What is the extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2 ?

The extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2 is Minimum(-2,-3/2)

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^2-6xy+5y^2-x+3y+2

extreme points

f (x, y) = 2 x^{2} - 6 x y + 5 y^{2} - x + 3 y + 2

Minimum (- 2, - \frac{3}{2})

Solution steps

Find the critical points:

(- 2, - \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 2, - \frac{3}{2}) :

Minimum

Minimum (- 2, - \frac{3}{2})

f (x) = 2 x^{3} - x^{2} - 4 x + 2, - 1 \leq x \leq 0

(x - 1)^{3}

f (x) = - x^{2} + 12, - 3 \leq x \leq 4

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

2 x^{3} - 24 x + 1

What is the extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2 ?
The extreme f(x,y)=2x^2-6xy+5y^2-x+3y+2 is Minimum(-2,-3/2)