Explanation Given: The linear programming problem is, Maximize : z = 5 x 1 − 3 x 2 subject t o : x 1 + x 2 ≤ 100 2 x 1 + 3 x 2 ≤ 75 x 1 + 4 x 2 ≥ 50 with x 1 ≥ 0 , x 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 calculation of x 1 ≥ 0 and x 2 ≥ 0 using simplex method is as follows. The linear programming problem is, Maximize : z = 5 x 1 − 3 x 2 subject t o : x 1 + x 2 ≤ 100 2 x 1 + 3 x 2 ≤ 75 x 1 + 4 x 2 ≥ 50 with x 1 ≥ 0 , x 2 ≥ 0 . Note that the above linear programming problem is maximization type. So it is in standard form. Add a slack variable s 1 in first constraint, s 2 in second constraint, subtract a surplus variable s 3 in third constraint and rewrite the linear programming problem as follows. Maximize : z = 5 x 1 − 3 x 2 subject t o : x 1 + x 2 + s 1 = 100 2 x 1 + 3 x 2 + s 2 = 75 x 1 + 4 x 2 − s 3 = 50 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 (1) The system (1) can be written in the system linear of equation as, x 1 + x 2 + s 1 = 100 2 x 1 + 3 x 2 + s 2 = 75 x 1 + 4 x 2 − s 3 = 50 − 5 x 1 + 3 x 2 + z = 0 . Construct the initial simplex tableau form the above linear system as follows. [ x 1 x 2 s 1 s 2 s 3 z 1 1 1 0 0 0 100 2 3 0 1 0 0 75 1 4 0 0 − 1 0 50 − 5 3 0 0 0 1 0 ] Look the 3 × 3 identity matrix. So, the basic variables are s 1 , s 2 and s 3 . Thus, the basic solution is x 1 = 0 , x 2 = 0 , s 1 = 100 , s 2 = 75 , s 3 = − 50 and z = 0

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Chapter 4.4, Problem 13E

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The value of x 1 ≥ 0 , and x 2 ≥ 0 using simplex method.

Solution Summary: The author explains how to solve a linear programming problem using the simplex method.

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To find: The value of x1≥0 , and x2≥0 using simplex method.

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