Extremum of ln(x)/x = C (analytical)

Question

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)?

Answer 1

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.

Answer 2

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 $y=0.3672692115$