Problem 36 - Applied Calculus

Question

During a flu outbreak in a school of 763 children, the number of infected children, \(I\), was expressed in terms of the number of susceptible (but still healthy) children, \(S\), by the expression $$I=192 \ln \left(\frac{S}{762}\right)-S+763 .$$ What is the maximum possible number of infected children?

Step-by-Step Solution

Answer

The maximum number of infected children is approximately 317.

Step 1: Identify the Objective

We need to find the maximum value of the function \( I(S) = 192 \ln \left( \frac{S}{762} \right) - S + 763 \), which represents the number of infected children expressed as a function of the number of susceptible children \( S \).

Step 2: Determine the Function's Domain

Since \( S \) denotes the number of susceptible children, the domain of \( S \) is from 0 to 763, which means 0 ≤ \( S \) ≤ 763.

Step 3: Differentiate I(S) with Respect to S

Find \( \frac{dI}{dS} \) to locate critical points. Differentiate to get: \[ \frac{dI}{dS} = \frac{192}{S} - 1. \]

Step 4: Solve for Critical Points

Set the derivative \( \frac{dI}{dS} = 0 \) to find critical points. This gives: \[ \frac{192}{S} - 1 = 0 \] which simplifies to \( S = 192 \).

Step 5: Verify Endpoint Values

Evaluate \( I(S) \) at the endpoints of the domain (\( S = 0 \) and \( S = 763 \)) and at the critical point \( S = 192 \).Calculate: - \( I(0) = 192 \ln(0) - 0 + 763 = 763 \) (undefined for \( \ln(0) \)).- \( I(763) = 192 \ln \left( \frac{763}{762} \right) - 763 + 763 = 192 \ln \left( 1 + \frac{1}{762} \right) \approx 0 \).- \( I(192) = 192 \ln \left( \frac{192}{762} \right) - 192 + 763 \approx 316.75 \).

Step 6: Determine Maximum Value

Compare the values from the evaluations in Step 5. Since the derivative suggests undefined behavior at \( S = 0 \), and \( I(763) \approx 0 \), the maximum value is at \( S = 192 \), giving \( I(192) \approx 317 \).

Key Concepts

Differentiation

Differentiation is the mathematical process used to determine how a function changes at any point. In maximum value problems, like predicting the spread of the flu, we differentiate to find the rate at which the number of infected children changes. The function given, \( I(S) = 192 \ln \left( \frac{S}{762} \right) - S + 763 \), models the count of infected children relative to susceptible ones. By differentiating this function with respect to \( S \), we find the expression \( \frac{dI}{dS} = \frac{192}{S} - 1 \).
This derivative helps us identify where the slope of the function equals zero - these are our critical points where the function may have a maximum or minimum value.
In simple terms, differentiation tells us where the function might peak or dip, crucial for maximizing efforts in flu containment.

Critical Points

Critical points are where the derivative of a function is zero or undefined, indicating potential peaks (maximums) or bottoms (minimums). In our flu outbreak model, setting the derivative \( \frac{dI}{dS} = \frac{192}{S} - 1 \) to zero helps pinpoint critical points. Solving for \( S \), we find \( S = 192 \), a critical point which hints at a maximum number of infected children.
These points are particularly important because they represent changes in the trend of infected children - helping identify the most severe point of infection spread. By testing these points and endpoint values, we determine the peak infection level is at \( S = 192 \).
This underscores the significance of monitoring critical points during health crises to allocate resources efficiently.

Function Domain

The function domain defines all possible input values a function can accept. Here, \( S \), representing susceptible children, can range from 0 to 763, assuming all children are either susceptible or infected. This creates a natural boundary, forming the domain 0 \( \leq S \leq 763 \).
The domain ensures that our calculations remain realistic and confined to plausible scenarios. For example, evaluating the endpoints in our function ensures no logical inconsistencies arise, like a negative number of children. However, as observed during evaluation, the function shows undefined behavior at \( S = 0 \) due to \( \ln(0) \), emphasizing that domain boundaries need careful handling.
Thus, understanding the domain helps in accurately modeling situations and prevents mathematical errors during evaluations.