Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 32

The equation $\displaystyle \frac{dP}{dt} = r_0 \left( 1 - \frac{P(t)}{p_c} \right)P(t) - \beta P(t)$ represents the rate of change of the fish population present in the fish pond.

$
\begin{equation}
\begin{aligned}
\text{where }& r_0 \text{ is the birth rate of the fish.}\\
& P_c \text{ is the maximum population that the pond can sustain.}\\
& \beta \text{ is the percentage of the population that is harvested.}
\end{aligned}
\end{equation}
$


a.) At what value $\displaystyle\frac{dP}{dt}$ corresponds to a stable population?
b.) Find the stable population level. Suppose that a pond can sustain 10,000 fish, the birth rate is 4%, and the harvesting rate is 4%.
c.) What happens if $\beta$ is raised to 5%.

a.) If the population is stable, then rate of change is neither decreasing nor increasing, that is, $\displaystyle \frac{dP}{dt} = 0$


$
\begin{equation}
\begin{aligned}
\frac{dP}{dt} = 0 &= r_0 \left( 1 - \frac{P(t)}{p_c} \right)P(t) - \beta P(t)\\
\\
\beta \cancel{P(t)} &= r_0 \left( 1 - \frac{P(t)}{p_c} \right)\cancel{P(t)}\\
\\
\beta & = r_0 \left( 1 - \frac{P(t)}{p_c} \right)
\end{aligned}
\end{equation}
$


b.) $\displaystyle \frac{dP}{dt} = 0$, for the condition of population stability.
From part(a),

$
\begin{equation}
\begin{aligned}
\beta & = r_0 \left( 1 - \frac{P(t)}{p_c} \right) && \text{where } r_0 = \frac{5}{100}, \quad P_c = 10,000, \quad \beta = \frac{4}{100}\\
\\
\frac{4}{\cancel{100}} &= \frac{5}{\cancel{100}} \left( 1 - \frac{P(t)}{10,000} \right)\\
\\
\frac{4}{5} &= 1 - \frac{P(t)}{10,000}\\
\\
P(t) & = \left( 1 - \frac{4}{5} \right) 10,000\\
\\
P(t) &= 2,000
\end{aligned}
\end{equation}
$


c.) If $\displaystyle\beta = \frac{5}{100}$, from part (a)


$
\begin{equation}
\begin{aligned}
\cancel{\frac{5}{100}} &= \cancel{\frac{5}{100}} \left( 1 - \frac{P(t)}{10,000}\right)\\
\\
1 &= 1 - \frac{P(t)}{10,000}\\
\\
P(t) &= (1-1) (10,000)\\
\\
P(t) &= 0
\end{aligned}
\end{equation}
$