Fish Farms

In a fish farm, a population of fish is introduced into a pond and harvested regularly. A model for the rate of change of the fish population is given by the following equation, which represents dP/dt:

ratechangepop[r0_,Pc_,beta_] = r0 (1 - P/Pc) P - beta P

where r0 is the birth rate of the fish, Pc is the maximum population that the pond can sustain (called the carrying capacity, hence the subscript c), beta is the percentage of the population that is harvested, t is the time, and P is the population at any time t, which we could write as P(t).

We don't know what P(t) is, but we do know that it is a function of time.

What value of dP/dt corresponds to a stable population?

A stable population means the rate of  change of the population is zero; that is, the population is constant. This means that dP/dt = 0. A stable population P will therefore satisfy the equation:

eqstable[r0_,Pc_,beta_] := ratechangepop[r0,Pc,beta] == 0

Here is the equation, where I have used some of Mathematica's special formatting abilities for subscripts, superscripts and special characters.

eqstable[(Subscript[r, 0]),(Subscript[P, c]),β]

If the pond can sustain 10 000 fish, the birth rate is 5%, and the harvesting rate is 4%, find the stable population level.

Translating this information into our variables, we have:
r0 = birth rate = 5% = 0.05,
Pc = carrying capacity = 10 000,
beta = percentage of fish harvested = 4% = 0.04,
P = fish population.

To obtain the stable population we must solve the following equation for P:

Solve[ eqstable[0.05,10000,0.04],P]

So we see that for the quantities given, a stable population of fish would be 0 (a degenerate solution with no fish in the pond), or 2000.

What happens to the stable population if the harvesting percentage is raised to 5%?

The harvesting percentage is beta, so for these new conditions we must solve the equation:

Solve[eqstable[0.05,10000,0.05],P]

Oh dear! We see that if we raise the harvesting percentage from 4% to 5% we end up with no fish in the pond as the stable fish population. This is a case where we are overharvesting the resource.