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

The extreme F(x,y)=3x^2y+y^3-3x^2-3y^2+2 is Minimum(0,2),Maximum(0,0),Saddle(1,1),Saddle(-1,1)

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = (6 y - 6)^{2} - 36 x^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, 0) :

Maximum

Check critical point

(1, 1) :

Saddle

Check critical point

(- 1, 1) :

Saddle

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

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