The problem can be solved using the logistic growth model. The logistic growth model is given by the equation:

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

Where:

• $P(t)$ is the population at time $t$,
• $K$ is the carrying capacity (200 deer),
• $P_0$ is the initial population (50 deer),
• $r$ is the growth rate (2).

Step-by-step Solution:

1. Substitute the given values into the equation: $P(t) = \frac{200}{1 + \left( \frac{200 - 50}{50} \right) e^{-2t}}$

2. Simplify the expression: $P(t) = \frac{200}{1 + 3 e^{-2t}}$

This is the population $P(t)$ at any given time $t$.

Let me know if you'd like further details or have any questions.

---

Here are 5 related questions to expand on this problem:

1. What happens to the population as $t \to \infty$?
2. How would a different growth rate affect the solution?
3. Can you calculate $P(t)$ for a specific $t$ (e.g., $t = 5$)?
4. How does the logistic growth model differ from exponential growth?
5. What is the significance of the carrying capacity in this model?

Tip: Logistic growth models are useful when population growth is limited by resources.