What is the extreme f(x,y)=2x^2+4xy-x^2y-4x ?

The extreme f(x,y)=2x^2+4xy-x^2y-4x is Saddle(0,1),Saddle(4,3)

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extreme f(x,y)=2x^2+4xy-x^2y-4x

extreme

Saddle (0, 1), Saddle (4, 3)

Solution steps

Find the critical points:

(0, 1), (4, 3)

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

D (x, y) = - 4 x^{2} + 16 x - 16

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(4, 3) :

Saddle

Saddle (0, 1), Saddle (4, 3)

What is the extreme f(x,y)=2x^2+4xy-x^2y-4x ?
The extreme f(x,y)=2x^2+4xy-x^2y-4x is Saddle(0,1),Saddle(4,3)