Tailoring unit: corner points, a redundant constraint and a bad rental offer
A self-help group at a senior secondary school runs a tailoring unit making two items: shirts (x per week) and trousers (y per week). The weekly limits are 20 machine-hours (2 hours per shirt, 1 hour per trousers), 16 labour-hours (1 hour per shirt, 2 hours per trousers), and 14 hours of packing-table time (1 hour per item). Profit is ₹40 per shirt and ₹50 per trousers. The treasurer's draft plan reads: “Packing is our real bottleneck, so spend packing capacity carefully”; he proposes making 10 shirts and no trousers, since that uses the machine fully. Separately, a neighbouring shop offers the group 4 extra machine-hours per week at ₹50 per hour, and the treasurer wants to accept because “machine time is what limits the shirts”.
1. Formulate the linear programme, find the corner points of the feasible region, and state the optimal weekly plan and profit.
Read the worked answer
Step 1 - formulation. Maximise Z = 40x + 50y subject to
2x + y ≤ 20 (machine), x + 2y ≤ 16 (labour), x + y ≤ 14 (packing), x ≥ 0, y ≥ 0.
Step 2 - corner points of the region bounded by the first two constraints, x ≥ 0, y ≥ 0.
- x = 0 with x + 2y = 16 gives (0, 8).
- y = 0 with 2x + y = 20 gives (10, 0).
- Intersection of 2x + y = 20 and x + 2y = 16: doubling the second, 2x + 4y = 32; subtracting the first gives 3y = 12, y = 4, then x = 8. So (8, 4).
- Origin (0, 0).
Check packing at all four: 0, 10, 12, 0 - all ≤ 14, so no corner is cut off by the packing line.
Step 3 - evaluate Z.
(0, 0) → 0; (10, 0) → 400; (8, 4) → 320 + 200 = 520; (0, 8) → 400.
Step 4 - conclusion. Maximum Z = ₹520 at x = 8 shirts, y = 4 trousers per week. The treasurer's plan (10 shirts, 0 trousers) earns only ₹400 - ₹120 short - and the reason is that the intersection point uses machine capacity efficiently and keeps labour fully employed, whereas 10 shirts leaves 6 labour-hours idle.
2. Is the packing constraint active in the optimal plan, and is it binding in the sense the treasurer means? Justify your answer rigorously rather than by inspecting one point.
Read the worked answer
Step 1 - check the optimum. At (8, 4), packing use is 8 + 4 = 12 hours against a limit of 14, so 2 hours are idle. The treasurer's claim that packing is the bottleneck is false at the optimum.
Step 2 - check all corners, not just the optimum. Corner values of x + y: (0,0) → 0; (10,0) → 10; (8,4) → 12; (0,8) → 8. The largest is 12 < 14.
Step 3 - extend from corners to the whole region. The feasible region is the convex polygon with those four vertices, and x + y is linear. A linear function on a convex polygon attains its maximum at a vertex, so x + y ≤ 12 everywhere in the region. Hence every feasible point satisfies x + y ≤ 14: the packing constraint is redundant - it can be deleted without changing the feasible region or the optimum.
Step 4 - what is actually binding. At (8, 4) both the machine constraint (16 + 4 = 20) and the labour constraint (8 + 8 = 16) hold with equality; there is no slack in either. These two, not packing, control the answer. The treasurer's plan misreads the structure of the problem: a constraint being “the tightest-sounding” in conversation says nothing; only the slack at the optimum does.
3. Should the group accept 4 extra machine-hours per week at ₹50 per hour? Estimate the worth of one extra machine-hour and decide.
Read the worked answer
Step 1 - re-solve with 24 machine-hours. Constraints become 2x + y ≤ 24, x + 2y ≤ 16, x + y ≤ 14.
Corners: (0, 8) [labour and y-axis]; (12, 0) [machine and x-axis]; and the machine–labour intersection: y = 24 − 2x into x + 2y = 16 gives x + 48 − 4x = 16, so x = 32/3 ≈ 10.67 and y = 8/3 ≈ 2.67. Packing check: 32/3 + 8/3 = 40/3 ≈ 13.33 ≤ 14, so this corner is feasible (the packing constraint stays redundant but only just).
Step 2 - profits. (0, 8) → 400; (12, 0) → 480; (32/3, 8/3) → 40(32/3) + 50(8/3) = (1280 + 400)/3 = ₹560.
Step 3 - value of the extra hours. Profit rises from ₹520 to ₹560, a gain of ₹40 for 4 machine-hours, i.e. about ₹10 per extra machine-hour. The rental price is ₹50 per hour.
Step 4 - integer sanity check. Shirts and trousers are whole units, so also test whole-number plans under 24 hours: (10, 3) uses 23 machine-hours, 16 labour-hours and 13 packing-hours, giving ₹550. The gain is ₹30 over 4 hours = ₹7.50 per hour - still far below ₹50.
Step 5 - decision. Decline. Whether or not fractional units are allowed, the marginal value of machine time (₹7.50–₹10 per hour) is well under the ₹50 asking price, so the rental would reduce profit. A useful extra conclusion: only the machine and labour constraints have positive marginal value here, and once the machine constraint is relaxed labour becomes the tighter of the two.