What is the extreme f(x,y)=x^2+xy+y^2+4x-7y+6 ?

The extreme f(x,y)=x^2+xy+y^2+4x-7y+6 is Minimum(-5,6)

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+4x-7y+6

extreme points

f (x, y) = x^{2} + x y + y^{2} + 4 x - 7 y + 6

Minimum (- 5, 6)

Solution steps

Find the critical points:

(- 5, 6)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 5, 6) :

Minimum

Minimum (- 5, 6)

f (x) = - 3 x^{2} - 5, - 4 \leq x \leq 3

f (x, y) = x^{2} + y^{2} + 8 x - 10 y

f (x) = ln (2 x^{2} - 2 x + 3), 0 \leq x \leq 2

f (x) = 5 - 6 x^{2}, - 4 \leq x \leq 2

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

What is the extreme f(x,y)=x^2+xy+y^2+4x-7y+6 ?
The extreme f(x,y)=x^2+xy+y^2+4x-7y+6 is Minimum(-5,6)