I cant get the third question correct. One question for fishery management is ho

Question

I cant get the third question correct.

One question for fishery management is how to control fishing to optimize profits for the fishermen. We will soon study the continuous logistic growth equation for populations. One differential equation describing the population dynamics for a population of fish F| with harvesting is given by the equation. dF/dt = rF(1 - F/K) - xF,| where r| is the growth rate of this species of fish at low density. K| is the carrying capacity of this population, and xi is the harvesting effort of the fishermen. We will show that the nonzero equilibrium of this equation is given by F_e = K (r-x)/r.| One formula for profitability is computed by the equation P = xF_e,| so P(x) = Kx (r - x)/r.| Find the level of harvesting ad that produces the maximum profit possible with this dynamics. X_max = where x_maxmight depend on any of the parameters listed in this problem. What is the equilibrium population at this optimal profitability? Also, determine the maximum possible fish population for this model and at what harvesting level this occurs. (Clearly, this is a grossly oversimplified model, but can give some estimates for long range management.)

Explanation / Answer

Given

dF/dt = rF(1 - F/K) - xF

and the zero equillibrium equation given by

F = K (r - x)/r

which can be written as

F/K = (r - x)/r = y(say)

Dividing the differential equation by K,

dy/dt = ry(1 - y) - xy

The solution to the differential equation can be written as

y(t) = 1/(1 + (1/x0 - 1)ert) + 1/ext

The function can be given as a sigmoid function.

So, the level of harvesting that produces maximum profit with this dynamics

xmax = r/2

and the equillibrium population at this level can be given as

F = K/2

The maximum possible fish population for this model can be given as

F = 1 - F/K

which will take place when x = 0

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