What is the extreme f(x)=ln(9-x^2-9y^2) ?

The extreme f(x)=ln(9-x^2-9y^2) is Maximum(0,0)

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extreme f(x)=ln(9-x^2-9y^2)

extreme points

f (x) = ln (9 - x^{2} - 9 y^{2})

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{18 ( 9 - x ^{2} + 9 y ^{2} )}{( 9 - x ^{2} - 9 y ^{2} ) ^{2}}

D (x, y) = - \frac{36 ( x ^{2} + 9 y ^{2} + 9 )}{( x ^{2} + 9 y ^{2} - 9 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

x^{4} + y^{4} - 4 x y + 5

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

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

f (x) = x^{2} + 6 x + 9

f (x) = x + 5

What is the extreme f(x)=ln(9-x^2-9y^2) ?
The extreme f(x)=ln(9-x^2-9y^2) is Maximum(0,0)