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प्रश्न
A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer's profit on an item of model A is ₹ 15 and on an item of model B is ₹ 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.
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उत्तर
Let x articles of model A and y articles of model B be made.
A number of articles cannot be negative.
Therefore, x, y ≥ 0
According to the question, the making of a model A requires 2 hrs. work by a skilled man and model B requires 1 hr by a skilled man
2x + y ≤ 40
The making of a model A requires 2 hrs. work by a semi-skilled man model B requires 3 hrs. work by a semi-skilled man.
2x + 3y ≤ 80
Total profit = Z = 15x + 10y which is to be maximised
Thus, the mathematical formulation of the given linear programming problem is Max Z = 15x + 10y
subject to
2x + y ≤ 40
2x + 3y ≤ 80
x ≥ 0
y ≥ 0
The feasible region determined by the system of constraints is

The corner points are `"A" (0, 80/3)`, B(10,20), C(20,0)
The values of Z at these corner points are as follows
| Corner point | Z= 15x+10y |
| A | `800/3` |
| B | 350 |
| C | 300 |
The maximum value of Z is 300 which is attained at C(20, 0)
Thus, the maximum profit is Rs 300 obtained when 10 units of a deluxe model and 20 unit of the ordinary model is produced.
संबंधित प्रश्न
Minimum and maximum z = 5x + 2y subject to the following constraints:
x-2y ≤ 2
3x+2y ≤ 12
-3x+2y ≤ 3
x ≥ 0,y ≥ 0
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Solve the following Linear Programming Problem graphically.
Maximise Z = 5x + 2y subject to:
x – 2y ≤ 2,
3x + 2y ≤ 12,
– 3x + 2y ≤ 3,
x ≥ 0, y ≥ 0
