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

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

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

extreme points

U (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)

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