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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

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

f (x) = x^{4} - 2 x^{3}, - 2 \leq x \leq 2

f (x) = \frac{7 x}{x ^{2} - 49}

f (x, y) = 9 x^{3} + \frac{y ^{3}}{3} - 4 x y

f (x, y) = 2 x y^{4} - 5 x^{2} y^{3} - 3 x^{3} y

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