Explanation Given information: The objective function which has to maximize is: M = 2 x + 3 y which is subjected to the constraints: { x + 2 y ≤ 14 x + y ≤ 9 3 x + 2 y ≤ 24 x ≥ 0 y ≥ 0 Formula used: The maximization problem of simplex can be solved as below: Step 1: Write the initial tableau and find the feasible solution. Step 2: Find the pivot column, by finding maximum negative number in the last row of the tableau. Step 3: Now divide the all elements which are in the last column by their corresponding elements of pivot column to decide pivot elements. Step 4: 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 5: 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: Write the provided constraints in the form of linear equations by introducing slack variables, u , v and w which are: x + 2 y + u = 14 x + y + v = 9 3 x + 2 y + w = 24 − 2 x − 3 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 [ 1 2 1 0 0 0 14 1 1 0 1 0 0 9 3 2 0 0 1 0 24 − 2 − 3 0 0 0 1 0 ] So, the above simplex tableau is initial tableau, therefore one feasible solution is: x = 0 , y = 0 , u = 14 , v = 9 , w = 24 and M = 0 Step 2: Find the pivot column, by finding maximum negative number in the last row of the tableau. From the above table it can be observed that in the last row the maximum negative number is − 3 so the 2 nd column is pivot column. Step 3: Now divide the all elements which are in the last column by their corresponding elements of pivot column to decide pivot elements. x y u v w M u v w M [ 1 2 1 0 0 0 14 1 1 0 1 0 0 9 3 2 0 0 1 0 24 − 2 − 3 0 0 0 1 0 ] 14 2 = 7 9 1 = 9 24 2 = 12 It can be observed that the minimum positive ratio is 7 which is 1 st row, therefore the pivot element is 1 st row and 2 nd column, which is 2. 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 [ 1 2 1 1 2 0 0 0 7 1 1 0 1 0 0 9 3 2 0 0 1 0 24 − 2 − 3 0 0 0 1 0 ] R 2 − 1 × R 1 R 3 − 2 × R 1 R 4 + 3 × R 1 → [ 1 2 1 1 2 0 0 0 7 1 2 0 − 1 2 1 0 0 2 2 0 − 1 0 1 0 10 − 1 2 0 3 2 0 0 1 21 ] So, the next tableau for pivot 2 is: y v w M [ 1 2 1 1 2 0 0 0 7 1 2 0 − 1 2 1 0 0 2 2 0 − 1 0 1 0 10 − 1 2 0 3 2 0 0 1 21 ] The feasible solution for this tableau is: y = 7 , v = 2 , w = 10 and M = 21

Finite Mathematics & Its Applications (12th Edition)

12th Edition

ISBN: 9780134507125

Author: Goldstein

Publisher: PEARSON

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Chapter 4, Problem 3RE

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To calculate: The maximum value of objective function: M=2x+3y which is subjected to the constraints:

{x+2y≤14x+y≤93x+2y≤24x≥0y≥0

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Chapter 4, Problem 2RE

Chapter 4, Problem 4RE

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Finite Mathematics & Its Applications (12th Edition)

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The maximum value of objective function: M = 2 x + 3 y which is subjected to the constraints: { x + 2 y ≤ 14 x + y ≤ 9 3 x + 2 y ≤ 24 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the maximization problem of simplex can be solved by writing the initial tableau and finding the feasible solution.

Videos

To calculate: The maximum value of objective function: M=2x+3y which is subjected to the constraints:

{x+2y≤14x+y≤93x+2y≤24x≥0y≥0

Want to see the full answer?

Chapter 4 Solutions

The maximum value of objective function: M = 2 x + 3 y which is subjected to the constraints: { x + 2 y ≤ 14 x + y ≤ 9 3 x + 2 y ≤ 24 x ≥ 0 y ≥ 0

Solution Summary: The author explains how the maximization problem of simplex can be solved by writing the initial tableau and finding the feasible solution.

Videos

To calculate: The maximum value of objective function: M=2x+3y which is subjected to the constraints: {x+2y≤14x+y≤93x+2y≤24x≥0y≥0

Want to see the full answer?

Chapter 4 Solutions

To calculate: The maximum value of objective function: M=2x+3y which is subjected to the constraints:

{x+2y≤14x+y≤93x+2y≤24x≥0y≥0