# Rectangle and Ellipse

• Jun 16th 2009, 11:02 AM
osodud
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$$\displaystyle = 1 \displaystyle a and \displaystyle b > \displaystyle 1. Calculate the dimensions of the rectangle with maximum area. Thanks • Jun 16th 2009, 11:08 AM skeeter Quote: Originally Posted by osodud We have a Rectangle with sides parallel to axes (x and y). The rectangle is inside of the Ellipse \displaystyle x^2/a^2 \displaystyle + \displaystyle y^2/b^2$$\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 ...

$\displaystyle A = x \cdot b\sqrt{1 - \frac{x^2}{a^2}}$
• Jun 16th 2009, 11:20 AM
pankaj
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=$\displaystyle 2b\cos\frac{\pi}{4}=b\sqrt{2}$