How to calculate the maximum value of: $\frac{25x}{x^2+1600x+640000}$? Wolfram says it's 800, but how to calculate it?
$$
\frac{25x}{x^2+1600x+640000}
$$
 A: Note that
$$
\frac{25x}{x^2+1600x+640 000}=\left(\frac{5}{\sqrt{x}+\frac{800}{\sqrt{x}}}\right)^2
$$
So maximum value is attained when denominator attains it minimal value. Now we use the following trick
$$
\sqrt{x}+\frac{800}{\sqrt{x}}=\left(\sqrt[4]{x}-\frac{\sqrt{800}}{\sqrt[4]{x}}\right)^2+2\sqrt{800}
$$
to see that minimal value of denominator attained when
$$
\sqrt[4]{x}-\frac{\sqrt{800}}{\sqrt[4]{x}}=0
$$
i.e. when $x=800$.
A: 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 $$\frac{25\cdot800}{1600^2}=\frac{25}{3200}=\frac1{128}\;.$$
A: By AM-GM inequality you have
$$\frac{x+800}{2} \geq \sqrt{800x}$$
Thus
$$\frac{1}{3200x} \geq \frac{1}{(x+800)^2}$$
Multiplying by $25x$ you get
$$\frac{25}{3200} \geq \frac{25}{(x+800)^2}$$
Equality is only when we have equality in AM-GM, that is when $x=800$.
A: 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
