# What is the area of the largest rectangle that can be inscribed in the ellipse: 9(x^2) + 4(y^2) = 36?

Jun 16, 2016

$A = 12$

#### Explanation:

$9 \left({x}^{2}\right) + 4 \left({y}^{2}\right) = 36 \equiv {x}^{2} / 4 + {y}^{2} / 9 = 1$

The problem can be posed as:

Find Max $x y$ or equivalently Max ${x}^{2} {y}^{2}$ such that

${x}^{2} / 4 + {y}^{2} / 9 = 1$

Making now $X = {x}^{2} , Y = {y}^{2}$ the problem is equivalent to

Find $\max \left(X \cdot Y\right)$ subject to $\frac{X}{4} + \frac{Y}{9} = 1$

The lagrangian for determination of stationary points is

$L \left(X , Y , \lambda\right) = X \cdot Y + \lambda \left(\frac{X}{4} + \frac{Y}{9} - 1\right)$

The stationarity conditions are

$\nabla L \left(X , Y , \lambda\right) = \vec{0}$

or

 { (lambda/2 + Y = 0),( lambda/9 + X = 0),( X/2 + Y/9 - 1= 0) :}

Solving for $X , Y , \lambda$ gives

$\left\{{X}_{0} = 2 , {Y}_{0} = \frac{9}{2} , {\lambda}_{0} = - 18\right\}$

so $\left\{{x}_{0} = \sqrt{2} , {y}_{0} = \frac{3}{\sqrt{2}}\right\}$

$A = 4 {x}_{0} {y}_{0} = 4 \times 3 = 12$ 