Explanation Method used: “Simplex method: 1. Determine the objective function. 2. Write all the necessary constraints. 3. Convert each constraint into an equation by adding a slack variable in each. 4. Set up the initial simplex tableau. 5. Locate the most negative indicator. If there are two such indicators, choose the one farther to the left. 6. 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. 7. 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. 8. If the indicators are all positive or 0, this is the final tableau. If not, go back to step 5 and repeat the process until a tableau with no negative indicators is obtained. 9. Read the solution from the final tableau.” Calculation: The solution of the maximization linear programming problem with initial simplex tableau is as follows. The initial tableau of the maximization linear programming problem is, [ x 1 x 2 s 1 s 2 z 2 7 1 0 0 14 2 3 0 1 0 10 − 2 − 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 x 1 = 0 , x 2 = 0 , x 3 = 0 , s 1 = 14 , s 2 = 10 and z = 0 . Thus, the bottom row is, z = 2 x 1 + 4 x 2 . The value of z increases as the value of variables increases. 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. Compute the ratios to obtain the pivot element as follows. The value of b 1 a 12 is, b 1 a 13 = 14 7 = 2 The value of b 2 a 22 is, b 2 a 22 = 10 3 = 3.33 The value of the smallest ratio is 2. Thus, the pivot element is a 12 = 7 . Perform the row operation − 3 R 1 + 7 R 2 → R 2 in second row and rewrite second row with a 22 = 0 . Perform the row operation 4 R 1 + 7 R 3 → R 3 in third row and rewrite third row with a 32 = 0 . Thus, the second tableau is as follows. [ x 1 x 2 s 1 s 2 z 2 7 1 0 0 14 8 0 − 3 7 0 28 − 6 0 4 0 7 56 ] Here notice that the bottom row has one negative element indicated that this tableau is not final simplex tableau. Look in the bottom row, the coefficient of x 1 is most negative, so choose first column as a pivot column and bring x 1 into the solution

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