Question

Suppose a logistic fish growth function, where the specie's intrinsic growth rate is r=1 and the carrying capacity of the fishery is K=100. Give some level of stock X, the growth next of that stock is F(X)=rX(1-X/K).

A. Suppose there are 80 fish in year 0. No harvest occurs. How many fish will there be in year 1?

B. Suppose there are 80 fish in year 0. No harvest occurs. What is the equilibrium stock level?

C. Suppose there are 40 fish in year 0, and each year 20 fish are harvested. What is the equilibrium stock level?

D. Suppose there are 20 fish in year 0, and each year 20 fish are harvested. What is the equilibrium stock level?

E. Suppose there are 180 fish in year 0, and each year 20 fish are harvested. What is the equilibrium stock level?

Answer 1

We are given r=1 k=100 Х F(X) = 0X(-) eden - Oxf1- Xo = 80 n Ao dt 100 We know the solution XLE) = 100 +(100-) et UU as x (0) = 80 2 x (1) - 100 1 +0.250- Ñ 9165776 ~ 92

Xo =80 K=100 F(x) = xx(-* ) alone = YX(13 l. For equilibrium 3 rca (1-2) = 0 a x=0 since x=0 or Not n=k possible as Xo=80 : x=k : x= 100 equilibrium It is the Stock level.

Xo = 40 Y K = 100 h = 2:0/year. dx = xr-h dt for eaudibrium put dx .20 dt xe-n=0 x - и - 2o This is the equilibrium Stock level. d. Xo = 20 h= 20/year - dit - & * de -20 -h put da so sa-in=0 a x= n = 20 = 20 equilibrium Stock level this is the e. The equilibrium level will not change for Xo = 180