A manufacturer must decide how many chairs and tables to produce in a given week. It takes 2 hours to construct a chair and 3 hours to construct a table. The finishing process requires 2 hours for each chair and 2 hours for each table. There are 36 work hours available for construction and 28 work hours available for finishing. If the profit is $15 per chair and $25 per table, how many of each should be produced to maximize profit? Define your variables. y for chairs for tables
A manufacturer must decide how many chairs and tables to produce in a given week. It takes 2 hours to construct a chair and 3 hours to construct a table. The finishing process requires 2 hours for each chair and 2 hours for each table. There are 36 work hours available for construction and 28 work hours available for finishing. If the profit is $15 per chair and $25 per table, how many of each should be produced to maximize profit? Define your variables. y for chairs for tables
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
A manufacturer must decide how many chairs and tables to produce in a given week. It takes 2 hours to construct a chair and 3 hours to construct a table. The finishing process requires 2 hours for each chair and 2 hours for each table. There are 36 work hours available for construction and 28 work hours available for finishing. If the profit is $15 per chair and $25 per table, how many of each should be produced to maximize profit?
### Variables
- \( x \): Number of chairs
- \( y \): Number of tables
### Profit or Objective Function
\[ P(x, y) = 15x + 25y \]
### Constraints
1. Construction:
\[ 2x + 3y \leq 36 \]
2. Finishing:
\[ 2x + 2y \leq 28 \]
3. Non-negativity:
\[ x \geq 0 \]
\[ y \geq 0 \]
### Graph Description
The graph is a coordinate plane grid. Points and lines depict the constraints. The lines divide the plane into regions, showing feasible solutions for \( x \) and \( y \). The line \( 2x + 3y = 36 \) and \( 2x + 2y = 28 \) are plotted.
The equation for one of the lines is given as:
\[ y = -\frac{2}{3}x + 12 \]
### Vertices of the Feasible Region
- Points where the constraint lines intersect are crucial for determining potential maximum profit.
- The exercise asks to determine these vertices, but they are left blank.
#### Note
The task is to evaluate these points to find which gives the highest value for the objective function (profit equation).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffac1bf48-50df-4cbd-860a-5a4ad873c531%2Fe5315efd-cab4-4694-a877-6175954ce05f%2Fygj0jv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A manufacturer must decide how many chairs and tables to produce in a given week. It takes 2 hours to construct a chair and 3 hours to construct a table. The finishing process requires 2 hours for each chair and 2 hours for each table. There are 36 work hours available for construction and 28 work hours available for finishing. If the profit is $15 per chair and $25 per table, how many of each should be produced to maximize profit?
### Variables
- \( x \): Number of chairs
- \( y \): Number of tables
### Profit or Objective Function
\[ P(x, y) = 15x + 25y \]
### Constraints
1. Construction:
\[ 2x + 3y \leq 36 \]
2. Finishing:
\[ 2x + 2y \leq 28 \]
3. Non-negativity:
\[ x \geq 0 \]
\[ y \geq 0 \]
### Graph Description
The graph is a coordinate plane grid. Points and lines depict the constraints. The lines divide the plane into regions, showing feasible solutions for \( x \) and \( y \). The line \( 2x + 3y = 36 \) and \( 2x + 2y = 28 \) are plotted.
The equation for one of the lines is given as:
\[ y = -\frac{2}{3}x + 12 \]
### Vertices of the Feasible Region
- Points where the constraint lines intersect are crucial for determining potential maximum profit.
- The exercise asks to determine these vertices, but they are left blank.
#### Note
The task is to evaluate these points to find which gives the highest value for the objective function (profit equation).
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