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

The extreme f(x,y)=ln(sqrt(36-4x^2-9y^2)) is Maximum(0,0)

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - \frac{36 ( 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) :

Maximum

Maximum (0, 0)

f (x) = \frac{1}{3} x^{3} - x^{2} + 3

f (x) = \frac{1}{3} x^{3} - x^{2} + 1

(x + 2)^{2} (x - 3)

f (x) = 3 cos (2 x) + 4

f (x) = 4 x^{3} - 12 x + 7

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