What is the extreme f(x,y)=x^3-y^3+3xy ?

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

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extreme f(x,y)=x^3-y^3+3xy

extreme

Saddle (0, 0), Minimum (1, - 1)

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 y

D (x, y) = - 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, - 1) :

Minimum

Saddle (0, 0), Minimum (1, - 1)

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