Appendix A

If a population is growing at per capita rate $r(1-N/K)$, then the number of fish being produced is $rN(1-N/K)$. If the same number of fish is removed, the population size will not change over time. So the maximum sustained yield is attained at the value of $N$ that maximizes $f(N) = rN(1-N/K)$. Remembering a little calculus, this corresponds to the value of $N$ for which $df/dN = 0$ and $d^2f/dN^2 < 0$.

$$\begin{eqnarray*} \frac{df}{dN} &=& r\left(1 - \frac{2N}{K}\right) \\ \frac{d^2f}{dN^2} &=& -\left(\frac{2}{K}\right) \quad . \end{eqnarray*}$$

So $f(N)$ is maximized at $N=K/2$, and the maximum sustained yield is

$$\begin{eqnarray*} f\left(\frac{K}{2}\right) &=& r\left(\frac{K}{2}\right) \left(... ...ft(\frac{K/2}{K}\right)\right) \\ &=& r\left(\frac{K}{4}\right) \end{eqnarray*}$$