- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
Logistic growth and phase lines practice question 5
The number of fish in a lake, years after it is stocked, satisfies the equation below. At what population is it growing fastest, and what is that fastest rate of growth?
Give the fastest rate of growth, in fish per year.
The growth rate is an ordinary function of . Where is that function largest, and what is its value there?
The growth rate is a function of the population: . To find where it is largest, set , not .
. Solve for , then substitute that population back into .
A whole number.
Theory UBC Math 100 §13.2Key results
Worked solution
Strategy
The rate of growth depends only on P, so maximise it as an ordinary function of P. For logistic growth the maximum is always at half the carrying capacity.
Working
Treat the growth rate as a function of P and differentiate.
A maximum, at half the carrying capacity K = 0.5/0.0001 = 5000.
Substitute into the whole expression to get the rate itself.
Answer
The answer is growth is fastest at 2500 fish, where the population grows by 625 fish per year.
Check
Nearby populations grow more slowly: at 2400 and at 2600 fish the rate is 624 fish per year, just under 625.
Where this goes wrong
Giving 2500, the population at which growth is fastest, as the answer. The question asks for the rate there: G(2500) = 625 fish per year.
Leans on
Finding a maximum with the derivative, from Optimization
Saved in this browser only.