When a population (such as a fish species) is harvested, the maximal sustainable yield, or MSY, is the harvesting rate that gives the greatest yield, without driving the population to extinction. If you take out more than the MSY, then the population crashes; if you take out less than the MSY than your harvesting operation is less efficient.

One of the most famous population crashes due to overharvesting is the collapse of the Atlantic northwest cod fishery, which happened in 1992. The cod population has still not recovered. It’s worth adding that the concept of maximal sustainable yield is controversial, and many ecologists believe that implementation of this idea can cause severe damage to a population.

In this question we’ll work out the MSY for a simple population model with harvesting.

a. We begin by looking at a model of population growth without harvesting. A common model of population growth is the logistic equation, which says that the growth rate, R, of a population is given by We talk about this model in more detail in Chapter 25, Section 25.1.

R = rN 

1 −

N K



, for some constants r and K, where N is the number in the population.

i. What are the units of K and r? (Hint: first choose sensible units for R and N.)

ii. Now plot R(N) for a variety of values of r and K.
iii. Which values of N give zero population growth?
iv. Show that the maximal growth rate of the population is rK/4 and that this rate occurs when N = K/2.
v. Based on your answers to the previous three parts, what do K and r represent scientifically?

b. Now modify the logistic growth model to include a harvesting term. Assume that the population is harvested at the rate hN, i.e., the rate of harvesting (fishing, hunting, whatever) is proportional to the number in the population.
This gives the equation for the rate of population growth This is a very simple model of harvesting, but good enough for now. as R = rN 
1 −
N K 
− hN.

i. Show that, in order to get a population that has zero growth rate at some positive N, you must have h < r.
Show also that this population with zero growth rate Of course, h > 0 otherwise it’s not harvesting.
(called the steady state population, Ns) is Ns = K 
1 −
h r 
.
ii. Now set N to be the steady state population (i.e., put N = Ns). The rate of harvesting is then hNs. Show that the maximal rate of harvesting is rK/4 and that this occurs when h = r/2.
In other words, the MSY is the same as the maximal growth rate of the population, which is rather a neat result.