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

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

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+xy+y^2+3x-3y+4

extreme points

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

Minimum (- 3, 3)

Solution steps

Find the critical points:

(- 3, 3)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 3, 3) :

Minimum

Minimum (- 3, 3)

f (x, y) = 6 - 2 x + 4 y - x^{2} - 4 y^{2}

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

f (x) = 5 x^{2} - 10 x + 2, - 4 \leq x \leq 2

f (x) = x + sin (x), - \frac{π}{2} \leq x \leq \frac{π}{2}

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

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