Extremum of ln(x)/x = C (analytical) I want to find the maximum value of y of the following equation:
$A\cdot \sqrt{x^2 +y^2} = e^{\frac{\sqrt{x^2 + y^2}-x}{B}}~(1)$
I tried to use polar coordinates with $x = r \cos(\phi)$ and $y=r \sin(\phi)$ to solve this problem. This yields to
$A \cdot r = e^{\frac{r ( 1-\cos(\phi))}{B}}~(2)$
with $\partial y /\partial x = \frac{\partial y}{\partial\phi} \frac{\partial\phi}{\partial x} = \frac{\partial (r(\phi) \sin(\phi))}{\partial\phi} \frac{\partial\phi}{\partial (r(\phi) \cos(\phi))} = 0  ~(3)$
I am stuck with the term originating from (2) $ln(r) = r\cdot C$ or $r=e^r\cdot C$. I cannot wrap my head around it.
Any other ideas how to approach finding the maximum value of y in (1)?
 A: Implicitly differentiate (1) with respect to $x$,
$$
A\frac{\mathrm{d}r}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(e^{(r-x)/B})=\frac1Be^{(r-x)/B}\frac{\mathrm{d}}{\mathrm{d}x}(r-x)=\frac1BAr\left(\frac{\mathrm{d}r}{\mathrm{d}x}-1\right)
$$
Simplifying,
$$
\frac{\mathrm{d}r}{\mathrm{d}x}=\frac{r}{r-B}
$$
and using $r^2=x^2+y^2$,
$$
y\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{r^2}{2(r-B)}-x.
$$
At the maximum of $y$, we have $\frac{\mathrm{d}y}{\mathrm{d}x}=0$, so
$$
\frac{r^2}{r-B}=2x.
$$
So you need to solve
$$
\left\{
\begin{aligned}
Ar&=e^{(r-x)/B}\\
r^2 &=2x(r-B)
\end{aligned}
\right.
$$
which I doubt a closed form of the solution exists.  Your best bet is probably some numerical methods.
A: Starting from user10354138's answer, it is possible to reduce the problem to a single equation since
$$Ar=e^{\frac{r-x}B}\implies r=-B W\left(-\frac{e^{-\frac{x}{B}}}{A B}\right)$$ where appears Lambert function.
This means that we need to solve for $x$ the remaining equation which becomes
$$2 x
   \left(W\left(-\frac{e^{-\frac{x}{B}}}{A B}\right)+1\right)+B\, W\left(-\frac{e^{-\frac{x}{B}}}{A B}\right)^2=0$$ which, for sure, requires a numerical method.
Trying for $A=\pi$ and $B=e$, the function looks almost as a straight line and a series expansion around $x=0$ gives, as an estimate, 
$x_0=-0.0288123$ while the exact solution is $-0.0288782$.
Using Newton method, the iterates would be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0 \\
 1 & -0.02881229451 \\
 2 & -0.02887821498 \\
 3 & -0.02887821533
\end{array}
\right)$$ leading to $r=0.3684028027$ and finally $y=0.3672692115$
