Harvesting a Renewable Resource. Suppose that the population \(y\) of a certain
species of fish (for example, tuna or halibut) in a given area of the ocean is
described by the logistic equation
$$
d y / d t=r(1-y / K) y .
$$
While it is desirable to utilize this source of food, it is intuitively clear
that if too many fish are caught, then the fish population may be reduced
below a useful level, and possibly even driven to extinction. Problems 20 and
21 explore some of the questions involved in formulating a rational strategy
for managing the fishery.
In this problem we assume that fish are caught at a constant rate \(h\)
independent of the size of the fish population. Then \(y\) satisfies
$$
d y / d t=r(1-y / K) y-h
$$
The assumption of a constant catch rate \(h\) may be reasonable when \(y\) is
large, but becomes less so when \(y\) is small.
(a) If \(hy_{0}>y_{1},\) then \(y \rightarrow y_{2}\) as \(t \rightarrow \infty,\) but
that if \(y_{0}r K / 4,\) show that \(y\) decreases to zero as \(l\) increases
regardless of the value of \(y_{0}\).
(c) If \(h=r K / 4\), show that there is a single cquilibrium point \(y=K / 2\)
and that this point is semistable (see Problem 7 ). Thus the maximum
sustainable yield is \(h_{m}=r K / 4\) corresponding to the equilibrium value
\(y=K / 2 .\) Observe that \(h_{m}\) has the same value as \(Y_{m}\) in Problem
\(20(\mathrm{d})\). The fishery is considered to be overexploited if \(y\) is
reduced to a level below \(K / 2\).