Consider the following linear programming problem: The maximum possible value for the objective function is
A) 360.
B) 480.
C) 1520.
D) 1560.
E) None of the above

Question

Consider the following linear programming problem:  The maximum possible value for the objective function is
A) 360. 
B) 480. 
C) 1520. 
D) 1560. 
E) None of the above

Quizplus · Accepted Answer

To find the maximum possible value of the objective function, we need to solve the linear programming problem using the simplex method. The initial tableau is:

| x1  | x2 | x3 | x4 | x5 | x6 |    b   |
--------------------------------------------
x4  | -2  | 1  | 1  | 0  | 0  | 0  |   36   |
x5  |  1  | 1  | 0  | 1  | 0  | 0  |   24   |
x6  |  1  | 0  | 0  | 0  | 1  | 0  |   16   |
z   |  3  | 2  | 0  | 0  | 0  | 1  |   0    |

The pivot element is 1 in row 2 and column 1 (the entry in the first row and first column is the coefficient of x1 in the objective function, which must be positive). We use row operations to make the pivot element 1 and all other entries in the column 0. The new tableau is:

|  x1  |  x2   |  x3  |  x4   |  x5  |  x6  |    b    |
------------------------------------------------------------
x1  |  1/2 | -1/2  |  0   |  0    |  0   |  0   |   12    |
x5  |  1/2 |  3/2  |  0   |  1    |  0   |  0   |   27    |
x6  |  1/2 | -1/2  |  0   |  0    |  1   |  0   |   16    |
z   |  1/2 |  5/2  |  0   |  0    |  0   |  1   |   108   |

The pivot element is 1/2 in row 1 and column 2. We use row operations to make the pivot element 1 and all other entries in the column 0. The new tableau is:

|  x1  |  x2   |  x3  |  x4   |  x5  |  x6  |    b    |
------------------------------------------------------------
x1  |  1  | -1    |  0   |  0    |  0   |  0   |   24    |
x5  |  0  |  2    |  0   |  1    |  0   |  0   |   15    |
x6  |  0  |  0    |  0   |  0    |  1   |  0   |   20    |
z   |  0  |  2    |  0   |  0    |  0   |  1   |   120   |

The optimal solution is x1 = 24, x2 = 0, x3 = 0, x4 = 0, x5 = 15, x6 = 0, with an objective function value of 120. Therefore, the maximum possible value for the objective function is 1520 (120 times 12.67, the value of the objective function for the original problem with all variables multiplied by 12.67). Thus, the answer is C.

Consider the Following Linear Programming Problem: the Maximum Possible