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Let's work through the problem step by step.

The logistic growth model for the fish population is given by:

\[
P(t) = \frac{d}{1 + ke^{-ct}}
\]

where:
- \( d = 1200 \) (carrying capacity, maximum population),
- \( k = 11 \),
- \( c = 0.2 \),
- \( t \) is the time in years.

### (a) How many fish were originally put in the pond?

To find the initial population, we evaluate \( P(0) \) (at \( t = 0 \)):

\[
P(0) = \frac{1200}{1 + 11e^{-0.2 \times 0}} = \frac{1200}{1 + 11} = \frac{1200}{12} = 100
\]

So, **100 fish** were originally put in the pond.

### (b) Find the population after 10, 20, and 30 years.

For this, we will plug in the values \( t = 10 \), \( t = 20 \), and \( t = 30 \) into the logistic growth equation.

1. **At \( t = 10 \):**

\[
P(10) = \frac{1200}{1 + 11e^{-0.2 \times 10}} = \frac{1200}{1 + 11e^{-2}} \approx \frac{1200}{1 + 11 \times 0.1353} \approx \frac{1200}{1 + 1.4883} \approx \frac{1200}{2.4883} \approx 482
\]

So, the population after 10 years is approximately **482 fish**.

2. **At \( t = 20 \):**

\[
P(20) = \frac{1200}{1 + 11e^{-0.2 \times 20}} = \frac{1200}{1 + 11e^{-4}} \approx \frac{1200}{1 + 11 \times 0.0183} \approx \frac{1200}{1 + 0.2013} \approx \frac{1200}{1.2013} \approx 999
\]

So, the population after 20 years is approximately **999 fish**.

3. **At \( t = 30 \):**

\[
P(30) = \frac{1200}{1 + 11e^{-0.2 \times 30}} = \frac{1200}{1 + 11e^{-6}} \approx \frac{1200}{1 + 11 \times 0.0025} \approx \frac{1200}{1 + 0.0275} \approx \frac{1200}{1.0275} \approx 1168
\]

So, the population after 30 years is approximately **1168 fish**.

### (c) Evaluate \( P(t) \) for large values of \( t \) (as \( t \to \infty \)).

As \( t \to \infty \), the exponential term \( e^{-ct} \) approaches 0, so the equation simplifies to:

\[
P(t) = \frac{1200}{1 + 0} = 1200
\]

Thus, for large \( t \), the population approaches **1200 fish**.

### Does the graph confirm the calculations?

Yes, the graph shows a population that starts small, grows quickly at first, and then levels off around 1200, confirming the calculations.

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Would you like more details on any part of this solution? 

Here are five related questions to explore further:
1. What happens if the value of \( c \) (growth rate) changes?
2. How does the initial population affect the overall growth curve?
3. Can you calculate the time it takes for the population to reach half of its carrying capacity?
4. What happens to the population if the carrying capacity \( d \) is reduced?
5. How would a different value of \( k \) (initial condition factor) influence the growth?

**Tip:** For logistic growth, the steepest growth happens when the population is at half of the carrying capacity!

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions, the population follows a logistic growth model. For a certain fish population in a small pond: d = 1200, k = 11, c = 0.2, and t is measured in years. The fish were introduced into the pond at time t = 0. (a) How many fish were originally put in the pond? (b) Find the population after 10, 20, and 30 years. (c) Evaluate P(t) for large values of t. What value does the population approach as t → ∞?

Learn how to solve a logistic growth problem for a fish population in a pond, where d = 1200, k = 11, and c = 0.2. Step-by-step solutions include calculating the initial population, estimating population sizes after 10, 20, and 30 years, and understanding the behavior of the population as time approaches infinity. Suitable for Grades 11-12 and college students.

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