College Algebra, Chapter 2, 2.5, Section 2.5, Problem 48

According to a scientist, the rate $r$ at which a disease spreads in a population of size $P$ is jointly proportional to the number $x$ of infected people and the number $P-x$ who are not infected. An infection erupts in a small town that has population $P =5000$.
a.) Write an equation that expresses $r$ as a function of $x$

$
\begin{equation}
\begin{aligned}
r &= k x ( P - x ) && \text{Model}\\
\\
r &= k x (5000 -x) && \text{Substitute } P = 5000
\end{aligned}
\end{equation}
$




b.) Compare the rate of spread of this infection when 10 people are infected to the rate of spread when 1000 people are infected. Which rate is larger? By what factor?

$
\begin{equation}
\begin{aligned}
r_1 &= k(10)(5000 -10)\\
\\
r_1 &= k (49900)\\
\\
k &= \frac{r_1}{49900} && \Longleftarrow \text{Equation 1}\\
\\
\\
\\
\\
r_2 &= k(100)(5000-100)\\
\\
r_2 &= k(49000)\\
\\
k &= \frac{r_2}{49000} && \Longleftarrow \text{Equation 2}\\
\end{aligned}
\end{equation}
$

By using Equations 1 and 2

$
\begin{equation}
\begin{aligned}
\frac{r_1}{49900} &= \frac{r_2}{49000} && \text{Multiply both sides by } 49900\\
\\
r_1 &= \frac{499}{490} r_2\\
\\
r_1 &= 1.02 r_2
\end{aligned}
\end{equation}
$

It shows that the first rate is larger by a factor of $0.02$

c.) Calculate the rate of spread when the entire population is infected. Why does this answer make intuitive sense?
When the entire population is infected, then $x = 5000$

$
\begin{equation}
\begin{aligned}
r &= k x(P-x)\\
\\
r &= (5000)(5000-5000)\\
\\
r &= 0
\end{aligned}
\end{equation}
$

It shows that there is no need for the disease to spread on since the entire population is already infected.