A Two-Product Workshop: Contracts, Breakdowns and Whole Units
A small furniture workshop in Hubballi makes chairs and tables. Each chair needs 2 hours of carpentry and 1 hour of finishing; each table needs 1 hour of carpentry and 3 hours of finishing. In a normal week the workshop has 40 carpentry hours and 45 finishing hours available. A chair earns a profit of ₹300 and a table ₹400, and the owner can sell everything produced. The owner has always built roughly equal numbers of both and has never done a formal optimisation. Two pressures now arrive at once: a bulk customer will sign a weekly contract only if at least 12 tables are delivered every week, and the carpentry machine is due for servicing, which would cut carpentry capacity from 40 hours to 26 hours. The owner wants to know the best weekly plan before the contract is signed, what the contract is really costing, and whether the arithmetic answer can actually be built and sold.
1. Formulate the normal-week problem and find the optimal weekly plan and profit.
Read the worked answer
Let x = chairs and y = tables per week.
Maximise P = 300x + 400y subject to 2x + y ≤ 40 (carpentry), x + 3y ≤ 45 (finishing), x, y ≥ 0.
The boundary intersections with the axes give (0, 0), (20, 0) and (0, 15). The two constraint lines meet where y = 40 − 2x and x + 3(40 − 2x) = 45, i.e. x + 120 − 6x = 45, giving 5x = 75, so x = 15 and y = 10.
Evaluate P at each corner: (0,0) → 0; (20,0) → 6000; (0,15) → 6000; (15,10) → 300(15) + 400(10) = 4500 + 4000 = ₹8,500.
Judgement: the optimum is 15 chairs and 10 tables, ₹8,500 per week. Both constraints are tight there (2 × 15 + 10 = 40 and 15 + 30 = 45), so there is no spare capacity in either department - the owner cannot expand output without buying hours in both. Note that the two axis corners tie at ₹6,000; that tie is a distraction, not the answer.
2. Impose the contract requirement of at least 12 tables and quantify what it costs the owner.
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Add y ≥ 12. The old optimum (15, 10) is now infeasible. The feasible corners become (0, 12), (9, 12) and (0, 15).
(0, 12): carpentry 12 ≤ 40, finishing 36 ≤ 45 → P = ₹4,800.
(9, 12): carpentry 18 + 12 = 30 ≤ 40, finishing 9 + 36 = 45 (tight) → P = 2,700 + 4,800 = ₹7,500.
(0, 15): carpentry 15 ≤ 40, finishing 45 (tight) → P = ₹6,000.
Maximum is ₹7,500 at 9 chairs and 12 tables.
Judgement: the contract costs ₹8,500 − ₹7,500 = ₹1,000 per week, i.e. it is worth quoting a surcharge of at least ₹1,000 per week (₹500 per guaranteed table above the previous 10). Observe that carpentry now has 10 idle hours while finishing is fully used, so the binding resource has switched: any extra capacity bought should be in finishing, not carpentry. A plan that simply adds two tables to the old optimum, 15 chairs and 12 tables, would need 42 carpentry hours and 51 finishing hours and is not buildable.
3. With the machine serviced, carpentry capacity falls to 26 hours while the contract still applies. Find the optimum, then check whether whole units change the answer.
Read the worked answer
Constraints: 2x + y ≤ 26, x + 3y ≤ 45, y ≥ 12.
Solving y = 26 − 2x with x + 3y = 45: x + 78 − 6x = 45, so 5x = 33, x = 6.6 and y = 12.8 (which satisfies y ≥ 12). P = 300(6.6) + 400(12.8) = 1,980 + 5,120 = ₹7,100.
But 0.6 of a chair cannot be built, so test whole-number neighbours: (6, 13): carpentry 12 + 13 = 25 ≤ 26; finishing 6 + 39 = 45 ≤ 45; P = 1,800 + 5,200 = ₹7,000. (7, 12): carpentry 14 + 12 = 26; finishing 7 + 36 = 43; P = 2,100 + 4,800 = ₹6,900. (6, 14): finishing 6 + 42 = 48 > 45, infeasible. (5, 13): P = ₹6,700.
Judgement: the realisable optimum is 6 chairs and 13 tables for ₹7,000, only ₹100 below the continuous bound of ₹7,100. Naively rounding both variables down to (6, 12) would give ₹6,600 and lose ₹400 per week, so the integer check is not optional. Report the plan as ₹7,000 with the bound ₹7,100, and note that carpentry has 1 idle hour while finishing is again the binding constraint.