What is the extreme f(x,y)=x^3-6xy+y^3+4 ?

The extreme f(x,y)=x^3-6xy+y^3+4 is Saddle(0,0),Minimum(2,2)

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extreme f(x,y)=x^3-6xy+y^3+4

extreme

Saddle (0, 0), Minimum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Minimum

Saddle (0, 0), Minimum (2, 2)

What is the extreme f(x,y)=x^3-6xy+y^3+4 ?
The extreme f(x,y)=x^3-6xy+y^3+4 is Saddle(0,0),Minimum(2,2)