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

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

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

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