Question

Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of H. If N(t) is the population size at time t, the population dynamics are dN dt = r 1 − N K N − H. Different values of H will result in different equilibrium population sizes, and if H is large enough we might expect extinction.(a) Suppose r = 2, K = 1000, and H = 100. Find all equilibria. (Round your answers to the nearest integer. Enter your answers as a comma-separated list.) N hat =(b) Determine whether each of the equilibria found in part is locally stable or unstable. (Round your answers to the nearest whole number. Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) stable N hat = unstable N hat =Is the population predicted to go extinct?YesNo

Answer 1

Answer:

a) Equilibrium point : [ 947, 53 ]

b) N = 947 is stable equilibrium, N = 53  is unstable equilibrium

c) N0, the population will not go extinct

Step-by-step explanation:

a)

Given that;

r = 2, k = 1000, H = 100

dN/dT = R(1 - N/k)N - H

so we substitute

dN/dt = 2( 1 - N/1000)N - 100

now for equilibrium solution, dN/dt = 0

so

2( 1 - N/1000)N - 100 = 0

((1000 - N)/1000)N = 50

N^2 - 1000N + 50000 = 0

N = 1000 ± √(-1000)² - 4(1)(50000)) / 2(1)

N = 947.213 OR 52.786

approximately

N = 947 OR 53

Therefore Equilibrium point : [ 947, 53 ]

b)

g(N) = 2( 1 - N/1000)N - 100

= 2N - N²/500 - 100

g'(N) = 2 - N/250

SO AT 947

g'(N) = g'(947) =  2 - 947/250 = -1.788 which is less than (<) 0

so N = 947 is stable equilibrium

now AT 53

g'(N) = g"(53) = 2 - 53/250 = 1.788 which is greater than (>) 0

so N = 53  is unstable equilibrium

The capacity k=1000

If the population is less than 53 then the population will become extinct but since the capacity is equal to 1000 then the population will not go extinct.