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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Minimum (2, - \frac{3}{2})

Solution steps

Find the critical points:

(2, - \frac{3}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, - \frac{3}{2}) :

Minimum

Minimum (2, - \frac{3}{2})

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