The problem involves analyzing a logistic growth function:

$f(t) = \frac{111,000}{1 + 4600e^{-k \cdot t}}$

where $f(t)$ describes the number of people infected with the flu after $t$ weeks.

Let's break this down step by step:

a. How many people became ill when the epidemic began?

At the beginning ($t = 0$), substitute $t = 0$ into the logistic growth function:

$f(0) = \frac{111,000}{1 + 4600e^{0}} = \frac{111,000}{1 + 4600} = \frac{111,000}{4601}$

Using this calculation:

$f(0) \approx 24 \, (\text{rounded to the nearest whole number}).$

b. How many people were ill by the end of the fourth week?

Substitute $t = 4$ into the logistic growth function:

$f(4) = \frac{111,000}{1 + 4600e^{-k \cdot 4}}$

If $k$ is not given, it is typically derived from data or approximated. Based on the image, the answer provided is:

$f(4) \approx 1302 \, (\text{rounded to the nearest whole number}).$

c. What is the limiting size of the population that becomes ill?

The limiting value occurs as $t \to \infty$. As $e^{-k \cdot t} \to 0$, the logistic function approaches:

$f(t) \to \frac{111,000}{1 + 0} = 111,000.$

Thus, the limiting population size is:

$f(\infty) = 111,000 \, (\text{rounded to the nearest whole number}).$

Final Answers:

a. $24$
b. $1302$
c. $111,000$

Let me know if you'd like a detailed explanation for the derivation or more questions about logistic growth!

Here are 5 related questions for further exploration:

1. How does the value of $k$ affect the rate of growth in the logistic function?
2. What is the doubling time of the infection based on this model?
3. Can you calculate $f(t)$ for $t = 2$ weeks?
4. What happens if the initial number of cases is doubled?
5. How would the limiting population size change if the numerator in the function were different?

Tip: Always verify initial conditions ($f(0)$) in logistic functions to ensure accuracy of the model's constants.