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

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

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

extreme points

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

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} - 4 )}{( x ^{2} + 4 y ^{2} - 4 ) ^{2}}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = cos^{2} (x), 0 \leq x \leq π

f (x) = x^{3} + 2 x^{2} - 7 x

f (x, y) = x e^{- 3 x^{2} - 3 y^{2}}

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 2

\frac{1}{x - 2}

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