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I'm finding local extremes on this function $f(x,y)=x\cdot y\cdot \ln(x^{2}+y^{2})$ where $x,y$ is from $\mathbb{R}^2-$$ \{(0,0)\}$ . When i derivative it, i got a gradient:

$\nabla f(x,y)=\Biggl( y\cdot \biggl(\ln(x^2+y^2)+\frac{2x^2}{x^2+y^2}\biggr), x\cdot\biggl(\ln(x^2+y^2)+\frac{2y^2}{x^2+y^2}\biggr)\Biggr)$

Then i solve system of equations:

$\ y\cdot\biggl(\ln(x^2+y^2)+\frac{2x^2}{x^2+y^2}\biggr)=0$

$x\cdot\biggl(\ln(x^2+y^2)+\frac{2y^2}{x^2+y^2}\biggr)=0$

From this i found critical points:

$(0,1), (0,-1), (1,0), (-1,0)$

and when i solve case where $x\neq0$ and $y\neq0$ so i got stuck.

$\ln(x^2+y^2)+\frac{2y^2}{x^2+y^2}=0$

$\ln(x^2+y^2)+\frac{2x^2}{x^2+y^2}=0$

Please do you know someone how to solve $x,y$ from this system of equations? I know that there are $4$ other critical points.

Ennar
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S.Roman
  • 3

1 Answers1

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From $\ln(x^2+y^2)+\frac{2y^2}{x^2+y^2}=0$ you get

$(x^2+y^2) \ln (x^2+y^2)=-2y^2$

and from $\ln(x^2+y^2)+\frac{2x^2}{x^2+y^2}=0$ you get $(x^2+y^2) \ln (x^2+y^2)=-2x^2$.

Hence $x^2=y^2$.

This gives

$2x^2 \ln(2x^2)=-2x^2$.

Since $x \ne 0$, we derive $\ln(2x^2)=-1$.

Can you proceed ?

Fred
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