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Let's analyze and solve each part of this question.

The function for the percentage of infected population \( p(t) \) over time \( t \) days is given by:
\[
p(t) = 13t e^{-t/7}
\]

### Part (a): Finding the Maximum Point
To find when \( p(t) \) is at its maximum, we need to calculate the derivative \( p'(t) \) and set it equal to zero.

1. **Differentiate \( p(t) \):**
   Using the product rule for differentiation:
   \[
   p(t) = 13t \cdot e^{-t/7}
   \]
   Let \( u = 13t \) and \( v = e^{-t/7} \).

   Then:
   \[
   u' = 13 \quad \text{and} \quad v' = -\frac{1}{7} e^{-t/7}
   \]

   Applying the product rule \( p'(t) = u'v + uv' \):
   \[
   p'(t) = 13 \cdot e^{-t/7} + 13t \cdot \left(-\frac{1}{7}\right) e^{-t/7}
   \]
   \[
   = 13 e^{-t/7} - \frac{13t}{7} e^{-t/7}
   \]
   \[
   = \frac{13 e^{-t/7}}{7} (7 - t)
   \]

2. **Set \( p'(t) = 0 \):**
   \[
   \frac{13 e^{-t/7}}{7} (7 - t) = 0
   \]
   Since \( e^{-t/7} \neq 0 \), we only need to solve:
   \[
   7 - t = 0
   \]
   \[
   t = 7
   \]

Thus, the maximum percentage of infected people occurs at \( t = 7 \) days.

### Part (b): Calculating the Maximum Percentage
To find the maximum percentage, substitute \( t = 7 \) into \( p(t) \):

\[
p(7) = 13 \cdot 7 \cdot e^{-7/7}
\]
\[
= 13 \cdot 7 \cdot e^{-1}
\]
\[
= 91 \cdot \frac{1}{e}
\]
Using \( e \approx 2.718 \):
\[
p(7) \approx 91 \cdot \frac{1}{2.718} \approx 91 \cdot 0.3679 \approx 33.48
\]

So, the maximum percentage of the population infected is approximately **33.48%**.

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Would you like further details on this solution or help with similar problems? 

### Additional Related Questions
1. How do you determine the critical points of a function like \( p(t) \) in other contexts?
2. What is the significance of the value of \( e \) in exponential decay functions?
3. Can you explain how to interpret and use the derivative of a function in practical applications?
4. How would this function change if the infection spread was faster or slower?
5. Why does the infection rate eventually decrease even if the population is susceptible?

### Tip
To maximize or minimize functions, always consider both the derivative and the endpoints of the interval in practical problems.

A disease has hit a city. The percentage of the population infected t days after the disease arrives is approximated by p(t) = 13t * e^(-t/7) for 0 ≤ t ≤ 49.
(a) After how many days is the percentage of infected people a maximum?
(b) What is the maximum percent of the population infected?

Learn how to determine the maximum percentage of a population infected by a disease using calculus and exponential decay. This problem covers differentiation, optimization techniques, and applying exponential functions in a practical context. Suitable for high school or introductory college-level calculus students.

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