Questions: Suppose 200 trout are seeded into a lake. Absent constraint, their population will grow by 50% a year. If the lake can sustain a maximum of 2300 trout, use a logistic growth model to estimate the number of trout after 2 years. □ trout

Solution Steps

To solve this problem, we need to use the logistic growth model, which is given by the formula:

\[ P(t) = \frac{K}{1 + \left(\frac{K - P_0}{P_0}\right) e^{-rt}} \]

where:

• \( P(t) \) is the population at time \( t \),
• \( K \) is the carrying capacity (maximum population the environment can sustain),
• \( P_0 \) is the initial population,
• \( r \) is the growth rate,
• \( t \) is the time.

Given:

• \( P_0 = 200 \) (initial population),
• \( K = 2300 \) (carrying capacity),
• \( r = 0.5 \) (growth rate per year),
• \( t = 2 \) (time in years).

We will plug these values into the logistic growth formula to find the population after 2 years.

We start with the following parameters for the logistic growth model:

• Initial population \( P_0 = 200 \)
• Carrying capacity \( K = 2300 \)
• Growth rate \( r = 0.5 \)
• Time \( t = 2 \)

The logistic growth model is given by the formula:

\[ P(t) = \frac{K}{1 + \left(\frac{K - P_0}{P_0}\right) e^{-rt}} \]

Substituting the known values into the formula:

\[ P(2) = \frac{2300}{1 + \left(\frac{2300 - 200}{200}\right) e^{-0.5 \cdot 2}} \]

Calculating the expression step-by-step:

1. Calculate \( K - P_0 = 2300 - 200 = 2100 \).
2. Calculate \( \frac{K - P_0}{P_0} = \frac{2100}{200} = 10.5 \).
3. Calculate \( e^{-0.5 \cdot 2} = e^{-1} \approx 0.3679 \).
4. Now substitute these values back into the formula:

\[ P(2) = \frac{2300}{1 + 10.5 \cdot 0.3679} \approx \frac{2300}{1 + 3.866} \approx \frac{2300}{4.866} \approx 472.9849 \]

Final Answer