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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - \frac{8 ( 2 x ^{2} + y ^{2} + 1 )}{( 2 x ^{2} + y ^{2} - 1 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = 2 x^{2} - 8 x y + 10 y^{2} + 10 x - 4 y - 5

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

f (x, y) = 448 x + 233 y + x y - 2 x^{2} - 3 y^{2} - 15438

f (x) = \frac{2 x}{x ^{2} + 3}, - 4 \leq x \leq 0

f (x) = \frac{x e ^{x}}{4 - x}

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