Explanation Given information: The objective function which has to minimize is: M = 5 x + 7 y which is subjected to the constraints: { 2 x + y ≥ 10 3 x + 2 y ≥ 18 x + 2 y ≥ 10 x ≥ 0 y ≥ 0 Formula used: The minimization problem of simplex can be solved as below: Step 1: The minima problem can be converted into maxima problem by multiplying ( − 1 ) on both sides Step 2: Write the initial tableau and find the feasible solutions. Step 3: Make the all elements of last column positive by row operations. Step 4: Find the pivot column, by finding maximum negative number in the last row of the tableau. Step 5: Now divide the all elements which are in the last column by their corresponding elements of pivot column to decide pivot elements. Step 6: Now to make the pivot element unity, divide all the element of pivot row by 1 and make all other elements of pivot column to zero by row operation. Step 7: If there is one or more than one negative number in the last row, then the above tableau is not optimal, it needs iteration. Calculation: Since, this is minima problem, so convert the minima problem into maxima problem by multiplying ( − 1 ) on both sides of objective function which is: ( − 1 ) × M = − 5 x − 7 y . So, the objective function which has to maximize is M = − 5 x − 7 y . And, revised constraints are: { − 2 x − y ≤ − 10 − 3 x − 2 y ≤ − 18 − x − 2 y ≤ − 10 x ≥ 0 y ≥ 0 Write the provided constraints in the form of linear equations by introducing slack variables, u , v and w which are: − 2 x − y + u = − 10 − 3 x − 2 y + v = − 18 − x − 2 y + w = − 10 5 x + 7 y + M = 0 Step 1: Write the initial tableau and find the feasible solution. In the matrix form it can be written as: x y u v w M [ − 2 − 1 1 0 0 0 − 10 − 3 − 2 0 1 0 0 − 18 − 1 − 2 0 0 1 0 − 10 5 7 0 0 0 1 0 ] Step 2: Make the all elements of last column positive till all the elements of last column become positive. The choices to make positive are − 10 , − 18 or − 10 . Step 3: Choose the 3 rd row as a pivot row, and pivot element can be any element which can make the element − 10 positive, so the choice can be − 2 . x y u v w M u v w M [ − 2 − 1 1 0 0 0 − 10 − 3 − 2 0 1 0 0 − 18 − 1 − 2 0 0 1 0 − 10 5 7 0 0 0 1 0 ] Step 4: Now to make the pivot element unity, divide all the element of pivot row by − 2 and make all other elements of pivot column to zero by row operation which is as below: x y u v w M [ − 2 − 1 1 0 0 0 − 10 − 3 − 2 0 1 0 0 − 18 1 2 1 0 0 − 1 2 0 5 5 7 0 0 0 1 0 ] R 1 + 1 × R 3 R 2 + 2 × R 3 R 4 − 7 × R 3 → [ − 3 2 0 1 0 − 1 2 0 − 5 − 2 0 0 1 − 1 0 − 8 1 2 1 0 0 − 1 2 0 5 3 2 0 0 0 7 2 1 − 35 ] So, the next tableau for pivot − 2 is: u v w M [ − 3 2 0 1 0 − 1 2 0 − 5 − 2 0 0 1 − 1 0 − 8 1 2 1 0 0 − 1 2 0 5 3 2 0 0 0 7 2 1 − 35 ] Now the choices are − 5 o r − 8 to make them positive. Take 2 nd row as a pivot row which has to make positive and the element of 5 th column which is − 1 as a pivot element and apply row operation to make it unity

Finite Mathematics & Its Applications (12th Edition)

12th Edition

ISBN: 9780134437767

Author: Larry J. Goldstein, David I. Schneider, Martha J. Siegel, Steven Hair

Publisher: PEARSON

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Chapter 4, Problem 8RE

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To calculate: The minimum value of objective function: M=5x+7y which is subjected to the constraints:

{2x+y≥103x+2y≥18x+2y≥10x≥0y≥0

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Chapter 4, Problem 7RE

Chapter 4, Problem 9RE

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Chapter 4 Solutions

Finite Mathematics & Its Applications (12th Edition)

Ch. 4.1 - 1. Determine by inspection a particular solution...Ch. 4.1 - Prob. 2CYU Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - For each of the following linear programming...Ch. 4.1 - 712For each of the linear programming problems in...Ch. 4.1 - 7–12 For each of the linear programming problems...

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The minimum value of objective function: M = 5 x + 7 y which is subjected to the constraints: { 2 x + y ≥ 10 3 x + 2 y ≥ 18 x + 2 y ≥ 10 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the minimization problem of simplex can be solved by multiplying (-1) on both sides.

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To calculate: The minimum value of objective function: M=5x+7y which is subjected to the constraints:

{2x+y≥103x+2y≥18x+2y≥10x≥0y≥0

Want to see the full answer?

Chapter 4 Solutions

The minimum value of objective function: M = 5 x + 7 y which is subjected to the constraints: { 2 x + y ≥ 10 3 x + 2 y ≥ 18 x + 2 y ≥ 10 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the minimization problem of simplex can be solved by multiplying (-1) on both sides.

Videos

To calculate: The minimum value of objective function: M=5x+7y which is subjected to the constraints: {2x+y≥103x+2y≥18x+2y≥10x≥0y≥0

Want to see the full answer?

Chapter 4 Solutions

To calculate: The minimum value of objective function: M=5x+7y which is subjected to the constraints:

{2x+y≥103x+2y≥18x+2y≥10x≥0y≥0