What is the extreme f(x,y)=x^2+y^2-2x-6y+14 ?

The extreme f(x,y)=x^2+y^2-2x-6y+14 is Minimum(1,3)

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^2-2x-6y+14

extreme points

f (x, y) = x^{2} + y^{2} - 2 x - 6 y + 14

Minimum (1, 3)

Solution steps

Find the critical points:

(1, 3)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Minimum (1, 3)

f (x, y) = x^{3} - y^{3} - 2 x y + 6

f (x) = 6 x^{2} - 3 x^{3}

f (t) = 2 x

f (x, y) = 12 x y^{2} - 16 x + 25 x y^{3} - 18 x^{2} y^{3}

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x y + 4

What is the extreme f(x,y)=x^2+y^2-2x-6y+14 ?
The extreme f(x,y)=x^2+y^2-2x-6y+14 is Minimum(1,3)