Question

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uniformly distributed throughout the year. It is seen that the fish population holds fairly constant at about 4,000. If you wanted to maximize the sustainable yield, what would you suggest in terms of population size and yield? Use the logistic growth equation to solve the problem

Answer 1

I would most likely have a bag limit on fish and certain seasons that some fish are allowed to be taken out and fished for.

Answer 2

The population size would be
`p' = 5000`

The yield would be
`MaxYield = 2082 \ fishes \ per \ year`

Step-by-step explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

`(dN)/(dt) = (2000)/(1 \ year )`

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

`r = ([(dN)/(dt)] )/(N[1-(N)/(K) ])`

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

`r = ([(2000)/(year)] )/(4000[1-(4000)/(10,000) ]) =0.833`

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

`Max Yield = (rK)/(4) = ((10,000)(0.833))/(4) = 2082 \ fishes \ per \ year`

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by
`p'` would be

`p' = (K)/(2) = (10,000)/(2) = 5000\ fishes`