This problem involves the logistic growth model. The logistic growth equation is:

$P_{n+1} = P_n + r P_n \left(1 - \frac{P_n}{K}\right)$

Where:

• $P_n$ is the population at year $n$,
• $r$ is the growth rate ($10\%$ or $0.1$),
• $K$ is the carrying capacity ($500$ in this case),
• $P_{n+1}$ is the population after one year.

Step 1: Calculate the population after one year ($P_1$):

Using the initial population ($P_0 = 40$): $P_1 = P_0 + r P_0 \left(1 - \frac{P_0}{K}\right)$ Substitute the values: $P_1 = 40 + 0.1 \cdot 40 \left(1 - \frac{40}{500}\right)$ $P_1 = 40 + 4 \left(1 - 0.08\right)$ $P_1 = 40 + 4 \cdot 0.92 = 40 + 3.68 = 43.68$

Step 2: Calculate the population after two years ($P_2$):

Using $P_1 = 43.68$: $P_2 = P_1 + r P_1 \left(1 - \frac{P_1}{K}\right)$ Substitute the values: $P_2 = 43.68 + 0.1 \cdot 43.68 \left(1 - \frac{43.68}{500}\right)$ $P_2 = 43.68 + 4.368 \left(1 - 0.08736\right)$ $P_2 = 43.68 + 4.368 \cdot 0.91264 = 43.68 + 3.986 = 47.67$

Final Answers:

• $P_1 = 43.68$
• $P_2 = 47.67$

Would you like a detailed breakdown of each step or to explore similar problems?

Related Questions:

1. How does the carrying capacity affect long-term population growth in logistic models?
2. What happens if the growth rate is increased to $20\%$?
3. How does the logistic model differ from exponential growth?
4. Can we calculate when the population will reach $250$ deer?
5. How can human activities impact carrying capacity in this context?

Tip:

For larger populations or time intervals, numerical methods or software tools can simplify solving iterative logistic equations.