Notebook Production Plan Under Machine-Hour Limits
A small stationery workshop in Birgunj makes two notebooks, Regular (x) and Premium (y). Each Regular notebook needs 2 hours on the cutting machine and 1 hour on the binding machine. Each Premium notebook needs 1 hour on cutting and 2 hours on binding. Each week the workshop has at most 100 cutting hours and 80 binding hours. A school contract requires at least 10 Regular notebooks per week, and the market can absorb at most 40 Premium notebooks per week. Profit is Rs 30 per Regular and Rs 40 per Premium. Last week the workshop made 30 Regular and 20 Premium notebooks and earned Rs 1,700. The owner suspects this was not the best plan and asks for a corner-point analysis. All production within the stated limits can be sold.
1. Write the inequalities that define the feasible region and find all corner points, showing how each intersection is obtained.
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Let x = Regular notebooks and y = Premium notebooks. The constraints are:
- Cutting: 2x + y ≤ 100
- Binding: x + 2y ≤ 80
- Contract: x ≥ 10
- Market limit: y ≤ 40
- Also x ≥ 0, y ≥ 0, though x ≥ 10 already handles the x lower bound.
The boundary lines are 2x + y = 100, x + 2y = 80, x = 10 and y = 40.
Corner points come from feasible intersections:
- x = 10 with y = 0 gives (10, 0); it satisfies 20 + 0 ≤ 100 and 10 + 0 ≤ 80.
- 2x + y = 100 with y = 0 gives x = 50, so (50, 0); it satisfies 50 + 0 ≤ 80.
- Solving 2x + y = 100 and x + 2y = 80: from the first, y = 100 − 2x. Substituting gives x + 200 − 4x = 80, so −3x = −120, x = 40 and y = 20. This gives (40, 20), which satisfies x ≥ 10 and y ≤ 40.
- x = 10 with x + 2y = 80 gives 10 + 2y = 80, y = 35, so (10, 35). It satisfies 2(10) + 35 = 55 ≤ 100.
The intersection (30, 40) from 2x + y = 100 and y = 40 fails x + 2y ≤ 80 because 30 + 80 = 110 > 80, so it is rejected. The intersection (0, 40) fails x ≥ 10, and (10, 80) fails both y ≤ 40 and binding. Thus the corner points are (10, 0), (50, 0), (40, 20) and (10, 35).
2. Evaluate the profit at each corner and state the optimal weekly plan. Compare it with last week’s plan of 30 Regular and 20 Premium notebooks.
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The profit is P = 30x + 40y. At the corner points:
- (10, 0): P = 30(10) + 40(0) = Rs 300
- (50, 0): P = 30(50) + 40(0) = Rs 1,500
- (40, 20): P = 30(40) + 40(20) = 1,200 + 800 = Rs 2,000
- (10, 35): P = 30(10) + 40(35) = 300 + 1,400 = Rs 1,700
The maximum is Rs 2,000 at 40 Regular and 20 Premium notebooks. At this plan, cutting hours used = 2(40) + 20 = 100, and binding hours used = 40 + 2(20) = 80, so both machine limits are fully used.
Last week’s plan of 30 Regular and 20 Premium gives P = 30(30) + 40(20) = 900 + 800 = Rs 1,700. It leaves cutting hours at 2(30) + 20 = 80 and binding hours at 30 + 40 = 70, so 20 cutting hours and 10 binding hours are unused. Moving from (30, 20) to (40, 20) adds 10 Regular notebooks, which uses exactly 20 cutting hours and 10 binding hours. The optimal plan improves profit by Rs 300.
3. A worker argues that because Premium earns Rs 40 per notebook, the workshop should make as many Premium notebooks as possible, that is, y = 40. Test this proposal against the constraints and the optimal profit.
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If y = 40, the contract still requires x ≥ 10. The binding constraint becomes x + 2(40) ≤ 80, so x + 80 ≤ 80, which gives x ≤ 0. This contradicts x ≥ 10. Therefore y = 40 is not feasible at all, despite the higher profit per Premium notebook.
The largest Premium level compatible with x = 10 comes from binding: 10 + 2y ≤ 80, so y ≤ 35. The point (10, 35) is feasible and earns Rs 1,700, which is Rs 300 less than the optimum of Rs 2,000 at (40, 20). The worker is comparing profit per notebook but ignoring that each Premium notebook uses 2 binding hours, while the binding capacity is the tighter resource at the optimum. The correct conclusion is that the optimal plan is still 40 Regular and 20 Premium notebooks.