What is the extreme f(x,y)=x^2-xy+y^2+9x-6y+10 ?

The extreme f(x,y)=x^2-xy+y^2+9x-6y+10 is Minimum(-4,1)

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+9x-6y+10

extreme points

f (x, y) = x^{2} - x y + y^{2} + 9 x - 6 y + 10

Minimum (- 4, 1)

Solution steps

Find the critical points:

(- 4, 1)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(- 4, 1) :

Minimum

Minimum (- 4, 1)

f (x, y) = x^{2} y + y^{3} - 48 y

f (x) = - 5 x^{4} - 5 x^{3} + 29

f (x) = 7 sin (x) + 7 cos (x), 0 \leq x \leq 2 π

x^{3} - 6 x^{2} + 1

f (x) = 2 x^{3} + 12 x^{2} + 18 x

What is the extreme f(x,y)=x^2-xy+y^2+9x-6y+10 ?
The extreme f(x,y)=x^2-xy+y^2+9x-6y+10 is Minimum(-4,1)