What is the extreme f(x,y)=2x^2+y^2+8x-6y-2xy+12 ?

The extreme f(x,y)=2x^2+y^2+8x-6y-2xy+12 is Minimum(-1,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=2x^2+y^2+8x-6y-2xy+12

extreme points

f (x, y) = 2 x^{2} + y^{2} + 8 x - 6 y - 2 x y + 12

Minimum (- 1, 2)

Solution steps

Find the critical points:

(- 1, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Minimum

Minimum (- 1, 2)

f (x) = x^{3} - 3 x^{2} + 24 x

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

f (x) = \frac{x ^{2}}{x + 4}

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

\frac{ln ( x ^{2} )}{x}

What is the extreme f(x,y)=2x^2+y^2+8x-6y-2xy+12 ?
The extreme f(x,y)=2x^2+y^2+8x-6y-2xy+12 is Minimum(-1,2)