How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$

It’s straightforward as a calculus problem. To solve it without calculus, note that 
$$\frac{25x}{(x^2+1600x+640000)}=\frac{25x}{(x+800)^2}\;,\tag{1}$$
so the denominator is always positive, the the function has its maximum at some positive value of $x$. That maximum will occur where
$$\frac{(x+800)^2}{25x}=\frac1{25}\left(x+1600+\frac{640000}x\right)=64+\frac1{25}\left(x+\frac{640000}x\right)\tag{1}$$
has its minimum (over the range $x>0$). This in turn occurs where $x+\dfrac{640000}x$ has its minimum.
Now $x$ and $\frac{640000}x$ are a pair of numbers whose product is $640000=800^2$; if we set $x=800$, their sum is $1600$. Suppose that we set $x=800+a$ for some $a>0$; then 
$$\begin{align*}
x+\frac{640000}x&=800+a+\frac{640000}{800+a}\\
&=\frac{1280000+1600a+a^2}{800+a}\\
&=1600+\frac{a^2}{800+a}\\
&>1600\;.
\end{align*}$$
Thus, $x=800$ gives us the minimum value of of $x+\frac{640000}x$, namely, $1600$, and hence the minimum value of $(2)$ and the maximum value of $(1)$. Substituting $x=800$ into $(1)$, we find that the maximum is $$\frac{25\cdot800}{1600^2}=\frac{25}{3200}=\frac1{128}\;.$$
