Expert: Sherman D. - 12/30/2004

Question
My problem is my understanding of the question posed since it seems vague.

Assume that a fish population grows according to the equation: p = m/(1+ ae^-kt)

p is the initial population, m is the carrying capacity of the pond (maximum number of fish it can sustain), k = 0.9, and t is the number of years.

a) given that you have 1000 fish initially, what is the value of a? (I obtained 24.40878378 by substituting 0 for t)

b) How many years will it take for your pond to reach the carrying capacity?
(Does this mean that p is now equal to m?)


Thank you so much for your help, and it would be wonderful if I could receive a quick response because very many friends are counting on it.

Answer
p = m/(1 + ae^(-kt))

a.)
1000 = m/(1 + ae^(-.09t))
1 + ae^(-.09t) = m/1000
ae^(-.09t) = (m/1000) - 1
a = ((m/1000) - 1)/(e^(-.09t))

b.)
If it is telling you to use the values you get in a.), then
1000 = m/(1 + ae^(-.09t))
m = 1000(1 + ae^(-.09t))
now you must plug in what value you got for "a"

Sorry, but thats all i could get. Since i don't have the maximum population or "a", i can't really do much.

Also in b.) How many fish you start out with is usally less than the maximum capacity. For example, if you start out with 1000, you could end up with 2000 in one year. The only time maximum capacity becomes less than the initial population is when you have all the same sex fish or polluted water. The only time they maximum capacity and intial population are the same, is when the pond is controlled or regulated to stay the same amount.