In a private lake, 200 fish are stocked and the population grows according to the logistic function: P(t) = 2000 / (1 + 9e^(-t/3)) where t is in months.
At what time is the population growing the fastest? How many fish are there at this time?
2007-01-07 16:00:23 · 2 answers · asked by Taryn 2 in Science & Mathematics ➔ Mathematics
So, we want to take the first derivative to get the rate of change, and then evaluate where it is the largest.
I'm sorry, it's been twenty years since I did a differentiation with that, and I'm not sure if I need to use the chain rule or not. If you can take the derivative, you then can get rate of change. Plot the rate of change over a few months - use a spreadsheet, perhaps, and you should be able to determine a maximum.
2007-01-07 16:15:14 · answer #1 · answered by John T 6 · 0⤊ 1⤋
The derivative is the rate of change.
P(t) = 2000 / (1 + 9e^(-t/3))
dp/dt = 6000e^(-t/3)/ {(1 + 9e^(-t/3)}²
The rate of change is fastest at t = 0. At that time there are 200 fish.
2007-01-08 00:25:24 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋