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Chapter 2: Problem 19

The farm population has declined dramatically in the years since World War II, and with that decline, rural school districts have been faced with consolidating in order to be economically efficient. One researcher studied data from the early 1960 s on expenditures for high schools ranging from 150 to 2400 in enrollment. \({ }^{34}\) He considered the cost per pupil as a function of the number of pupils enrolled in the high school, and he found the approximate formula $$ C=743-0.402 n+0.00012 n^{2} $$ where \(n\) is the number of pupils enrolled and \(C\) is the cost, in dollars, per pupil. a. Make a graph of \(C\) versus \(n\). b. What enrollment size gives a minimum per-pupil cost? c. If a high school had an enrollment of 1200 , how much in per-pupil cost would be saved by increasing enrollment to the optimal size found in part b?

Short Answer

Expert verified
The optimal enrollment size is 1675 pupils, which reduces the cost by approximately $49.8 per pupil compared to an enrollment of 1200.

Step by step solution

01

Understanding the Formula

The given formula is \( C = 743 - 0.402n + 0.00012n^2 \) where \( n \) is the number of pupils, and \( C \) is the cost per pupil. The formula includes a linear term \(- 0.402n\) and a quadratic term \(0.00012n^2\). This indicates that the relationship between cost and enrollment is quadratic, which can potentially have a minimum point.
02

Plotting the Graph

To make a graph of \( C \) versus \( n \), calculate \( C \) for various values of \( n \) ranging from 150 to 2400 and plot these points. The x-axis will represent the number of students \( n \), and the y-axis will represent the cost per pupil \( C \). A software or a graphing calculator can be used to visualize this quadratic function.
03

Finding the Minimum Cost

To find the enrollment size that gives the minimum per-pupil cost, determine the vertex of the quadratic function \( C = 743 - 0.402n + 0.00012n^2 \). The vertex form for a quadratic \( ax^2 + bx + c \) gives the vertex at \( n = -\frac{b}{2a} \). Here, \( a = 0.00012 \) and \( b = -0.402 \), so \( n = \frac{-(-0.402)}{2 \times 0.00012} = \frac{0.402}{0.00024} \approx 1675 \). Thus, the enrollment size that gives the minimum cost is approximately 1675 students.
04

Calculating Cost for Enrollment of 1200 and 1675

Calculate \( C \) for both \( n = 1200 \) and \( n = 1675 \). For \( n = 1200 \), \( C = 743 - 0.402\times1200 + 0.00012\times1200^2 \). Compute similarly for \( n = 1675 \). Using these calculations, determine \( C_{1200} \approx 455.6 \) and \( C_{1675} \approx 405.8 \).
05

Calculating the Savings

The savings per pupil by increasing enrollment from 1200 to 1675 is \( C_{1200} - C_{1675} = 455.6 - 405.8 = 49.8 \) dollars. This is the amount saved in per-pupil cost by increasing to the optimal size.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cost Analysis
Cost analysis in the context of school expenditure is about evaluating how costs change with varying enrollment sizes. In the given exercise, the cost per student is expressed as a quadratic function: \[ C = 743 - 0.402n + 0.00012n^2 \] Here, \( C \) represents the cost per pupil, and \( n \) is the number of enrolled students. This formula incorporates two components: a linear term \(-0.402n\) which generally predicts a decrease in costs as more students enroll, and a quadratic term \(0.00012n^2\) which reveals that beyond a certain point, costs start increasing again as enrollment increases. The analysis helps school administrators comprehend how reallocating resources in response to enrollment changes can optimize spending. By understanding these costs, they can make informed decisions about resource distribution and manage expenses effectively. This makes evaluating the cost function critical for any planning regarding school budgets.
Enrollment Optimization
Optimization in this scenario involves finding the student number that yields the lowest possible cost per pupil. Knowing this optimal size can guide decisions on whether or not to change the school's enrollment capacity. To achieve this, the given quadratic function needs to be analyzed to identify the point at which costs are minimized. This point, in a quadratic function, is known as the vertex. By substituting the values given into the vertex formula \( n = \frac{-b}{2a} \), where \( a = 0.00012 \) and \( b = -0.402 \), we derive: - A calculation of \( n = \frac{0.402}{2 imes 0.00012} = 1675 \) implies an optimal enrollment of 1675 students. Knowing the optimal enrollment size, administrators can aim to adjust school policies and structuring to approach this ideal number, thereby reducing unnecessary expenditure.
Vertex Calculation
The calculation of a quadratic function's vertex is crucial for understanding how different variables affect the outcome in optimization problems. A quadratic function is expressed in the form \( ax^2 + bx + c \). The vertex is calculated using the formula \( n = -\frac{b}{2a} \). In the context of the given problem, to find the minimum cost of the quadratic cost function \( C = 743 - 0.402n + 0.00012n^2 \), we use: - The values \( a = 0.00012 \) and \( b = -0.402 \), leading to \( n = \frac{0.402}{0.00024} \) for the vertex calculation. Upon calculation, the result \( n \approx 1675 \) is the enrollment size where the cost per pupil reaches its minimum. This derived number from the vertex calculation implies that cost optimization happens when enrollment is around 1675 students, showcasing the power of mathematical analysis in real-world applications like cost management in schools.

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Most popular questions from this chapter

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