Let x1 = number of units of A
Let x2 = number of units of B
Maximize Z = 4x1 + 6x2
Subject to:
1x1 + 3x2 ≤ 240
3x1 + 4x2 ≤ 370
2x1 + x2 ≤ 180
x1, x2 ≥ 0
The initial tableau is:
| | x1 | x2 | s1 | s2 | s3 | Z |
| - | - | - | - | - | - | - |
| s1 | 1 | 3 | 1 | 0 | 0 | 0 |
| s2 | 3 | 4 | 0 | 1 | 0 | 0 |
| s3 | 2 | 1 | 0 | 0 | 1 | 0 |
| Z | 4 | 6 | 0 | 0 | 0 | 0 |
The first pivot is (s1, x1):
| | x1 | x2 | s1 | s2 | s3 | Z |
| - | - | - | - | - | - | - |
| s1 | 1 | 3 | 1 | 0 | 0 | 0 |
| s2 | 3 | 4 | 0 | 1 | 0 | 0 |
| s3 | 2 | 1 | 0 | 0 | 1 | 0 |
| Z | 4 | 6 | 0 | 0 | 0 | 0 |
| | x1 | x2 | s1 | s2 | s3 | Z |
| - | - | - | - | - | - | - |
| s1 | 1 | 0 | 1/3 | -1 | 0 | 80 |
| s2 | 3 | 4 | 0 | 1 | 0 | 0 |
| s3 | 2 | 1 | -2/3 | 0 | 1 | 40 |
| Z | 4 | 6 | -4/3 | 0 | 0 | 160 |
The second pivot is (s3, x2):
| | x1 | x2 | s1 | s2 | s3 | Z |
| - | - | - | - | - | - | - |
| s1 | 1 | 0 | 1/3 | -1 | 0 | 80 |
| s2 | 3 | 4 | 0 | 1 | 0 | 0 |
| s3 | 2 | 1 | -2/3 | 0 | 1 | 40 |
| Z | 4 | 6 | -4/3 | 0 | 0 | 160 |
| | x1 | x2 | s1 | s2 | s3 | Z |
| - | - | - | - | - | - | - |
| s1 | 1 | 0 | 1/3 | -1 | 0 | 80 |
| s2 | 3 | 0 | -1/3 | 1 | -1 | 60 |
| s3 | 2 | 1 | 0 | 0 | 1 | 40 |
| Z | 4 | 6 | -1/3 | 0 | -1 | 140 |
The optimal solution is x1 = 80 and x2 = 60, which yields a maximum gross income of 140.
firm has available 240, 370 and 180 kg of wood, plastic and steel respectively. The firm produces two products A and B. Each unit of A requires 1, 3 and 2 kg of wood, plastic and steel, respectively. The corresponding requirement for each unit of B are 3, 4 and 1, respectively. If A sells for 4 and B for 6, determine how many units of A and B should be produced in order to obtain the maximum gross income. Use the simplex method.
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