Logistic growth and phase lines, question 8

Topic: Logistic growth and phase lines
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §13.1Key results

Left alone, a fish stock of $N(t)$ tonnes would grow logistically with $r = 0.3$ per year and carrying capacity 4000 tonnes. A fishery removes a constant $h$ tonnes per year, so the stock satisfies the equation below. Find the maximum sustainable yield: the largest $h$ for which the stock has a steady state. For any larger $h$ the stock collapses, whatever its starting size.

\frac{dN}{dt} = 0.3N\left(1 - \frac{N}{4000}\right) - h

Give h in tonnes per year.

Your answer

tonnes/year

A whole number.

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Theory UBC Math 100 §13.1Key results

Worked solution

Strategy

A steady state is a stock whose natural growth exactly replaces the catch. That is possible only while the catch is no more than the largest growth the stock can produce, so the critical harvest is the maximum of the growth function.

Working

$\frac{dN}{dt} = 0 \iff G(N) = h, \qquad G(N) = 0.3N - 0.000075N^{2}$
At a steady state the catch equals the natural growth.
$G'(N) = 0.3 - 0.00015N = 0 \iff N = 2000$
The peak of the parabola is at half the carrying capacity.
$G(2000) = 0.3(2000)\left(1 - \tfrac{1}{2}\right) = 300$
The largest natural growth the stock can produce: rK/4.
$h \le 300:\ G(N) = h \text{ has a solution}; \qquad h > 300:\ \frac{dN}{dt} = G(N) - h < 0 \text{ for every } N$
Above 300 the stock falls whatever its size, so it collapses.

Answer

h = \frac{rK}{4} = 300 \text{ tonnes per year,} \quad \text{from a stock held at } 2000 \text{ tonnes}

Check

The steady-state equation 0.000075N² − 0.3N + h = 0 has discriminant 0.09 − 0.0003h, which is negative exactly when h > 300.

Where this goes wrong

Giving 2000, the stock at which the yield is largest, instead of the yield. The catch that stock can sustain is its natural growth there, G(2000) = 300 tonnes per year.