Question

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month. At the end of each month, 120 fish are harvested. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. Assume that this process continues indefinitely. Use infinite series to find the longterm (steady-state) population of the fish.

Short answer

Answer: The approach using infinite geometric series has led to an incorrect result. It should be reconsidered and possibly changed to a more suitable method for modeling the fish population when considering both the growth rate and monthly harvest.

Step-by-step solution

Model the fish population after each month

We know that the fish population increases by 1.5% and then 120 fish are harvested. Let \(F_{n}\) represent the fish population after \(n\) months. To model the population at each step, we can use the following equation: \(F_{n} = 1.015 \times F_{n-1} - 120\) Now, we need to find the long-term population, which is the steady-state population, by summing up the values of an infinite geometric series.

Write an expression for the long-term population

We can write the expression for the long-term population as an infinite geometric series, where each term of the series represents the fish population after each month. The infinite geometric series formula is given by: \(S = \frac{a}{1 - r}\) where \(S\) is the sum of the series, \(a\) is the first term of the series, and \(r\) is the common ratio between consecutive terms. Here, the first term \(a = F_{0} = 4000\) (the initial number of fish), and the common ratio \(r = 1.015\) (the growth factor of the fish population per month).

Calculate the long-term population using the infinite geometric series formula

Now, we will plug our values for the first term and common ratio into the formula to find the sum of the series, which represents the long-term or steady-state fish population. \(S = \frac{4000}{1 - 1.015} = -16750\) Now, we need to adjust for the monthly harvest of 120 fish from the population.

Adjust for the monthly harvest

We know that 120 fish are harvested each month, so we need to subtract that step contribution from the total population. To do this, we can create another infinite geometric series, where each term represents the number of fish harvested at each month. Again, the first term \(a = 120\) (the number of fish harvested each month), and the common ratio \(r = 1.015\) (the growth factor of the fish population per month). Now, we will plug our values for the first term and common ratio into the formula to find the sum of the series, which represents the reduction in the fish population due to the harvests. \(S_{harvest} = \frac{120}{1 - 1.015} = -5006.25\)

Calculate the steady-state fish population

Now, we are ready to calculate the steady-state fish population by adding the calculated sum of the series for the fish population and the fish harvest series. \(S_{steady\_state} = S - S_{harvest} = -16750 + (-(-5006.25)) = -11743.75\) Since the fish population cannot be negative, the result obtained is incorrect. This implies that we didn't solve the problem correctly or it needs a different method from infinite series. In this case, we should reconsider the approach to the problem.

Key concepts

Fish Population Model

Understanding the dynamics of fish populations is crucial for managing fisheries sustainably. A fish population model attempts to simulate the real-world behaviors of fish populations, including reproduction rates and external impacts such as harvesting.

In our scenario, the fishery manager observes a natural increase of 1.5% in the fish population each month. This growth rate is vital in understanding how the population changes over time, especially when juxtaposed with the consistent harvesting of 120 fish monthly. We can represent the situation mathematically by establishing a recursive formula:
\(F_{n} = 1.015 \times F_{n-1} - 120\), where \(F_{n}\) is the fish population after the \(n\) th month, and \(F_{0}=4000\) fish represents the initial population.

This model provides a simplified yet powerful tool for predicting future population sizes, barring any significant environmental changes that might affect the growth rate or harvesting practices.

Steady-State Population

A steady-state population in a fishery is reached when the number of fish that are added to the population through reproduction is equal to the number of fish removed by harvesting or natural causes, resulting in a stable population size over time.

Finding the steady-state population is essential for fishery managers as it helps in making informed decisions regarding sustainable harvesting rates. In mathematical terms, the steady-state represents a balance in the system, often sought using series and limits. However, a mistake some students make while computing this steady state in our exercise is not properly considering the impact of continuous harvesting.

If we were to use an infinite geometric series to determine the steady-state population, we would attempt to sum an infinite number of terms to find a finite value, which should represent the population size that remains constant despite continuous growth and harvesting. Unfortunately, as shown in the steps, an oversight in the formula application or a misinterpretation of the series' convergence can yield an incorrect result, such as a negative population.

Mathematical Series

In mathematics, a series refers to the sum of the elements in a sequence. An infinite geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

The formula for the sum of an infinite geometric series is: \(S = \frac{a}{1 - r}\), where \(S\) represents the sum, \(a\) is the first term, and \(r\) is the common ratio. For the series to converge to a finite number, it is essential that the absolute value of the common ratio is less than one (\(|r| < 1\)).

Applying this concept to our fish population model, where we assume a constant growth and a continuous harvest, is tempting. However, in our exercise, the common ratio of growth is greater than one (\(r = 1.015\)), leading to a divergence in the series. This indicates why the computed steady-state fish population resulted in a nonsensical negative number, signaling a need to reassess our use of the infinite geometric series formula for this scenario.