Problem 75 - A Graphical Approach to College Algebra

Question

Solve each problem. Suppose the average number of vehicles arriving at the main gate of an amusement park is equal to 10 per minute, while the average number of vehicles being admitted through the gate per minute is equal to \(x .\) Then the average waiting time in minutes for each vehicle at the gate is given by $$f(x)=\frac{x-5}{x^{2}-10 x}$$ where \(x>10 .\) (Source: Mannering, \(\mathrm{F}\). and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, 2d. ed., John Wiley and Sons.) (a) Estimate the admittance rate \(x\) that results in an average wait of 15 seconds. (b) If one attendant can serve 5 vehicles per minute, how many attendants are needed to keep the average wait to 15 seconds or less?

Step-by-Step Solution

Answer

(a) The admittance rate is approximately 12.385 vehicles per minute. (b) 3 attendants are needed.

Step 1: Understand the Problem Requirements

The problem provides a function for the average waiting time for vehicles at an amusement park gate. We need to estimate the admittance rate \(x\) that reduces this wait time to 15 seconds (or 0.25 minutes), given by the function \(f(x) = \frac{x-5}{x^2-10x}\) where \(x > 10\). The second part asks us to determine the number of attendants needed based on a serving capacity of 5 vehicles per minute.

Step 2: Convert Time from Seconds to Minutes

Since the given waiting time is 15 seconds, we convert it into minutes because the function \(f(x)\) is in terms of minutes. So, 15 seconds is equivalent to \(0.25\) minutes.

Step 3: Set up the Equation

Set \(f(x) = 0.25\) using the formula provided: \[ \frac{x-5}{x^2-10x} = 0.25 \] We need to solve this equation for \(x\), keeping in mind that \(x > 10\).

Step 4: Solve the Equation

First, cross-multiply to eliminate the fraction: \[ x - 5 = 0.25(x^2 - 10x) \] Next, distribute the 0.25: \[ x - 5 = 0.25x^2 - 2.5x \] Rearrange the terms to form a quadratic equation: \[ 0.25x^2 - 3.5x + 5 = 0 \] Multiply through by 4 to simplify coefficients: \[ x^2 - 14x + 20 = 0 \] Use the quadratic formula where \(a=1\), \(b=-14\), \(c=20\): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{14 \pm \sqrt{196 - 80}}{2} \] \[ x = \frac{14 \pm \sqrt{116}}{2} \] \[ x = \frac{14 \pm 10.77}{2} \] Calculate both possible values for \(x\). Only consider \(x > 10\).

Step 5: Calculate Possible Values for \(x\)

Calculate the two potential solutions: - \(x = \frac{14 + 10.77}{2} = 12.385\)- \(x = \frac{14 - 10.77}{2} = 1.615\)Since \(x > 10\), the only valid solution is \(x = 12.385\).

Step 6: Determine the Number of Attendants

Given each attendant can serve 5 vehicles per minute, we need to adjust the admittance rate \(x = 12.385\) to meet this requirement. Calculate the number of attendants: \[ \frac{12.385}{5} = 2.477 \] Since the number of attendants must be whole, round up to the nearest whole number, resulting in 3 attendants.

Key Concepts

understanding quadratic equations

Quadratic equations are one of the fundamental aspects of algebra. They are equations that can be rearranged into the standard form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants. A quadratic equation features the variable \( x \) raised to the power of two, which is the highest degree in the equation.
These equations typically have two solutions and are represented graphically as parabolas. Solving quadratic equations can be approached through various methods:

• Factoring, when the equation neatly separates into products of binomials.
• Completing the square, a method to convert the equation into a perfect square trinomial.
• Using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the precise values of \( x \) that satisfy the equation by providing two potential solutions. In situations where the quadratic cannot be easily factored, the quadratic formula becomes particularly useful.

Applying these techniques allows us to solve complex problems involving quadratic equations, such as determining optimal vehicle admission rates at an amusement park gate.

average waiting time

Average waiting time is a central concept in any scenario involving queues, such as at amusement parks. It represents the mean time a vehicle or person waits before being served. In the given problem, the average waiting time is defined with a function:

• \(f(x) = \frac{x-5}{x^2-10x}\)

where \( x \) signifies the rate of vehicles admitted each minute.
To find our target waiting time, we converted 15 seconds to 0.25 minutes for consistency, since the function's outputs are in minutes. The approach involves setting \( f(x) = 0.25 \) and solving for \( x \) to determine the necessary vehicle entry rate that ensures the wait time does not exceed this limit.
This concept illustrates essential ideas in operations research and queue management, highlighting how controlling entry rates can minimize delays and enhance efficiency.

vehicle admittance rate

The vehicle admittance rate is the number of vehicles allowed through the gate per unit of time. It plays a crucial role in calculating average waiting times and optimizing service.
In this exercise, the goal was to find an admittance rate \( x \) that results in a wait time of less than or equal to 15 seconds. We calculated:

• The ideal vehicle entry rate \( x \) to be approximately 12.385 vehicles per minute, based solely on solving the quadratic equation derived from the waiting time formula.

Moreover, knowing each attendant can process 5 vehicles a minute helps calculate how many are needed to maintain efficiency:

• Using simple division \( \frac{12.385}{5} \), we found that 2.477 attendants are required, whereby rounding up gives us 3 attendants to ensure the wait time doesn't exceed the target.

Understanding and managing the admittance rate are vital for efficient operations, ensuring customer satisfaction by reducing delays, and properly utilizing available resources.