What is the extreme f(x,y)=11x^2-2xy+2y^2+3 ?

The extreme f(x,y)=11x^2-2xy+2y^2+3 is Minimum(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)=11x^2-2xy+2y^2+3

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 84

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

What is the extreme f(x,y)=11x^2-2xy+2y^2+3 ?
The extreme f(x,y)=11x^2-2xy+2y^2+3 is Minimum(0,0)