Question

Suppose a population of fish at a farm follows logistic growth. The per capita birth rate is 0.7 fish per day, the per capita natural death rate is 0.2 fish per day, and the carrying capacity is 20000. Fish are harvested at a constant rate of 7000 fish per week.

Find the equilibrium solutions to the equation.

Answer 1

Given:

Per capita birth rate = 0.7 fish per day

Per capita death rate = 0.2 fish per day

Harvesting rate = 7000 fish per week

Let at any time Day D : Population (P)

0.7*(P) = New born fish on that day

0.2*(P) = Dead Fishes

Total population Next day Day (D+1) : P + 0.7*(P) - 0.2*(P) = 1.5*(P) =

on (D+2) = (1.5)^2 P

So in 7 days Final Population = (1.5)^7P

For equillibrium New fishes = harvested fishes

(1.5)^7P - P = 7000

16.0859 P = 7000

Around 436 fishes at time of equillibrium

(1.003505 -1)P = 7000

Then

P = 7000/0.003505 = 1997002 (Approx)

I hope help you.