What is the extreme f(x,y)=x^2+y^2-6x-2y+12 ?

The extreme f(x,y)=x^2+y^2-6x-2y+12 is Minimum(3,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=x^2+y^2-6x-2y+12

extreme points

f (x, y) = x^{2} + y^{2} - 6 x - 2 y + 12

Minimum (3, 1)

Solution steps

Find the critical points:

(3, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(3, 1) :

Minimum

Minimum (3, 1)

f (x) = x ln (\frac{x}{6})

f (x) = - x^{2} + 3 x^{2} + 9 x + 1

f (x) = \frac{t}{t - 1} (2.3)

f (x) = (x^{2} + 1) (4 x^{2} - 1)

72 y^{2} + x^{2} - x^{2} y

What is the extreme f(x,y)=x^2+y^2-6x-2y+12 ?
The extreme f(x,y)=x^2+y^2-6x-2y+12 is Minimum(3,1)