Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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
Transcribed Image Text:**Transcription for Educational Website**
The table represents a simplex tableau used in linear programming to find the optimal solution for a linear optimization problem.
| | x | y | z | a | b | c | p |
|---|---|---|---|---|---|---|---|
| a | 5 | 0 | 0 | 6 | -2| 4 | 0 | 24 |
| z | 2 | 0 | 11| 0 | 1 | 3 | 0 | 44 |
| y |-1 | 3 | 0 | 0 | 0 | 7 | 0 | 12 |
| p | 3 | 0 | 0 | 0 | 6 | 9 | 1 | 52 |
Variables:
- Decision variables (x, y, z)
- Slack/surplus variables (a, b, c)
- Objective function value (p)
**Questions:**
a) What is the optimal solution?
b) What is the optimal value?
**Explanation:**
To answer the questions, the tableau undergoes a series of pivot operations to reach the optimal solution. The basic feasible solution can be read from the tableau as the coefficients of the identity matrix (the columns containing mostly zeros and a single '1' per column), and the final column provides the solution values for these basic variables. The row p shows the coefficients of the objective function.
To answer part a, interpret the solution by reading off the values assigned to variables aligning with identity columns, considering pivot operations if required.
For part b, the optimal value is the entry in the bottom-right corner of the tableau, which represents the optimized value of the objective function given the constraints.
**Note:**
The specific operations depend on the problem's context, which involves optimizing a linear function subject to constraints represented by the rows.
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