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

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Answer 1 · 2016-06-16T19:02:25

$A = 12$

$9(x^2) + 4(y^2) = 36 equiv x^2/4+y^2/9=1$

The problem can be posed as:

Find Max $xy$ or equivalently Max $x^2y^2$ such that

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

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

Find $max(X*Y)$ subject to $X/4+Y/9=1$

The lagrangian for determination of stationary points is

$L(X,Y,lambda) = X*Y+lambda(X/4+Y/9-1)$

The stationarity conditions are

$grad L(X,Y,lambda) = vec 0$

or

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

Solving for $X,Y,lambda$ gives

${X_0 = 2, Y_0 = 9/2, lambda_0 = -18}$

so ${x_0 = sqrt(2), y_0 = 3/sqrt(2)}$

$A = 4 x_0 y_0 = 4 xx3 = 12$

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What is the area of the largest rectangle that can be inscribed in the ellipse: 9(x^2) + 4(y^2) = 369(x^2) + 4(y^2) = 36?