(i) Variables :
Let x1 and x2 denote the number of ordinary and auto-cut voltage stabilized.
(ii) Objective function:
Profit on x1 units of ordinary stabilizers = 100x1
Profit on x2 units of auto-cut stabilized = 150x2
Total profit = 100x1 + 150x2
Let Z = 100x1 + 150x2, which is the objective function.
Since the profit is to be maximized. We have to
Maximize, Z = 100x1 + 15x2
(iii) Constraints:
The assembling and testing time required for x1 units of ordinary stabilizers = 0.8x1 and for x2 units of auto-cut stabilizers = 1.2x2
Since the manufacturing capacity is 720 hours per week.
We get 0.8x1 + 1.2x2 ≤ 720
Maximum weekly sale of ordinary stabilizer is 600
i.e., x1 ≤ 600
Maximum weekly sales of auto-cut stabilizer is 400
i.e., x2 ≤ 400
(iv) Non-negative restrictions:
Since the number of both the types of stabilizer is nonnegative, we get x1, x2 ≥ 0.
Thus, the mathematical formulation of the LPP is,
Maximize Z = 100x1 + 150x2
Subject to the constraints
0.8x1 + 1.2x2 ≤ 720, x1 ≤ 600, x2 ≤ 400, x1, x2 ≥ 0