Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**The Decline and Consolidation of Rural High Schools: Analyzing Enrollment's Impact on Costs**
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 1960s on expenditures for high schools ranging from 150 to 2400 in enrollment. He considered the cost per pupil as a function of the number of pupils enrolled in the high school, and he found the following approximate formula, where \( n \) is the number of pupils enrolled and \( C \) is the cost, in dollars, per pupil.
\[ C = 743 - 0.402n + 0.00012n^2 \]
**Task (a): Plot the Graph**
(a) Make a graph of \( C \) versus \( n \).
*Explanation of Graphs*:
The image contains four different graphs, each representing a relationship between the cost per pupil (\( C \)) and the number of students (\( n \)). However, without specific axis labels, the appropriate selection can be inferred from accurate mathematical understanding of the quadratic equation provided.
**Tasks (b) and (c)**:
(b) What enrollment size gives a minimum per pupil cost?
(c) If a high school had an enrollment of 1300, how much in per pupil cost would be saved by changing enrollment to the optimal size found in part (b)?
(Round your answer to two decimal places.)
\[ \text{Savings per pupil: } \boxed{\$} \]
By calculating, the vertex of the quadratic function can be found using the vertex formula, \( n = -\frac{b}{2a} \). Here, \( a = 0.00012 \) and \( b = -0.402 \).
\[ n = -\frac{-0.402}{2 \times 0.00012} = \frac{0.402}{0.00024} = 1675 \]
Thus, the enrollment size giving the minimum cost per pupil is approximately \(\boxed{1675}\) students.
Calculating the minimum cost per pupil at optimal enrollment size:
\[ C(1675) = 743 - 0.402(1675) + 0.00012(1675^2) \]
\[ C(1675) = 743 - 673.](https://content.bartleby.com/qna-images/question/8d91cf51-36b6-472d-8082-aa48d6da434d/71547b9d-c696-4d48-83d8-5e9700068444/wjzl5p_processed.png)
Transcribed Image Text:**The Decline and Consolidation of Rural High Schools: Analyzing Enrollment's Impact on Costs**
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 1960s on expenditures for high schools ranging from 150 to 2400 in enrollment. He considered the cost per pupil as a function of the number of pupils enrolled in the high school, and he found the following approximate formula, where \( n \) is the number of pupils enrolled and \( C \) is the cost, in dollars, per pupil.
\[ C = 743 - 0.402n + 0.00012n^2 \]
**Task (a): Plot the Graph**
(a) Make a graph of \( C \) versus \( n \).
*Explanation of Graphs*:
The image contains four different graphs, each representing a relationship between the cost per pupil (\( C \)) and the number of students (\( n \)). However, without specific axis labels, the appropriate selection can be inferred from accurate mathematical understanding of the quadratic equation provided.
**Tasks (b) and (c)**:
(b) What enrollment size gives a minimum per pupil cost?
(c) If a high school had an enrollment of 1300, how much in per pupil cost would be saved by changing enrollment to the optimal size found in part (b)?
(Round your answer to two decimal places.)
\[ \text{Savings per pupil: } \boxed{\$} \]
By calculating, the vertex of the quadratic function can be found using the vertex formula, \( n = -\frac{b}{2a} \). Here, \( a = 0.00012 \) and \( b = -0.402 \).
\[ n = -\frac{-0.402}{2 \times 0.00012} = \frac{0.402}{0.00024} = 1675 \]
Thus, the enrollment size giving the minimum cost per pupil is approximately \(\boxed{1675}\) students.
Calculating the minimum cost per pupil at optimal enrollment size:
\[ C(1675) = 743 - 0.402(1675) + 0.00012(1675^2) \]
\[ C(1675) = 743 - 673.
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