The bakery club's rounding problem
A school club runs a small bakery and must decide each morning how many of two products to bake. Loaf A uses 300 g of flour, 100 g of butter and 2 units of oven time; loaf B uses 200 g of flour, 200 g of butter and 3 units of oven time. The day's supplies are 3.6 kg of flour, 2.0 kg of butter and 30 units of oven time. Only whole loaves can be made and sold, and unsold loaves cannot be kept for the next day. A sells for 500 yen, B for 700 yen, and the club wants the greatest possible revenue.
A member treats the problem as continuous, finds the best corner of the feasible region at A = 9.6, B = 3.6, and proposes baking 10 loaves of A and 4 of B, calling this "the rounded optimum" and valuing it at 7 800 yen. The president, who must place the flour order this afternoon, is uneasy and asks the science teacher to check the plan before anything is bought.
1. Check the proposed plan (10 A, 4 B) against all three constraints, then find a plan in whole loaves that earns more, if one exists. Show the arithmetic for every plan you compare.
Read the worked answer
The proposal is not feasible. Flour for 10 A and 4 B is 10 × 0.3 + 4 × 0.2 = 3.0 + 0.8 = 3.8 kg, but only 3.6 kg is available. The 7 800 yen figure therefore describes loaves that cannot be baked, and no rounding argument can rescue it.
Working in units of 100 g, the constraints are 3A + 2B ≤ 36, A + 2B ≤ 20 and 2A + 3B ≤ 30. Checking whole-loaf plans near the continuous corner:
- (9, 4): flour 27 + 8 = 35 ≤ 36; butter 9 + 8 = 17 ≤ 20; oven 18 + 12 = 30 ≤ 30. Revenue 4 500 + 2 800 = 7 300 yen.
- (8, 5): oven 16 + 15 = 31 > 30 - infeasible.
- (10, 3): revenue 5 000 + 2 100 = 7 100 yen.
- (6, 6): revenue 3 000 + 4 200 = 7 200 yen.
- (7, 5): revenue 7 000 yen; (3, 8): 7 100 yen.
The best feasible plan is 9 loaves of A and 4 of B, 7 300 yen, and it uses every unit of oven time exactly.
2. The member objects that checking plans one by one is not a proof. Without listing every feasible whole-loaf plan, show that 7 300 yen cannot be beaten.
Read the worked answer
Use the continuous problem as a bound. The continuous optimum lies where the flour and oven constraints meet: 3A + 2B = 36 and 2A + 3B = 30. Eliminating B gives 5A = 48, so A = 9.6, B = 3.6, with revenue 500(9.6) + 700(3.6) = 4 800 + 2 520 = 7 320 yen. Because the feasible region contains every whole-loaf plan, no whole-loaf plan can earn more than 7 320 yen.
Now use the arithmetic of the prices: every revenue figure is a whole number of hundreds of yen, since 500A + 700B = 100(5A + 7B). Any plan better than 7 300 yen would therefore earn at least 7 400 yen. But 7 400 > 7 320, which contradicts the bound.
Hence 7 300 yen is optimal. The logic matters: rounding 9.6 and 3.6 to 10 and 4 was invalid as a construction, but the continuous maximum is still a legitimate certificate of optimality once a feasible plan comes within one lattice step of it.
3. The flour delivery arrives 200 g short, so only 3.4 kg is available; butter and oven time are unchanged. Is the club's best response simply to keep the same plan with slightly smaller loaves? Find the new best plan and revenue, and comment on the size of the change.
Read the worked answer
Smaller loaves are not an option: the recipe fixes the size, so the change must be in the numbers baked. The plan (9, 4) needs 3.5 kg of flour and is now infeasible, so the optimum genuinely moves.
With 3A + 2B ≤ 34 (plus A + 2B ≤ 20 and 2A + 3B ≤ 30), scanning the whole-loaf points:
- (6, 6): flour 18 + 12 = 30 ≤ 34; butter 18 ≤ 20; oven 30 ≤ 30 → revenue 3 000 + 4 200 = 7 200 yen.
- (3, 8): revenue 7 100 yen; (0, 10): 7 000 yen; (9, 3): 6 600 yen; (8, 4): 6 800 yen.
- (7, 6) needs 32 oven units and (5, 7) needs 31 - both infeasible.
The new optimum is 6 A and 6 B for 7 200 yen. The interesting point for the club is that a 5.7 % cut in flour changes the production plan completely (9/4 to 6/6) yet costs only 100 yen: near a corner, the value of a resource moves in small jumps because the answer must sit on a lattice of whole loaves, not on the smooth line the continuous model draws.