What is the extreme P(x,y)=0.3x^2+0.2y^2+0.1xy-14x-10y+2000 ?

The extreme P(x,y)=0.3x^2+0.2y^2+0.1xy-14x-10y+2000 is Minimum(20,20)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme P(x,y)=0.3x^2+0.2y^2+0.1xy-14x-10y+2000

extreme

Minimum (20, 20)

Solution steps

Find the critical points:

(20, 20)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0.4

D (x, y) = 0.23

Check the sign of at each critical point

Check critical point

(20, 20) :

Minimum

Minimum (20, 20)

What is the extreme P(x,y)=0.3x^2+0.2y^2+0.1xy-14x-10y+2000 ?
The extreme P(x,y)=0.3x^2+0.2y^2+0.1xy-14x-10y+2000 is Minimum(20,20)