Logistic growth and phase lines, question 9

Topic: Logistic growth and phase lines
Difficulty: Challenging
Suggested time: 8 min
Marks: 4
Theory: UBC Math 100 §13.1Key results

A fishery works with a fixed effort $E$ , so its catch is proportional to the stock: $EN$ tonnes per year. The stock $N(t)$ , in tonnes, then satisfies the equation below, with $0 < E < 0.8$ . Find the positive steady state $N^{*}$ in terms of $E$ , and the catch $EN^{*}$ taken each year once the stock is there. Which effort $E$ makes that sustained catch as large as possible?

\frac{dN}{dt} = 0.8N\left(1 - \frac{N}{1000}\right) - EN

Give the effort E, per year.

Your answer

per year

A decimal or a fraction.

How sure are you?

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Theory UBC Math 100 §13.1Key results

Worked solution

Strategy

For a given effort the stock settles at its positive steady state, which is stable. Multiply that stock by the effort to get the yearly catch, then maximise the catch as a function of E.

Working

$0.8\left(1 - \frac{N}{1000}\right) = E \iff N^{*} = 1000\left(1 - \frac{E}{0.8}\right) = 1000 - 1250E$
Dividing by N is safe because the steady state wanted has N > 0; the other one is N = 0.
$Y(E) = E\,N^{*} = 1000E - 1250E^{2}$
Once the stock sits at N*, the catch is the same every year.
$Y'(E) = 1000 - 2500E = 0 \iff E = 0.4, \qquad Y''(E) = -2500 < 0$
A downward parabola, so this critical point is the maximum.
$N^{*} = 1000 - 1250(0.4) = 500, \qquad Y(0.4) = 0.4(500) = 200$
The stock is held at K/2 and the catch is rK/4, the same maximum sustainable yield as with a fixed catch.

Answer

E = 0.4 \text{ per year} \quad (N^{*} = 500 \text{ tonnes,} \quad \text{catch } 200 \text{ tonnes per year})

Check

Nearby efforts catch less: E = 0.3 and E = 0.5 each give 187.5 tonnes a year, against 200 at E = 0.4.

Where this goes wrong

Fishing as hard as the stock will bear, with E close to 0.8. At E = 0.8 the steady stock is zero, and so is the catch; the sustained catch peaks halfway, at E = 0.4.