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

Let  $$\frac{25x}{(x^2+1600x+640000)}=y$$
or $$x^2y+x(1600y-25)+640000y=0$$
This is a quadratic equation in $x$.
As $x$ is real, the discriminant $(1600y-25)^2-4\cdot y\cdot 640000y\ge 0$
On simplification, $-128y+1\ge 0\implies 128y\le 1\implies y\le \frac1 {128}$
So, the maximum value of  $y=\frac{25x}{(x^2+1600x+640000)}$ is $\frac1 {128}$
The value of $x$ for the maximum value of $y$ is $-\frac{1600y-25}{2y}$ where $y=\frac1 {128}$, 
so $x$ will be $\frac{25-1600\cdot \frac 2{128} }{\frac2{128}}=\frac{25\cdot 128-1600}2=800$
Observe that $y$ does not have any lower limit\minimum value.
This approach can be applied to the expression like $\frac{ax^2+bx+c}{Ax^2+Bx+C}$
Reference: Minimum value of given expression
