Study-Table Mix with a Rate Contract
A village workshop in Barabanki makes two models of steel study table, Standard (x per day) and Compact (y per day). Each Standard needs 3 hours of cutting and 2 hours of assembly; each Compact needs 2 hours of cutting and 4 hours of assembly. The workshop has 60 cutting hours and 64 assembly hours available each day, and can sell every table it makes at ₹400 profit on a Standard and ₹500 on a Compact. The owner wants the daily plan that maximises profit and, being cautious, also wants to know which of the two sections is worth expanding if he can buy only one extra machine-hour a day. A school chain then offers a rate contract that requires at least 10 tables of each model every day, failing which the contract lapses. The owner will sign only if the contract costs him no more than ₹150 a day in lost profit; otherwise he wants the minimum quantities renegotiated.
1. Formulate the linear programme, evaluate every corner of the feasible region, and give the optimal daily plan with the status of each section.
Read the worked answer
Maximise Z = 400x + 500y subject to 3x + 2y ≤ 60, 2x + 4y ≤ 64, x ≥ 0, y ≥ 0.
Corners: (0,0), (20,0) from the cutting line, (0,16) from the assembly line, and the intersection of the two lines. Doubling the first, 6x + 4y = 120; subtracting 2x + 4y = 64 gives 4x = 56, x = 14, y = (60 − 42)/2 = 9.
Z: (0,0) = 0; (20,0) = 8000; (0,16) = 8000; (14,9) = 5600 + 4500 = 10100.
Optimum: 14 Standard and 9 Compact, profit ₹10,100 per day. Both constraints are equalities at this point - cutting uses 3(14) + 2(9) = 60 of 60 hours and assembly uses 2(14) + 4(9) = 64 of 64 hours - so both sections are fully loaded and neither has slack.
2. Add the contract condition x ≥ 10, y ≥ 10. Find the new optimum and decide whether to sign.
Read the worked answer
The feasible region shrinks. Vertices: • (10,10): 30 + 20 = 50 ≤ 60 and 20 + 40 = 60 ≤ 64, feasible; • (10,11): assembly 20 + 44 = 64 exactly, cutting 30 + 22 = 52 ≤ 60; • (12,10): assembly 24 + 40 = 64 exactly, cutting 36 + 20 = 56 ≤ 60. The pair (10,15) on the cutting line violates assembly, and (13.3,10) on the cutting line also violates assembly, so cutting is now slack everywhere and assembly alone binds.
Z: (10,10) = 9000; (10,11) = 4000 + 5500 = 9500; (12,10) = 4800 + 5000 = 9800.
New optimum ₹9800 at 12 Standard and 10 Compact, so the contract costs 10100 − 9800 = ₹300 a day. That exceeds the owner's ₹150 ceiling, so he should not sign as drafted; the analysis gives him the number to bargain with - the chain must either lift the price or drop the minimum Standard quantity.
3. With the original constraints, find the gain in profit from one extra hour of each section and advise which machine-hour to buy.
Read the worked answer
One extra cutting hour (3x + 2y = 61 with 2x + 4y = 64): doubling the first gives 6x + 4y = 122; subtracting, 4x = 58, x = 14.5, y = (61 − 43.5)/2 = 8.75. Z = 5800 + 4375 = ₹10,175, a gain of ₹75.
One extra assembly hour (2x + 4y = 65 with 3x + 2y = 60): doubling the first, 6x + 4y = 120; subtracting, 4x = 55, x = 13.75, y = (60 − 41.25)/2 = 9.375. Z = 5500 + 4687.5 = ₹10,187.50, a gain of ₹87.50.
So an assembly hour is worth ₹87.50 against ₹75 for a cutting hour: buy assembly capacity. Both figures are local rates - valid for small increments, because a large enough expansion eventually makes the other constraint bind and the rate changes.