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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=x^2y^3-(x-2y)^2+6x^4y=3xy

extreme

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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