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

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.

honeycrisps have skyrocketed to about $4.50 per pound

i think its a and d

step-by-step explanation: