Respuesta :
Answer:
The population size would be [tex]p' = 5000[/tex]
The yield would be [tex]MaxYield = 2082 \ fishes \ per \ year[/tex]
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
[tex]\frac{dN}{dt} = \frac{2000}{1 \ year }[/tex]
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
[tex]r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}[/tex]
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
[tex]r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833[/tex]
The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as
[tex]Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year[/tex]
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 [tex]p'[/tex] would be
[tex]p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes[/tex]
Answer:
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.
