Find extremas of $f(x,y) = xy \ln(x^2+y^2), x>0, y>0$ As the title says I need to find extreme values(maximum and minimum) of
$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$
I don't understand how to find critical points of this problem.
I start with finding partial derivative and set derivatives equal to zero. And that is where I am stuck currently. So any help would be appreciated.
So: 
$
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy* ln(x^2+y^2)) = y*ln(x^2+y^2) + \frac{2xy^2}{x^2+y^2}
$
$
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy* ln(x^2+y^2)) = x*ln(x^2+y^2) + \frac{2x^2y}{x^2+y^2}
$
So we have now:
$
\nabla f(x,y) = (0,0) 
$
$
 y*ln(x^2+y^2) + \frac{2xy^2}{(x^2+y^2)^2} = 0
$
And 
$
 x*ln(x^2+y^2) + \frac{2x^2y}{(x^2+y^2)^2} = 0
$
After trying to solve these equations I get that $x=y$ Is that correct?. So I don't understand what are then critical points as It can't be (0,0)?
 A: It's easier to work with polar coordinates $x = r \cos \theta, y = r \sin \theta$ where (in your case) $r > 0$ and $0 < \theta < \frac{\pi}{2}$. Then we have
$$ f(x,y) = xy \ln(x^2 + y^2) = r \cos \theta r \sin \theta \ln(r^2) = r^2 \ln r \sin(2 \theta)  = g(r,\theta). $$
The partial derivatives of $g$ are given by
$$ \frac{\partial g}{\partial r} = r (2 \ln r + 1) \sin(2 \theta), \\
\frac{\partial g}{\partial \theta} = 2 r^2 \ln r \cos(2 \theta). $$
Since $r > 0$ and $\sin 2\theta > 0$ in our range, the equation $\frac{\partial g}{\partial r} = 0$ implies that $2 \ln r + 1 = 0$ or $r = e^{-\frac{1}{2}}$. The second equation then implies that $\cos 2 \theta = 0$ so $\theta = \frac{\pi}{4}$. Hence, we have one critical point at $r = e^{-\frac{1}{2}}$ and $\theta = \frac{\pi}{4}$. The second partial derivatives of $g$ are given by
$$ \frac{\partial^2 g}{\partial r^2} = (2 \ln r + 3) \sin(2 \theta), \\
\frac{\partial^2 g}{\partial r \partial \theta} = 2 r(2 \ln r + 1) \cos(2 \theta), \\
\frac{\partial^2 g}{\partial \theta} = -4 r^2 \ln r \sin(2 \theta). $$
Hence, the Hessian at $r = e^{-\frac{1}{2}}, \theta = \frac{\pi}{4}$ is given by
$$ \begin{pmatrix} 2 & 0 \\ 0 & 2e^{-1} \end{pmatrix}. $$
Since the Hessian is positive definite, $(e^{-\frac{1}{2}}, \frac{\pi}{4})$ a local minimum point for $g$ (and so $\frac{e^{-\frac{1}{2}}}{\sqrt{2}} \left( 1, 1 \right)$ is a local minimum for $f$).
A: We look for points for wich $\nabla f=0$
The method you describe is this:
$\dfrac{\partial f}{\partial x}=y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}$
$\dfrac{\partial f}{\partial y}=x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}$
(You need to practice the derivatives :)
Setting the partials to zero we can get the coordinates for those points.
$y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}=0$
$x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}=0$
$\dfrac{y}{x}\ln(x^2+y^2)=\dfrac{x}{y}\ln(x^2+y^2)\implies x^2=y^2\;,x^2+y^2\neq1$
Then, $y=x$ or $y=-x$, dropping the second as $x\gt0$ and $y\gt0$
$x\ln(2x^2)=-\dfrac{2x^3}{2x^2}\implies \ln(2x^2)=-1$ or $x=\pm\dfrac{e^{-1/2}}{\sqrt{2}}$
Dropping the negative solution:
$x=y=\dfrac{e^{-1/2}}{\sqrt{2}}$
(We need to consider the case $x^2+y^2=1$. It's easily seen that the partials doesn't vanish: for those points, $2x^2y=0$, that is not possible with $x\gt0\;,y\gt0$)
A: In polar coordinates, you have $$\sin(\theta)\cos(\theta)r^2\ln(r^2)$$
$\sin(\theta)\cos(\theta)$ is maximal when $\theta=\pi/4$, and $r^2\ln(r^2)$ is unbounded above. So there is no maximum on the specified domain.
$\sin(\theta)\cos(\theta)$ is always positive on the specified domain, and $r^2\ln(r^2)$ does have a negative absolute minimum value. So by finding the minimum of $r^2\ln(r^2)$ and pairing that with $\theta=\pi/4$, you can find the absolute minimum using one-variable calculus.
With $\frac{d}{dr}\left(r^2\ln(r^2)\right)=2r\left(\ln(r^2)+1\right)$, find that the minimal $r$-value is $e^{-1/2}$.
