What is the extreme f(x,y)=ln(x^2+4y^2-1) ?

The extreme f(x,y)=ln(x^2+4y^2-1) is Maximum(0,0)

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extreme f(x,y)=ln(x^2+4y^2-1)

extreme points

f (x, y) = ln (x^{2} + 4 y^{2} - 1)

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{8 ( - 4 y ^{2} + x ^{2} - 1 )}{( x ^{2} + 4 y ^{2} - 1 ) ^{2}}

D (x, y) = - \frac{16 ( x ^{2} + 4 y ^{2} + 1 )}{( x ^{2} + 4 y ^{2} - 1 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

x^{3} - 6 x^{2} + 9 x + 5

f (x) = x^{4} - 4 x^{3} - 8 x^{2} + 8

f (x, y) = 6 - x - 2 y

2 x^{3} + 3 x^{2} - 36 x

What is the extreme f(x,y)=ln(x^2+4y^2-1) ?
The extreme f(x,y)=ln(x^2+4y^2-1) is Maximum(0,0)