What is the extreme f(x,y)=((2x-x^2)(2y-y^2))/(xy) ?

The extreme f(x,y)=((2x-x^2)(2y-y^2))/(xy) is Saddle(2,2)

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extreme f(x,y)=((2x-x^2)(2y-y^2))/(xy)

extreme points

f (x, y) = \frac{( 2 x - x ^{2} ) ( 2 y - y ^{2} )}{x y}

Saddle (2, 2)

Solution steps

Find the critical points:

(2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 1

Check the sign of at each critical point

Check critical point

(2, 2) :

Saddle

Saddle (2, 2)

f (x, y) = x^{3} - y^{2} - 12 x + 6 y

f (x) = x^{4} - 8 x^{3} + 4

y = x^{5} - x^{3}

x^{3} + 1

x 2^{- x}

What is the extreme f(x,y)=((2x-x^2)(2y-y^2))/(xy) ?
The extreme f(x,y)=((2x-x^2)(2y-y^2))/(xy) is Saddle(2,2)