### Harvest of natural populations exercise answers

Here we give answers to some of the exercises on harvest of natural populations.

#### Exercise 3

- \begin{align*}
F'(p) & = 1 + r -2 r p - h\\
F'(0) & = 1+r-h > 1
\end{align*}
Since $F'(0) > 1$, we can conclude that $p_e=0$ is an unstable equilibrium.
\begin{align*} F'(1-h/r) & = 1 + r -2 r (1-h/r) - h \\ &= 1 - r + h \end{align*} To determine that $p_e=1-h/r$ is stable we need to show that $|F'(1-h/r)| \lt 1$ or that \begin{align*} |1-r+h| \lt 1. \end{align*} This inequality is equivalent to \begin{gather*} -1 \lt 1-r+h \lt 1 \end{gather*} or \begin{gather*} 0 < r-h < 2 \end{gather*} Because we are give that $h < r$, we can conclude that $0 < r-h$. In addition, because $r<2$ and $h$ is positive, we know that $r-h < r < 2$. Indeed $0 < r - h < 2$ and $1-h/r$ is a stable equilibrium.

#### Exercise 4

- Since $h=r/2$ when the harvest is maximized $p_e = 1-h/r=1/2$.
- At this equilibrium, $P_t = p_t M = \frac{1}{2}M$, or one-half of the maximum supportable population.

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