# Math Insight

### Harvest of natural populations exercises

In the page on harvesting natural populations, we introduced the logstic equation with a harvesting rate of the form \begin{align} P_{t+1}-P_{t} = r P_t\left(1-\frac{P_t}{M}\right) -h_t P_t, \label{harvest_fraction} \end{align} and analyzed a specific example with $r=0.2$. In these exercises, you will analyze the general from with constant $h_t=h$ and $0 < r < 2$.

#### Exercise 1

Divide each term of equation \eqref{harvest_fraction} by $M$ to obtain \begin{align} p_{t+1} - p_t = r p_t \left(1 - p_t \right) - h p_t \label{harvest_fraction_normalized} \end{align} where $p_t = P_t/M$.

#### Exercise 2

1. Show that the equilibria of equation \eqref{harvest_fraction_normalized} are $p_{e} = 0$ and $p_{e} = 1 - h/r$.
2. Show that in order for there to be a positive equilibrium, the fractional harvest rate, $h$, must be less than the low density growth rate, $r$.

#### Exercise 3

1. Convert equation \eqref{harvest_fraction_normalized} (which is in difference form) to function iteration form $p_{t+1}=F(P_t)$. Show that \begin{align} p_{t+1} = F(p_t) = p_t + r p_t (1 - p_t) - h p_t. \label{harvest_fraction_iteration} \end{align}
2. Show that the equilibria of iteration \eqref{harvest_fraction_iteration} are $p_{e} = 0$ and $p_{e} = 1 - h/r$.
3. Assume that $h < r < 2$. Compute $F'(p)$ and evaluate $F'(0)$ and $F'(1-h/r)$. Conclude that $0$ is an unstable equilibrium and $1-h/r$ is a stable equilibrium.

#### Exercise 4

Assume that $h < r < 2$ so that $1-h/r$ is a positive, stable equilibrium of the iteration of equation \eqref{harvest_fraction_iteration}.

The harvest at that equilibrium will be $h (1 - h/r)$.
1. Find the value of $h$ for which $h (1-h/r)$ is the largest. Conclude that in order to maximize harvest, the harvest fraction $h$ should be set at one-half the low density growth rate, $r$.
2. Find the positive equilibrium $p_e$ for this value of $h$.
3. The variable $p_t$ was the actual population size $P_t$ normalized by the carrying capacity $p_t=P_t/M$. What is the equilibrium value of the population size $P_t$ when the harvest rate is set at the value that maximizes harvest? Your answer will be in terms of $M$.