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

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

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

extreme points

f (x, y) = ln (x^{2} + y^{2} - 4)

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( - y ^{2} + x ^{2} - 4 )}{( x ^{2} + y ^{2} - 4 ) ^{2}}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = ln (x + y)

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

f (x, y) = 4 - x^{2} - 4 y^{2}

\frac{ln ( x )}{x ^{2}}

f (t) = e^{- a t}

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