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Thread: 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 $\displaystyle x^2/a^2$ $\displaystyle +$ $\displaystyle y^2/b^2$$\displaystyle = 1$
    $\displaystyle a $and $\displaystyle b $> $\displaystyle 1$.

    Calculate the dimensions of the rectangle with maximum area.

    Thanks
    Last edited by osodud; Jun 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 $\displaystyle x^2/a^2$ $\displaystyle +$ $\displaystyle y^2/b^2$$\displaystyle = 1$
    $\displaystyle a $and $\displaystyle b $> $\displaystyle 1$.

    Calculate the dimensions of the rectangle with maximum area.

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

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

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

    $\displaystyle
    Area=4ab\sin\theta.\cos\theta=2ab\sin 2\theta\leq 2ab$.......(since $\displaystyle \sin2\theta\leq 1)$

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

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

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

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