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

The extreme f(x,y)=ln(2-x^2-y^2) is Maximum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme points

f (x, y) = ln (2 - x^{2} - y^{2})

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 ( 2 - x ^{2} + y ^{2} )}{( 2 - x ^{2} - y ^{2} ) ^{2}}

D (x, y) = - \frac{4 ( x ^{2} + y ^{2} + 2 )}{( x ^{2} + y ^{2} - 2 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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