The logistic growth model is commonly expressed in the form:

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

where:

• $P(t)$ is the population at time $t$,
• $P_0$ is the initial population,
• $r$ is the growth rate constant,
• $K$ is the carrying capacity (maximum population).

Given:

• $P_0 = 3600$,
• $P(22) \approx 5565$,
• $r = 0.056$.

Step 1: Substitute Known Values into the Logistic Model

1. Identify the model parameters $P_0$, $r$, and $P(22)$, with the goal to solve for $K$ (the carrying capacity).

   Since $P(22) = 5565$, we plug in $t = 22$ to solve for $K$:

   $P(22) = \frac{K}{1 + \left(\frac{K - 3600}{3600}\right)e^{-0.056 \cdot 22}}$

2. Simplify the exponent:

   • Calculate $e^{-0.056 \cdot 22}$.

   $e^{-0.056 \cdot 22} \approx e^{-1.232} \approx 0.292$

3. Plug in the values and solve for $K$:

   $5565 = \frac{K}{1 + \left(\frac{K - 3600}{3600}\right) \cdot 0.292}$

4. Isolate $K$: After rearranging terms and solving for $K$, we should arrive at an approximate value of $K$.

Would you like a detailed breakdown of each calculation, or do you have any other questions on logistic growth?

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Here are 5 questions related to this topic:

1. How is logistic growth different from exponential growth?
2. Why is the carrying capacity $K$ important in logistic growth?
3. Can the growth rate constant $r$ change over time in real-world applications?
4. How does one interpret the value $P(t)$ in the context of population dynamics?
5. What are common applications of the logistic growth model outside of population studies?

Tip: Logistic growth is typically more realistic than exponential growth for modeling populations because it accounts for environmental limitations that cap growth.