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

The extreme f(x,y)=4-x^2-y^2-y is Maximum(0,-1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=4-x^2-y^2-y

extreme

Maximum (0, - \frac{1}{2})

Solution steps

Find the critical points:

(0, - \frac{1}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - \frac{1}{2}) :

Maximum

Maximum (0, - \frac{1}{2})

What is the extreme f(x,y)=4-x^2-y^2-y ?
The extreme f(x,y)=4-x^2-y^2-y is Maximum(0,-1/2)