Question

Suppose we stock a pond with 100 fish and note that the population doubles in the first year (with no harvesting), but after a number of years, the population stabilizes at what we think must be the carrying capacity of the pond: 2,000 fish. Growth seems to have followed a logistic curve. a. What population size should be maintained to achieve maximum yield, and what would be the maximum sustainable fish yield? b. If the population is maintained at 1,500 fish, what would be the sustainable yield?

Answer 1

Answer:

Explanation:

From the given information,

The first process is to determine the initial growth rate by using the formula:

$R_o = \dfrac{In(2)}{t_d}$

where;

The initial growth rate(constant) = $R_o$ ; &

the doubling time = $t_d$ = 1 year

$R_o = \dfrac{In(2)}{1 \ year}$

$R_o$ = 0.693 / year

Now, we move up to the net stage to find the growth constant by using the formula:

$r = \dfrac{R_o}{(1 - \dfrac{N_o}{K})}$

where;

$N_o$ = population of fish in the pond = 100

$R_o$  = The initial growth rate(constant)  = 0.693 /year

K = carrying capacity = 2000

Then;

$r = \dfrac{0.693 \ /year }{(1 - \dfrac{100}{2000})}$

$r = \dfrac{0.693 \ /year }{(1 - 0.05)}$

$r = \dfrac{0.693 \ /year }{(0.95)}$

r = 0.730 / year

a)

Now, the maximum yield can be evaluated by using the expression:

$= \dfrac{rk}{4}$

$= \dfrac{0.730 \times 2000}{4}$

= 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.

$N' = \dfrac{1}{2} \times 2000$

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 = $r\times N (1 - \dfrac{N}{K})$

The Sustainable yield = $(0.730/ year ) (1500) (1 - \dfrac{1500}{2000})$

The Sustainable yield = 1095 × 0.25

The Sustainable yield = 273.75 fish per year

The Sustainable yield ≅ 274 fish per year