The problem involves modeling the spread of an epidemic using a logistic function:

$f(t) = \frac{120,000}{1 + A e^{-kt}},$ where $t$ is the time (in months), $A$ and $k$ are constants to be determined, and $f(t)$ represents the number of infected people at time $t$.

Problem Breakdown:

1. Part (a):

   • Initial conditions are:
     • At $t = 0$, $f(0) = 120$ (120 infected people at the start).
     • At $t = 1$, $f(1) = 1000$ (1000 infected people after 1 month).
   • Using these conditions, calculate $A$ and $k$.

2. Part (b):

   • Total population at risk is $120,000$.
   • Determine $t$ when half the population is infected ($f(t) = 60,000$).

3. Part (c):

   • Sketch the graph of $f(t)$, showing the logistic growth curve.

I will now solve this step by step.

Solution: Part (a)

Step 1: Use $t = 0$, $f(0) = 120$:

Substitute $t = 0$ into the equation: $f(0) = \frac{120,000}{1 + A e^{0}} = \frac{120,000}{1 + A}.$ Equating $f(0)$ to 120: $120 = \frac{120,000}{1 + A}.$ Solve for $A$: $1 + A = \frac{120,000}{120} \quad \Rightarrow \quad 1 + A = 1000 \quad \Rightarrow \quad A = 999.$

Step 2: Use $t = 1$, $f(1) = 1000$:

Substitute $t = 1$ into the equation: $f(1) = \frac{120,000}{1 + 999 e^{-k}}.$ Equating $f(1)$ to 1000: $1000 = \frac{120,000}{1 + 999 e^{-k}}.$ Rearrange to isolate $e^{-k}$: $1 + 999 e^{-k} = \frac{120,000}{1000} \quad \Rightarrow \quad 1 + 999 e^{-k} = 120.$ $999 e^{-k} = 119 \quad \Rightarrow \quad e^{-k} = \frac{119}{999}.$ Take the natural logarithm: $-k = \ln\left(\frac{119}{999}\right) \quad \Rightarrow \quad k = -\ln\left(\frac{119}{999}\right).$ Simplify: $k = \ln\left(\frac{999}{119}\right).$ Numerical approximation: $k \approx \ln(8.395) \approx 2.13.$

Thus, $A = 999$ and $k \approx 2.13$.

Solution: Part (b)

We need to find $t$ when $f(t) = 60,000$: $60,000 = \frac{120,000}{1 + 999 e^{-2.13 t}}.$ Simplify: $1 + 999 e^{-2.13 t} = \frac{120,000}{60,000} \quad \Rightarrow \quad 1 + 999 e^{-2.13 t} = 2.$ $999 e^{-2.13 t} = 1 \quad \Rightarrow \quad e^{-2.13 t} = \frac{1}{999}.$ Take the natural logarithm: $-2.13 t = \ln\left(\frac{1}{999}\right) \quad \Rightarrow \quad t = -\frac{\ln\left(\frac{1}{999}\right)}{2.13}.$ Simplify: $t = \frac{\ln(999)}{2.13}.$ Numerical approximation: $t \approx \frac{\ln(999)}{2.13} \approx \frac{6.91}{2.13} \approx 3.24.$ Thus, half the population is infected after approximately 3.24 months.

Solution: Part (c)

The graph of the logistic function is a sigmoid curve. Key points:

• Starts at $f(0) = 120$.
• Approaches $120,000$ as $t \to \infty$.
• Rapid growth around $t \approx 3.24$, when $f(t) = 60,000$.

I can create a sketch of the graph if you'd like. Let me know if you need a plotted graph.

Follow-Up Questions:

1. How does the value of $k$ influence the speed of infection spread in the model?
2. What happens to the function $f(t)$ as $t \to \infty$?
3. How is the logistic function used in other real-world scenarios beyond epidemiology?
4. What does the parameter $A$ represent in terms of the infection spread?
5. How would the model change if the population size was larger or smaller?

Tip:

In logistic models, the growth rate $k$ determines how quickly the population approaches its carrying capacity. A higher $k$ results in faster growth.