What is the extreme f(x,y)=x^2+y^2+8x-4y+7 ?

The extreme f(x,y)=x^2+y^2+8x-4y+7 is Minimum(-4,2)

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+8x-4y+7

extreme points

f (x, y) = x^{2} + y^{2} + 8 x - 4 y + 7

Minimum (- 4, 2)

Solution steps

Find the critical points:

(- 4, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 4, 2) :

Minimum

Minimum (- 4, 2)

f (x) = \frac{y - 1}{y ^{2} - y + 1}

F (x, y) = 15 x^{2} - 22 x y + 24 x + 8 y^{2} - 16 y

x^{6}

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

f (x) = x^{2} - 8 x + 40

What is the extreme f(x,y)=x^2+y^2+8x-4y+7 ?
The extreme f(x,y)=x^2+y^2+8x-4y+7 is Minimum(-4,2)