What is the extreme f(x,y)=ln(4x^2+18y^2-36) ?

The extreme f(x,y)=ln(4x^2+18y^2-36) is Maximum(0,0)

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extreme f(x,y)=ln(4x^2+18y^2-36)

extreme points

f (x, y) = ln (4 x^{2} + 18 y^{2} - 36)

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - \frac{72 ( 2 x ^{2} + 9 y ^{2} + 18 )}{( 2 x ^{2} + 9 y ^{2} - 18 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

F (x, y) = x^{3} - 3 x y + 3 y^{2} + 1

f (x) = x y + 5 x - 5

g (x) = \frac{x ^{5} + I n ( x )}{x ^{2} - x}

e^{x} (24 - x^{2})

x^{3} + 3 x y^{2} - 15 x - 12 y

What is the extreme f(x,y)=ln(4x^2+18y^2-36) ?
The extreme f(x,y)=ln(4x^2+18y^2-36) is Maximum(0,0)