36. Find the maximum value of the function $y=\frac{x}{a x^{2}+b}(a>0, b>0)$.
36. $\frac{a x^{2}+b}{2} \geqslant \sqrt{a x^{2} b}=x \sqrt{a b}$ for all $x>0$, therefore

$$
y=\frac{a}{a x^{2}+b} \leqslant \frac{x}{2 x \sqrt{a b}}=\frac{1}{2 \sqrt{a b}}
$$

The maximum value of $y$ is achieved when the denominator $a x^{2}+b$ is at its minimum, i.e., when $a x^{2}+b=2 x \sqrt{a b}$. It is equal to $\frac{1}{2 \sqrt{a b}}$.