Respuesta :
Answer:
Explanation:
From the given information,
The first process is to determine the initial growth rate by using the formula:
[tex]R_o = \dfrac{In(2)}{t_d}[/tex]
where;
The initial growth rate(constant) = [tex]R_o[/tex] ; &
the doubling time = [tex]t_d[/tex] = 1 year
[tex]R_o = \dfrac{In(2)}{1 \ year}[/tex]
[tex]R_o[/tex] = 0.693 / year
Now, we move up to the net stage to find the growth constant by using the formula:
[tex]r = \dfrac{R_o}{(1 - \dfrac{N_o}{K})}[/tex]
where;
[tex]N_o[/tex] = population of fish in the pond = 100
[tex]R_o[/tex] = The initial growth rate(constant) = 0.693 /year
K = carrying capacity = 2000
Then;
[tex]r = \dfrac{0.693 \ /year }{(1 - \dfrac{100}{2000})}[/tex]
[tex]r = \dfrac{0.693 \ /year }{(1 - 0.05)}[/tex]
[tex]r = \dfrac{0.693 \ /year }{(0.95)}[/tex]
r = 0.730 / year
a)
Now, the maximum yield can be evaluated by using the expression:
[tex]= \dfrac{rk}{4}[/tex]
[tex]= \dfrac{0.730 \times 2000}{4}[/tex]
= 365 fish per year
Also, the maximum sustainable yield is said to be half of the carrying capacity suppose that the population growth obeys logistic curve;
i.e.
[tex]N' = \dfrac{1}{2} \times 2000[/tex]
N' = 1000 fish
b)
If the population is maintained at 1500 fish;
The sustainable yield can be calculated by using the formula:
The Sustainable yield = [tex]r\times N (1 - \dfrac{N}{K})[/tex]
The Sustainable yield = [tex](0.730/ year ) (1500) (1 - \dfrac{1500}{2000})[/tex]
The Sustainable yield = 1095 × 0.25
The Sustainable yield = 273.75 fish per year
The Sustainable yield ≅ 274 fish per year
Population size should be maintained to achieve maximum yield and Sustainable yield if population is maintained at 1,500 fish are 364.7 , 273.525 fish per year
Sustainable yield:
Given that;
Stock in start = 100 fish
Double in 1 year
Capacity of pond = 2000 fish
Find:
Population size should be maintained to achieve maximum yield
Sustainable yield if population is maintained at 1,500 fish
Computation:
Growth rate = ln₂ / 1
Growth rate = 0.693 per year
With no growth constraint r = [tex]\frac{0.693}{1-\frac{100}{2000} }[/tex]
With no growth constraint r = 0.7294 per year
A. Population size should be maintained to achieve maximum yield = [tex]\frac{0.7294 \times 2000}{4}[/tex]
Population size should be maintained to achieve maximum yield = 364.7 fish per year
B. Sustainable yield if population is maintained at 1,500 fish = [tex][(0.7294))(1500)][1-\frac{1500}{2000} ][/tex]
Sustainable yield if population is maintained at 1,500 fish = 273.525 fish per year
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