Modeling population growth:
Being Po the initial population, K the carrying capacity (or maximum capacity), and r the growth rate, The model for the population at a time t is:
[tex]P(t)=\frac{K}{1+\left(\frac{K-P_o}{P_o}\right)e^{-rt}}[/tex]
For the problem, we are given Po = 300, r = 40% = 0.4, K = 3600. Substituting, we get the model:
[tex]P(t)=\frac{3600}{1+\left(\frac{3600-300}{300}\right)e^{-0.4t}}[/tex]
Operating:
[tex]P(t)=\frac{3600}{1+11e^{-0.4t}}[/tex]
For t = 4 years:
[tex]P(4)=\frac{3600}{1+11e^{-0.4\times4}}[/tex]
Calculating:
P(4) = 1118
Answer: 1118 trout
Note: The result above was rounded to the nearest whole number
