangle is inside of the Ellipse x^2/a^2 + y^2/b^2  = 1
    a and b > 1.

    Calculate the dimensions of the rectangle with maximum area.

    Thanks
    maximize the area of that section of the rectangle in quadrant I ...

    A = x \cdot b\sqrt{1 - \frac{x^2}{a^2}}<br />
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  3. #3
    Senior Member pankaj's Avatar
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    Let the vertices of the rectangle be (a\cos\theta,b\sin\theta),(-a\cos\theta,b\sin\theta),(-a\cos\theta,-b\sin\theta) and (a\cos\theta,-b\sin\theta).

    Length of the rectangle= 2a\cos\theta
    Breadth of the rectangle= 2b\sin\theta

     <br />
Area=4ab\sin\theta.\cos\theta=2ab\sin 2\theta\leq 2ab.......(since \sin2\theta\leq 1)

    Area of the largest rectangle = 2ab , for which \sin2\theta=1

    Thus 2\theta=\frac{\pi}{2} and therefore \theta=\frac{\pi}{4}

    Length= 2a\cos\frac{\pi}{4}=a\sqrt{2}

    Breadth= 2b\cos\frac{\pi}{4}=b\sqrt{2}
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