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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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)