What is the extreme f(x,y)=ln(10-x^2-y^2) ?

The extreme f(x,y)=ln(10-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

Popular >

extreme f(x,y)=ln(10-x^2-y^2)

extreme points

f (x, y) = ln (10 - x^{2} - y^{2})

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 ( 10 - x ^{2} + y ^{2} )}{( 10 - x ^{2} - y ^{2} ) ^{2}}

D (x, y) = - \frac{4 ( x ^{2} + y ^{2} + 10 )}{( x ^{2} + y ^{2} - 10 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{x ^{2} - 2 x - 3}{x + 5}, - 1 \leq x \leq 3

y = 10 cos (x) + 10 sin (x)

\frac{( - 6 x + 21 )}{( ( x + 1 ) ^{4} )}

\frac{- 5}{4 x - 8}

f (x) = f (x) = x^{3} - 4 x^{2} - 3 x + 1

What is the extreme f(x,y)=ln(10-x^2-y^2) ?
The extreme f(x,y)=ln(10-x^2-y^2) is Maximum(0,0)