What is the extreme f(x,y)=2x^2+3xy+4y^2-5x+2y ?

The extreme f(x,y)=2x^2+3xy+4y^2-5x+2y is Minimum(2,-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)=2x^2+3xy+4y^2-5x+2y

extreme points

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2 y

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 23

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

f (x) = 2 x^{2} + 8 x

f (x) = x^{\frac{3}{5}} (4 - x)

f (x) = x y + \frac{50}{x} + \frac{20}{y}

f (x, y) = (5 - x) (5 - y) (x + y - 5)

\frac{- 3 x + 66}{( x - 86 ) ^{5}}

What is the extreme f(x,y)=2x^2+3xy+4y^2-5x+2y ?
The extreme f(x,y)=2x^2+3xy+4y^2-5x+2y is Minimum(2,-1)