Explanation Given information: The objective function which has to maximize is: M = 3 x + 4 y + 5 z + 4 w which is subjected to the constraints: { 6 x + 9 y + 12 z + 15 w ≤ 672 x − y + 2 z + 2 w ≤ 92 5 x + 10 y − 5 z + 4 w ≤ 280 x ≥ 0 y ≥ 0 z ≥ 0 w ≥ 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, t , u , and v which are: 6 x + 9 y + 12 z + 15 w + t = 672 x − y + 2 z + 2 w + u = 92 5 x + 10 y − 5 z + 4 w + v = 280 − 3 x − 4 y − 5 z − 4 w + M = 0 Step 1: Write the initial tableau and find the feasible solution. In the matrix form it can be written as: x y z w t u v M t u v M [ 6 9 12 15 1 0 0 0 672 1 − 1 2 2 0 1 0 0 92 5 10 − 5 4 0 0 1 0 280 − 3 − 4 − 5 − 4 0 0 0 1 0 ] So, the above simplex tableau is initial tableau, therefore one feasible solution is: x = 0 , y = 0 , z = 0 , t = 672 , u = 92 , v = 280 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 − 5 so the 3 rd 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. The ratio is ( 672 12 = 56 ) , ( 92 2 = 46 ) and ( 280 − 5 = − 56 ) , so the minimum positive is 46 which is 2 nd row and 3 rd column. Hence, the pivot is 2. x y z w t u v M t u v M [ 6 9 12 15 1 0 0 0 672 1 − 1 2 2 0 1 0 0 92 5 10 − 5 4 0 0 1 0 280 − 3 − 4 − 5 − 4 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 z w t u v M [ 6 9 12 15 1 0 0 0 672 1 2 − 1 2 1 1 0 1 2 0 0 46 5 10 − 5 4 0 0 1 0 280 − 3 − 4 − 5 − 4 0 0 0 1 0 ] R 1 − 12 × R 2 R 3 + 5 × R 2 R 4 + 5 × R 2 → [ 0 15 0 3 1 − 6 0 0 120 1 2 − 1 2 1 1 0 1 2 0 0 46 15 2 15 2 0 9 0 5 2 1 0 510 − 1 2 − 13 2 0 1 0 5 2 0 1 230 ] So, the next tableau for pivot 2 is: x y z t u v w M t z v M [ 0 15 0 3 1 − 6 0 0 120 1 2 − 1 2 1 1 0 1 2 0 0 46 15 2 15 2 0 9 0 5 2 1 0 510 − 1 2 − 13 2 0 1 0 5 2 0 1 230 ] The feasible solution for this tableau is: t = 120 , z = 46 , v = 510 and M = 230 . 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. So, there is two negative elements in last row, choose the maximum negative element so the 2 nd column is pivot column. Step 6: Calculate the minimum positive ratio which is the ratio of all elements of last column to the all corresponding elements of pivot column, and select the minimum positive ratio which is: ( 120 15 = 8 ) The only choice which is minimum positive is 1 st row, so the pivot element is 15

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 10RE

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To calculate: The maximum value of objective function: M=3x+4y+5z+4w which is subjected to the constraints:

{6x+9y+12z+15w≤672x−y+2z+2w≤925x+10y−5z+4w≤280x≥0y≥0z≥0w≥0

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Chapter 4, Problem 9RE

Chapter 4, Problem 11RE

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Chapter 4 Solutions

Finite Mathematics & Its Applications (12th Edition)

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The maximum value of objective function: M = 3 x + 4 y + 5 z + 4 w which is subjected to the constraints: { 6 x + 9 y + 12 z + 15 w ≤ 672 x − y + 2 z + 2 w ≤ 92 5 x + 10 y − 5 z + 4 w ≤ 280 x ≥ 0 y ≥ 0 z ≥ 0 w ≥ 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.

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To calculate: The maximum value of objective function: M=3x+4y+5z+4w which is subjected to the constraints:

{6x+9y+12z+15w≤672x−y+2z+2w≤925x+10y−5z+4w≤280x≥0y≥0z≥0w≥0

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Chapter 4 Solutions

The maximum value of objective function: M = 3 x + 4 y + 5 z + 4 w which is subjected to the constraints: { 6 x + 9 y + 12 z + 15 w ≤ 672 x − y + 2 z + 2 w ≤ 92 5 x + 10 y − 5 z + 4 w ≤ 280 x ≥ 0 y ≥ 0 z ≥ 0 w ≥ 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=3x+4y+5z+4w which is subjected to the constraints: {6x+9y+12z+15w≤672x−y+2z+2w≤925x+10y−5z+4w≤280x≥0y≥0z≥0w≥0

Want to see the full answer?

Chapter 4 Solutions

To calculate: The maximum value of objective function: M=3x+4y+5z+4w which is subjected to the constraints:

{6x+9y+12z+15w≤672x−y+2z+2w≤925x+10y−5z+4w≤280x≥0y≥0z≥0w≥0