What is the extreme f(x,y)=ln(9x^2+3y^2-4) ?

The extreme f(x,y)=ln(9x^2+3y^2-4) is Maximum(0,0)

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extreme f(x,y)=ln(9x^2+3y^2-4)

extreme

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{6 ( - 3 y ^{2} + 9 x ^{2} - 4 )}{( 9 x ^{2} + 3 y ^{2} - 4 ) ^{2}}

D (x, y) = \frac{- 8748 x ^{4} - 5832 x ^{2} y ^{2} - 972 y ^{4} + 1728}{( - 4 + 9 x ^{2} + 3 y ^{2} ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

What is the extreme f(x,y)=ln(9x^2+3y^2-4) ?
The extreme f(x,y)=ln(9x^2+3y^2-4) is Maximum(0,0)