What is the extreme f(x,y)=x+y-3/2 (x^2+y^2+1) ?

The extreme f(x,y)=x+y-3/2 (x^2+y^2+1) is Maximum(1/3 , 1/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=x+y-3/2 (x^2+y^2+1)

extreme

Maximum (\frac{1}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(\frac{1}{3}, \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 3

D (x, y) = 9

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, \frac{1}{3}) :

Maximum

Maximum (\frac{1}{3}, \frac{1}{3})

What is the extreme f(x,y)=x+y-3/2 (x^2+y^2+1) ?
The extreme f(x,y)=x+y-3/2 (x^2+y^2+1) is Maximum(1/3 , 1/3)