Consider a child waiting at a street corner for a gap in traffic that is large enough so that he can safely cross the street. A mathematical model for traffic shows that if the expected waiting time for the child is to be at most 1​ minute, then the maximum traffic​ flow, in cars per​ hour, is given by

f left parenthesis x right parenthesis equals StartFraction 28 comma 548 left parenthesis 2.364 minus log x right parenthesis Over x EndFractionf(x)=28,548(2.364−log x)x​,

where x is the width of the street in feet. Complete parts **a.** and **b.** below.

Question content area bottom

Part 1

**a.** Find the maximum traffic flow and the rate of change of the maximum traffic flow with respect to street width for the street width of

3434

feet.

The maximum traffic flow is about

enter your response here

vehicles per hour.

​(Do not round until the final answer. Then round to the nearest integer as​ needed.)

To solve this problem, we need to evaluate the given function for the maximum traffic flow and its rate of change with respect to the street width when \( x = 34 \) feet. The function is defined as:

\[
f(x) = \frac{28,548 \cdot (2.364 - \log x)}{x}
\]

where \( x \) is the street width in feet, \( \log x \) is the natural logarithm (base \( e \)), and \( f(x) \) represents the maximum traffic flow in vehicles per hour. Here's how to proceed:

### 1. Finding the Maximum Traffic Flow:
We substitute \( x = 34 \) into the function:

\[
f(34) = \frac{28,548 \cdot (2.364 - \log 34)}{34}
\]

Let's calculate this value step-by-step.

### 2. Finding the Rate of Change with Respect to \( x \):
The rate of change of \( f(x) \) with respect to \( x \) is found by taking the derivative of \( f(x) \), denoted as \( f'(x) \). Using the quotient rule for differentiation, we can compute \( f'(x) \):

\[
f'(x) = \frac{d}{dx} \left( \frac{28,548 \cdot (2.364 - \log x)}{x} \right)
\]

After finding \( f'(x) \), we evaluate it at \( x = 34 \).

Let's perform these calculations.1. **Maximum Traffic Flow:**
   The maximum traffic flow for a street width of 34 feet is approximately **-976** vehicles per hour (rounded to the nearest integer).

2. **Rate of Change:**
   The rate of change of the maximum traffic flow with respect to street width at 34 feet is approximately **4.01** vehicles per hour per foot.

Let me know if you'd like further details or have any questions.

### Related Questions:
1. How would the maximum traffic flow change if the street width was 50 feet?
2. What would be the rate of change for a street width of 40 feet?
3. Can you find the critical points for the function \( f(x) \)?
4. What does a negative maximum traffic flow signify in the context of this problem?
5. How does the maximum traffic flow behave as \( x \) becomes very large?

### Tip:
To analyze how a function behaves as its input increases or decreases significantly, consider its limit properties to get a general understanding of long-term trends.

Consider a child waiting at a street corner for a gap in traffic that is large enough so that he can safely cross the street. A mathematical model for traffic shows that if the expected waiting time for the child is to be at most 1 minute, then the maximum traffic flow, in cars per hour, is given by f(x) = 28,548(2.364 - log x) / x, where x is the width of the street in feet. Find the maximum traffic flow and the rate of change of the maximum traffic flow with respect to street width for the street width of 34 feet.

This problem involves using a traffic flow model to calculate the maximum traffic flow and its rate of change with respect to street width. It uses calculus concepts such as logarithmic functions and the quotient rule to solve the problem for a street width of 34 feet. Suitable for high school students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation