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

The extreme f(x)=1-x^2-x-3y^2+y is Maximum(-1/2 , 1/6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Maximum (- \frac{1}{2}, \frac{1}{6})

Solution steps

Find the critical points:

(- \frac{1}{2}, \frac{1}{6})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, \frac{1}{6}) :

Maximum

Maximum (- \frac{1}{2}, \frac{1}{6})

What is the extreme f(x)=1-x^2-x-3y^2+y ?
The extreme f(x)=1-x^2-x-3y^2+y is Maximum(-1/2 , 1/6)