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) = - 2 x^{3} + 21 x^{2} - 36 x + 3

f (x) = 3 x^{2} + 5 y^{2} - 10 x y - 6 y + 1

9600 + 280 x - 10 x^{2}

4 x^{3} + 9 x^{2} - 54 x + 6

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)