Explanation Given: The linear programming problem is as follows, Minimize : w = 6 y 1 + 4 y 2 subject t o : 3 y 1 + 4 y 2 ≥ 10 9 y 1 + 7 y 2 ≤ 18 with y 1 ≥ 0 , y 2 ≥ 0. . Method used: “Solving a Nonstandard Problem: 1. If necessary, convert the problem to a maximization problem. (If the minimization problem has objective function w , rewrite as a maximization problem with the same constraints and objective function z = − w ). 2. Add slack variables and subtract surplus variables as needed. 3. Write the initial simplex tableau. 4. If any basic variable has a negative value, locate the nonzero number in that variable`s column, and note what row it is in. 5. In the row located in Step 4, find the positive entry that is farthest to the left, and note what column it is in. 6. In the column found in Step 5, choose a pivot by investigating quotients. 7. Use the row operations to change the other numbers in the pivot column to 0. 8. Continue Step 4through 7 until all basic variables are nonnegative. If it never becomes impossible to continue, then the problem has no feasible solution. 9. Once, a feasible solution has been found, continue to use the simplex method until the optimal solution is not found.” Calculation: The minimization linear programming problem is, Minimize : w = 6 y 1 + 4 y 2 subject t o : 3 y 1 + 4 y 2 ≥ 10 9 y 1 + 7 y 2 ≤ 18 with y 1 ≥ 0 , y 2 ≥ 0. . Here, the given linear programming is of minimization type, change the minimization objective function into maximization by taking z = − w . So, the objective function become, Maximize : z = − 6 y 1 − 4 y 2 . Note that the constraints in the maximization problem are same as minimization problem. Thus, the equivalent maximization problem is as follows. Maximize : z = − 6 y 1 − 4 y 2 subject t o : 3 y 1 + 4 y 2 ≥ 10 9 y 1 + 7 y 2 ≤ 18 with y 1 ≥ 0 , y 2 ≥ 0. Note that, the above linear programming problem is maximization type. So, it is in standard form. Subtract a surplus variable s 1 in first constraint, add a slack variable s 2 in second constraint and rewrite the linear programming problem as follows. Maximize : z = − 6 y 1 − 4 y 2 subject t o : 3 y 1 + 4 y 2 − s 1 = 10 9 y 1 + 7 y 2 + s 2 = 18 with y 1 ≥ 0 , y 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0. (1) The system (1) can be written in the system linear of equation as, 3 y 1 + 4 y 2 − s 1 = 10 9 y 1 + 7 y 2 + s 2 = 18 6 y 1 + 4 y 2 + z = 0 Construct the initial simplex tableau form the above linear system as follows. [ y 1 y 2 s 1 s 2 z 3 4 − 1 0 0 10 9 7 0 1 0 18 6 4 0 0 1 0 ] Look the 2 × 2 identity matrix. So the basic variables are s 1 and s 2 . Thus, the basic solution is y 1 = 0 , y 2 = 0 , s 1 = − 10 , s 2 = 18 and z = 0 . This solution is not feasible, as the basic variable s = − 10 is negative. To choose the pivot, locate the positive entry which is farthest to the left in row second

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

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

Publisher: PEARSON

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Chapter 4.4, Problem 1YT

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Finite Mathematics and Calculus with Applications (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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