What is the extreme f(x,y)=ln(4x^2+9y^2+36) ?

The extreme f(x,y)=ln(4x^2+9y^2+36) is Minimum(0,0)

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extreme f(x,y)=ln(4x^2+9y^2+36)

extreme points

f (x, y) = ln (4 x^{2} + 9 y^{2} + 36)

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{18 ( - 9 y ^{2} + 4 x ^{2} + 36 )}{( 4 x ^{2} + 9 y ^{2} + 36 ) ^{2}}

D (x, y) = - \frac{144 ( 4 x ^{2} + 9 y ^{2} - 36 )}{( 4 x ^{2} + 9 y ^{2} + 36 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

What is the extreme f(x,y)=ln(4x^2+9y^2+36) ?
The extreme f(x,y)=ln(4x^2+9y^2+36) is Minimum(0,0)