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

The extreme f(x,y)=ln(x^2+y^2)-x-y^2 is Saddle(1/2 ,-(sqrt(3))/2),Saddle(1/2 ,(sqrt(3))/2)

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

extreme points

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

Saddle (\frac{1}{2}, - \frac{3}{2}), Saddle (\frac{1}{2}, \frac{3}{2})

Solution steps

Find the critical points:

(\frac{1}{2}, - \frac{3}{2}), (\frac{1}{2}, \frac{3}{2})

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

D (x, y) = \frac{4 x ^{6} + 4 x ^{4} y ^{2} - 4 x ^{4} - 4 x ^{2} y ^{4} - 8 x ^{2} y ^{2} - 4 y ^{6} - 4 y ^{4}}{( x ^{2} + y ^{2} ) ^{4}}

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, - \frac{3}{2}) :

Saddle

Check critical point

(\frac{1}{2}, \frac{3}{2}) :

Saddle

Saddle (\frac{1}{2}, - \frac{3}{2}), Saddle (\frac{1}{2}, \frac{3}{2})

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

f (x, y) = 5 x^{2} + 5 y^{2} + 20 x - 10 y + 40

f (x) = x (27 - 41 + 2 x) (\frac{41}{2} - x)

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

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

What is the extreme f(x,y)=ln(x^2+y^2)-x-y^2 ?
The extreme f(x,y)=ln(x^2+y^2)-x-y^2 is Saddle(1/2 ,-(sqrt(3))/2),Saddle(1/2 ,(sqrt(3))/2)