What is the extreme f(x,y)=4x^2+y^2-8x+6y+8 ?

The extreme f(x,y)=4x^2+y^2-8x+6y+8 is Minimum(1,-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=4x^2+y^2-8x+6y+8

extreme points

f (x, y) = 4 x^{2} + y^{2} - 8 x + 6 y + 8

Minimum (1, - 3)

Solution steps

Find the critical points:

(1, - 3)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(1, - 3) :

Minimum

Minimum (1, - 3)

What is the extreme f(x,y)=4x^2+y^2-8x+6y+8 ?
The extreme f(x,y)=4x^2+y^2-8x+6y+8 is Minimum(1,-3)