What is the extreme P(xy)=x^3y+x^2y-2xy ?

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

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extreme P(xy)=x^3y+x^2y-2xy

extreme

Saddle (0, 0), Saddle (1, 0), Saddle (- 2, 0)

Solution steps

Find the critical points:

(0, 0), (1, 0), (- 2, 0)

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

D (x, y) = - (3 x^{2} + 2 x - 2)^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 0) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Saddle (0, 0), Saddle (1, 0), Saddle (- 2, 0)

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