Maximum

$\frac{A_n}{A_{n-1}}=\frac{30^n+15^n}{n!} \cdot \frac{(n-1)!}{30^{n-1}+15^{n-1}}$

$= \frac{1}{n} \cdot \frac{15^n (2^n+1)}{15^{n-1} (2^{n-1}+1)}$

$=\frac{15}{n} \cdot \frac{2 \cdot 2^{n-1} +2-1}{2^{n-1}+1}$

$=\frac{15}{n} \bigg( 2-\frac{1}{2^{n-1}+1} \bigg)$

$=\frac{30}{n}-\frac{15}{n(2^{n-1}+1)}$

Consider which $\frac{A_n}{A_{n-1}}$ will slightly lesser than 1 and slightly larger than 1.

When $n=29$ , term become $1.03448...- \delta$ , $\delta$ is a number very small, actually $\frac{15}{29(2^{28}+1)}=1.927e-9$ . so we can neglect $\delta$ .

When $n=30$ , term become $1- \delta$ , this $\delta$ will smaller than before, but even if $\delta$ is very small, term will always smaller than 1, slightly.

So we get

$\frac{A_{29}}{A_{28}}>1$

$\frac{A_{30}}{A_{29}}<1$

We know $\boxed {A_{29}}$ will be largest.