What is the extreme f(x,y)=ln(sqrt(x^2+y^2-64)) ?

The extreme f(x,y)=ln(sqrt(x^2+y^2-64)) is Maximum(0,0)

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extreme f(x,y)=ln(sqrt(x^2+y^2-64))

extreme points

f (x, y) = ln (x^{2} + y^{2} - 64)

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{- y ^{2} + x ^{2} - 64}{( x ^{2} + y ^{2} - 64 ) ^{2}}

D (x, y) = - \frac{x ^{2} + y ^{2} + 64}{( x ^{2} + y ^{2} - 64 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

(x + 6)^{\frac{2}{3}}

x^{2} y^{2} - x - y

\frac{x ^{2} + 2}{x ^{2} - 4 x}

f (x, y) = x^{2} + y^{2} - 5 x + 4 y + x y

f (x) = x^{2} + x y + \frac{1}{2} y^{2} - 2 x + y

What is the extreme f(x,y)=ln(sqrt(x^2+y^2-64)) ?
The extreme f(x,y)=ln(sqrt(x^2+y^2-64)) is Maximum(0,0)