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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 ( 4 - x ^{2} + y ^{2} )}{( 4 - x ^{2} - y ^{2} ) ^{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) = 6 y^{2} + 4 x^{2} - 12 x y - 4 y

f (x, y) = x^{2} + x y + y^{2} - 28 y + 261

f (x) = I x^{2} - 4 I

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x

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

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