What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?

The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-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+3y^2-2xy+10x-2y+4

extreme points

f (x, y) = x^{2} + 3 y^{2} - 2 x y + 10 x - 2 y + 4

Minimum (- 7, - 2)

Solution steps

Find the critical points:

(- 7, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- 7, - 2) :

Minimum

Minimum (- 7, - 2)

- x \cdot e^{- x}

\frac{7 x}{x ^{2} - 49}

f (x, y) = 5 x_{6 y}

10 + x^{3} - \frac{9 x ^{2}}{2}

\frac{1}{x ^{2} - 5 x - 24}

What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?
The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-2)