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प्रश्न
Raunak is a small-scale entrepreneur who sells sewing machines in a rural market. He wants to expand his business but has two main constraints: Capital and Storage.
He has a total capital of ₹ 5,760 to invest. The godown can store a maximum number of 20 sewing machines.
Raunak sells two types of machines:
- Electronic sewing machine, each costs him 360.
- Manually operated sewing machine, each costs him 240.
His wife Radhika suggests selling an electronic machine ata profit of ₹ 22 and a manually operated sewing machine at a profit of 18.
Using the concept of Linear Programming Problem, find the number of sewing machines of each type that Raunak should sell to maximise his profit
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उत्तर
Let the Variables be:
x = Number of Electronic sewing machines.
y = Number of manually operated sewing machines.
Objective Function:
The goal is to maximise the total profit (Z).
Maximize Z = 22x = 18y
Constraints:
Storage Constraints: x + y ≤ 20
Capital Constraint: 360x + 24y ≤ 5760
3x + 2y ≤ 48 ... [Divide the whole equation by 120 to simplify]
Non-negativity Constraints: x ≥ 0, y ≥ 0
Finding Points for the Graph
To draw the lines, we find the intercepts (x = 0 and y = 0)
| Equation | If x = 0 | If y = 0 | Points |
| x + y = 20 | y = 20 | x = 20 | (0, 20) and (20, 0) |
| 3x + 2y = 48 | y = 24 | x = 16 | (0, 24) and (16, 0) |
Intersection Point of the lines:
Solving x + y = 20 and 3x + 2y = 48 simulantanneosly:
3x + 2(20 − x) = 48
3x + 40 − 2x = 48
x = 8, y = 12
Profit Calculation at Corner Points
The feasible region is bounded by the points: O(0, 0), A(16, 0), B(8, 12), and C(0, 20).
| Corner Point (x, y) | Objective Function Z = 22x + 18y | Profit (₹) |
| O(0, 0) | 22(0) + 18(0) | 0 |
| A(16, 0) | 22(16) + 18(0) | 352 |
| B(8, 12) | 22(8) + 18(12) | 392(MAX) |
| C(0, 20) | 22(0) + 18(20) | 360 |
