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

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

extreme points

f (x, y) = x^{2} + y^{2} - 4 x y

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 12

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