AdvancedMathematicsHigh School (11–12)Functions and CalculusNumerical

A fisheries officer stocks a lake with 400 fish…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A fisheries officer stocks a lake with $400$ fish. The lake can support at most $5000$ fish, and she models the number $P$ of fish at time $t$ years by the logistic equation $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{5000}\right)$ . After $3$ years the count is $800$ fish. She writes the solution as $P = \dfrac{5000}{1 + Ae^{-kt}}$ , where $A$ comes from the starting number. From the reading at $3$ years she finds $k$ . She then wants two more numbers for the management plan: the time at which the population reaches $2500$ fish, which is half the capacity, and the largest rate of growth of the population over its whole history, in fish per year. Find $k$ in per year, the time in years to reach $2500$ fish, and the largest growth rate in fish per year.

Need a reminder? The key ideaShow

A rate proportional to the amount gives $\dfrac{dN}{dt} = kN$ and $N = N_0 e^{kt}$ . The doubling time or the half-life is $\dfrac{\ln 2}{|k|}$ . The logistic equation $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{M}\right)$ has the solution $P = \dfrac{M}{1 + Ae^{-kt}}$ with $A = \dfrac{M - P_0}{P_0}$ .

\displaystyle N = N_0 e^{kt}.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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