Bakery Product Mix: A Demand Constraint the Helper Missed
A small bakery makes two kinds of catering tin. Each Standard tin (x) uses 2 kg of flour and 1 kg of butter; each Deluxe tin (y) uses 1 kg of flour and 2 kg of butter. This week the bakery has 10 kg of flour and 14 kg of butter. A market survey says that at most 5 Deluxe tins can be sold this week, and the owner wants to use the stock to make as much profit as possible. Profit is ₹400 per Standard tin and ₹500 per Deluxe tin, so the objective is P = 400x + 500y. A part-time helper solves the two supply equations 2x + y = 10 and x + 2y = 14, obtains x = 2, y = 6, and reports a maximum profit of ₹3,800. The owner is delighted, but the baker points out that the helper never checked whether (2, 6) can actually be made and sold within this week's limits. Tins cannot be sold in fractions.
1. Write down the full constraint set including the survey limit, find the feasible corner points, and give the correct maximum profit.
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Constraints: 2x + y ≤ 10 (flour), x + 2y ≤ 14 (butter), y ≤ 5 (survey), x ≥ 0, y ≥ 0.
Feasible corners: • (0, 0) → P = 0 • (5, 0) → P = ₹2,000 • (0, 5) → P = ₹2,500 • Flour line with y = 5: 2x + 5 = 10 → x = 2.5 → (2.5, 5) → P = 400(2.5) + 500(5) = ₹3,500 • Butter line with y = 5 gives x = 4, but 2(4) + 5 = 13 > 10, so it is outside the flour limit. • The lines 2x + y = 10 and x + 2y = 14 meet at (2, 6), but y = 6 violates the survey limit.
Maximum: ₹3,500 at (2.5, 5).
Slack check: flour 2(2.5) + 5 = 10 (binding); butter 2.5 + 10 = 12.5, leaving 1.5 kg unused (not binding); survey limit y = 5 (binding).
2. Diagnose the helper's answer of ₹3,800 at (2, 6), and state the best whole-number production plan.
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Diagnosis. (2, 6) uses flour 2(2) + 6 = 10 kg and butter 2 + 12 = 14 kg, so it satisfies both raw-material limits. It fails only the survey limit: 6 Deluxe tins exceed the maximum of 5 that can be sold. The helper optimised the two supply equations as if both were equalities and never checked the third constraint or tested feasibility - the classic error of solving the boundary equations rather than the constrained problem.
Best integer plan. Since tins are whole numbers, evaluate near (2.5, 5): • (2, 5): flour 4 + 5 = 9 ≤ 10, butter 2 + 10 = 12 ≤ 14 → P = 800 + 2,500 = ₹3,300 • (3, 5): flour 6 + 5 = 11 > 10 → infeasible • (3, 4): flour 6 + 4 = 10, butter 3 + 8 = 11 → P = 1,200 + 2,000 = ₹3,200
So make 2 Standard and 5 Deluxe tins for ₹3,300, which is ₹500 below the fractional optimum and ₹500 below the helper's impossible figure. The 0.5 tin gap matters here: rounding down costs more than rounding up (which is infeasible).
3. If 5 kg of butter spoil overnight, leaving only 9 kg, how does the optimum change, which constraints bind, and what is the marginal value of the lost butter?
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New system: 2x + y ≤ 10, x + 2y ≤ 9, y ≤ 5, x, y ≥ 0.
Corners of the feasible region: (0, 0); (5, 0); (0, 4.5) since x + 2y ≤ 9 rules out (0, 5).
The two supply lines meet where y = 10 − 2x and x + 2(10 − 2x) = 9 → −3x = −11 → x = 11/3 ≈ 3.67, y = 10 − 22/3 = 8/3 ≈ 2.67. This point satisfies y ≤ 5, so it is feasible.
P at (3.67, 2.67) = 400(3.67) + 500(2.67) ≈ 1,467 + 1,333 = ₹2,800.
Compare: (5, 0) gives ₹2,000; (0, 4.5) gives ₹2,250.
Binding constraints: flour and butter are both binding (no slack); the survey limit is slack (2.67 < 5).
Whole-number plan: (3, 3) uses 9 kg flour and 9 kg butter → P = 1,200 + 1,500 = ₹2,700 (better than (4, 2) at ₹2,600).
Marginal value: profit falls from ₹3,500 to ₹2,800 (₹3,300 to ₹2,700 in whole tins) for 5 kg of butter lost, roughly ₹140 per kg at the margin. The lesson is that once butter is binding, extra flour alone buys nothing.