What is the extreme f(x,y)=x^3+y^3+3x^2-3y^2 ?

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

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extreme f(x,y)=x^3+y^3+3x^2-3y^2

extreme points

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

Minimum (0, 2), Saddle (0, 0), Saddle (- 2, 2), Maximum (- 2, 0)

Solution steps

Find the critical points:

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

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

D (x, y) = (6 x + 6) (6 y - 6)

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 2) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Minimum (0, 2), Saddle (0, 0), Saddle (- 2, 2), Maximum (- 2, 0)

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