Question

A fisheries biologist is maximizing her fishing yield by maintaining a population of lake trout at exactly 300 individuals in the lake. Predict the population growth rate if the lake is stocked with an additional 150 fish. Assume that r for the trout is 0.005 individuals per individual per day. Again, use the differential equation for logistic population growth that was presented in Lecture.

Answer 1

Maximized growth yield=300 individuals.

Since, in the logistic curve maximized growth rate is at K/2 (K= carrying capacity of the population) =300 individuals.

Therefore, K=3002=600.

Total population size after addition (N) = number of individuals=300+150=450

r is given in the question=0.005

The equation for logistic growth rate=dN/dt=rN[1-(N/K)] = 0.005450[1-(450/600)]

=2.25[1-0.75]

=2.250.25

=0.5625(Answer)