Explanation Given: Consider the linear programming problem as follows. Minimize w = 6 y 1 + 10 y 2 subject to 3 y 1 + 5 y 2 ≥ 15 4 y 1 + 7 y 2 ≥ 20 with y 1 ≥ 0 , y 2 ≥ 0 Method used: “Duals and Simplex method: 1. Obtain the augmented matrix and transpose it. 2. Produce the standard maximization problem from the transposed matrix. 3. Apply simplex method. 4. Determine the objective function. 5. The minimum value of the objective function w is the maximum value of the objective function. 6. Write all the necessary constraints. 7. Convert each constraint into an equation by adding a slack variable in each. 8. Set up the initial simplex tableau. 9. Locate the most negative indicator. If there are two such indicators, choose the one farther to the left. 10. Form the necessary quotients to find the pivot. Disregard any quotients with 0 or a negative number in the denominator. The smallest nonnegative quotient gives the location of the pivot. If all quotients must be disregarded, no maximum solution exists. If two quotients are both equal and smallest, choose the pivot in the row nearest the top of the matrix. 11. Use row operations to change all other numbers in the pivot column to zero by adding a suitable multiple of the pivot row to a positive multiple of each row. 12. If the indicators are all positive or 0, this is the final tableau. If not, go back to step 9 and repeat the process until a tableau with no negative indicators is obtained. 13. Read the optimal solution to the minimization problem from the final tableau. 14. The entries in the bottom row of the columns that corresponds to the slack variables are the optimal solution only after the z -column becomes 1.” Calculation: Obtain the augmented matrix, [ 3 5 15 4 7 20 6 10 0 ] . Observe that the last row is the coefficients of the objective function w = 6 y 1 + 10 y 2 and the upper two rows are the coefficients of the two linear constraints 3 y 1 + 5 y 2 ≥ 15 and 4 y 1 + 7 y 2 ≥ 20 with their constants. Obtain the transpose of the augmented matrix, [ 3 4 6 5 7 10 15 20 0 ] . Observe that the last row becomes the coefficients of the objective function that is z = 15 x 1 + 20 x 2 and the upper two rows becomes the coefficients as well as constant of the two linear constraints that are 3 x 1 + 4 x 2 ≤ 6 and 5 x 1 + 7 x 2 ≤ 10 . Hence, the dual of the linear programming problem is obtained as follows. Maximize z = 15 x 1 + 20 x 2 subject to 3 x 1 + 4 x 2 ≤ 6 5 x 1 + 7 x 2 ≤ 10 with x 1 ≥ 0 , x 2 ≥ 0 Add slack variable to the above linear programming problem. Maximize z = 15 x 1 + 20 x 2 subject to 3 x 1 + 4 x 2 + s 1 = 6 5 x 1 + 7 x 2 + s 2 = 10 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as, z − 15 x 1 − 20 x 2 = 0 . Now, obtain the simplex tableau as follows. x 1 x 2 s 1 s 2 z [ 3 4 1 0 0 6 5 7 0 1 0 10 − 15 − 20 0 0 1 0 ] Here, the basic variables are s 1 and s 2 . Thus, the basic solution is x 1 = 0 , x 2 = 0 , s 1 = 6 , s 2 = 10 and z = 0 . This is not the optimal solution as only the slack variables are nonzero. Look in the bottom row, the coefficient of x 2 is most negative, so choose second column as a pivot column and bring x 2 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 s 1 s 2 z [ 3 4 1 0 0 6 5 7 0 1 0 10 − 15 − 20 0 0 1 0 ] → 6 / 4 → 10 / 7 Observe that 10 7 is the least. Hence, the pivot element is chosen as the second element of the pivot column in the above tableau. x 1 x 2 s 1 s 2 z [ 3 4 1 0 0 6 5 7 0 1 0 10 − 15 − 20 0 0 1 0 ] Simplify the simplex tableau with the help of pivot method until the last row is completely filled with non-negative entries. Apply the operation 7 R 1 − 4 R 2 → R 1 in the above tableau as follows. x 1 x 2 s 1 s 2 z [ 1 0 7 − 4 0 2 5 7 0 1 0 10 − 15 − 20 0 0 1 0 ] Apply the operation 7 R 3 + 20 R 2 → R 3 in the above tableau as follows

Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

10th Edition

ISBN: 9780133935592

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Finite Mathematics and Calculus with Applications Books a la Carte Plus MyLab Math Package (10th Edition)

Ch. 4.1 - Prob. 1YT Ch. 4.1 - Prob. 2YT Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - For Exercises 58, (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E

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