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

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

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

extreme

Minimum (- \frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

Solution steps

Find the critical points:

(- \frac{2}{3}, \frac{1}{3}), (0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 x

D (x, y) = 8 x y - 4 x^{2} - 8 x - 16 y^{2} + 16 y - 4

Check the sign of at each critical point

Check critical point

(- \frac{2}{3}, \frac{1}{3}) :

Minimum

Check critical point

(0, 1) :

Saddle

Minimum (- \frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

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