Rectangle and Ellipse

**osodud** · June 16th 2009, 11:02 AM

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

**skeeter** · June 16th 2009, 11:08 AM

Originally Posted by osodud

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 ...

$A = x \cdot b\sqrt{1 - \frac{x^2}{a^2}}<br />$

**pankaj** · June 16th 2009, 11:20 AM

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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