$\max\limits_{x,y} ~~xy \exp(-Ax^2y),$ for $x > 0$ and $0 < y \leq 1$ Is there a closed form expressions for optimal $x$ and $y$ for
$$z = \max\limits_{x,y} ~~xy \exp(-Ax^2y),$$
for $x > 0$ and $0 < y \leq 1$? If yes, how can we obtain it? $A$ is a positive constant.
I can obtain optimal $x$ and $y$ numerically. I tried taking partial derivative with respect to $x$ and $y$ without any luck.

Hints for the case $A=\frac{1}{2}$:


*

*Let $f(x,y) = xy\exp\left(-\frac{1}{2}x^2 y \right)$.

*Fix a $y\in(0,1]$. From 
$$ \partial_x f(x,y) = (1-x^2 y) \underbrace{y \exp\left(-\frac{1}{2}x^2 y \right)}_{>0} $$
follows that the function $x \mapsto f(x, y)$ increases strictly on $(0, x_y]$ and decreases strictly on $[x_y, \infty)$ for $x_y = 1/\sqrt{y}$.

*Thus, we have 
$$ \max_{x,y} f(x,y) = \max_{y} f(x_y, y) = \max_y \frac{1}{\sqrt{y}} \exp\left(-\frac12 \right). $$

