Question 1175054: Assume there is a certain population of fish in a pond whose growth is described by the logistic equation. It is estimated that the carrying capacity for the pond is 1500 fish. Absent constraints, the population would grow by 200% a year.
If the population is given by po=600, then after one breeding season the population of the pond is given by
p1=
After two breeding seasons the population of the pond is given by
p2=
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if the growth is 200% per year, then the growth rate is 2 per year.
the formula for future value is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the rat4e of growth per year.
n is the number of years.
in your problem, the formula becomes f = 600 * (1 + 2) ^ n
p1 is the number at the end of 1 year.
p2 is the number at the end of 2 years.
p1 = 600 * 3^1 = 1800
p2 = 600 * 3^2 = 5400
how does this work?
the population is 600 in p0.
the growth is 200% per year, meaning the growth is 1200 from p0 to p1.
the population is 600 + 1200 = 1800 in p1.
the growth is 200% per year, meaning the growth is 3600 from p1 to p2.
the population is 1800 + 3600 = 5400 in p2.
since the carrying capacity of the pond is 1500 fish, it's pretty clear that there's not a lot of room for growth.
at the given growth rate of 200% per year, the pond is over capacity at the end of 1 year.
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
Assume there is a certain population of fish in a pond whose growth is described
by the logistic equation. It is estimated that the carrying capacity for the pond is 1500 fish.
Absent constraints, the population would grow by 200% a year.
If the initial population is given by po= 600, then estimate the fish population in the pond
after one and two years
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One good Internet source for a beginner to read and to learn about a logistic equation is Libre text
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/08%3A_Introduction_to_Differential_Equations/8.04%3A_The_Logistic_Equation
This my post is to overlay calculations by the other tutor, that are irrelevant.
The general solution for the standard logistic differential equation is so called logistic function
P(t) =  .
Here  is the initial population, P(t) is current population, K is the carrying capacity
and "r" is growth rate. Number "e" = 2.71828 is the base of natural logarithm
(shown here as an approximate value).
In this current problem in the post, we are given
= 600, K = 1500, r = 2.0.
They want you determine P(1) and P(2), the population after 1 and 2 years.
(a) For t = 1 year
P(1) =  = 1246.879478, or 1247 (rounded).
(b) For t = 2 years
P(2) =  = 1459.89162, or 1460 (rounded).
Solved.
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Post-solution notice
In my opinion, the question in the post is posed/worded INCORRECTLY.
It asks about the breeding season, but we are not given information
for a breeding season - we are given information, related to years - so,
the question should be about the times of years, not the breeding seasons.
The life of fish includes not only breeding - it includes also struggle for food,
survival from predators and diseases, natural death, that is, processes with a period of a year.
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