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%.

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.