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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