Quantitative - Two products, three ingredients: the bottleneck problem
A home bakery in Jeddah makes two products from the same three ingredients. A box of date cookies uses 1.5 kg of dates, 0.4 kg of flour and 0.3 kg of butter, and earns 35 SAR of profit. A box of date bars uses 1.0 kg of dates, 0.6 kg of flour and 0.2 kg of butter, and earns 30 SAR. At the start of the week the bakery holds 60 kg of dates, 24 kg of flour and 10 kg of butter, and every box produced can be sold. A customer has already asked for 20 boxes of each product. The owner usually decides by 'make more of whatever earns more per box', but this week she suspects that rule will leave money on the table.
1. Can the bakery fill the customer's request for 20 boxes of each product? What remains of each ingredient afterwards?
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Check each ingredient separately.
- Dates: 20 × 1.5 + 20 × 1.0 = 30 + 20 = 50 kg used, so 10 kg left of 60.
- Flour: 20 × 0.4 + 20 × 0.6 = 8 + 12 = 20 kg used, so 4 kg left of 24.
- Butter: 20 × 0.3 + 20 × 0.2 = 6 + 4 = 10 kg used, so 0 kg left of 10.
Yes, the order fits - but only just, and the binding ingredient is butter. Butter is the scarce resource that will control the week's profit, and the leftover 10 kg of dates and 4 kg of flour are wasted capacity unless more butter is available. That is the signal to look at how much profit each product earns per kilogram of butter, not per box.
2. Find the mix that maximises weekly profit, and explain why 'make the higher-profit box' is the wrong rule here.
Read the worked answer
Write the constraints (C = cookie boxes, B = bar boxes):
- Dates: 1.5C + B ≤ 60
- Flour: 0.4C + 0.6B ≤ 24
- Butter: 0.3C + 0.2B ≤ 10
Test the corners of the feasible region:
- Butter only, no cookies: B = 50 from butter, but flour caps B at 40 → (0, 40), profit 1,200 SAR.
- Flour and butter together: 0.4C + 0.6B = 24 and 0.3C + 0.2B = 10 → multiply the second by 3: 0.9C + 0.6B = 30; subtract: 0.5C = 6 → C = 12, B = 32. Dates used: 18 + 32 = 50 ≤ 60 ✓. Profit = 12 × 35 + 32 × 30 = 420 + 960 = 1,380 SAR.
- Cookies only: butter caps C at 33.3, profit ≈ 1,167 SAR.
Best mix: 12 cookie boxes and 32 bar boxes, 1,380 SAR.
Why the rule misleads: cookies earn more per box (35 vs 30) but each box eats 0.3 kg of butter against 0.2 kg for a bar. Per kilogram of the scarce input, cookies return 35 ÷ 0.3 ≈ 117 SAR and bars return 30 ÷ 0.2 = 150 SAR. When butter is the bottleneck, the lower-profit box is the better use of it.
3. A supplier offers extra butter at 40 SAR per kilogram, up to 4 kg. How much should the owner buy, and what is the net gain?
Read the worked answer
Extra butter relaxes only the butter constraint, so profit can rise until a different constraint binds.
Step 1 - where does butter stop mattering? Raise butter to 12 kg and solve flour with butter: 0.4C + 0.6B = 24 and 0.3C + 0.2B = 12 → 0.9C + 0.6B = 36; subtract: 0.5C = 12 → C = 24, B = 24. Dates check: 1.5(24) + 24 = 60 ✓ exactly; flour = 24 ✓. Profit = 24 × 35 + 24 × 30 = 840 + 720 = 1,560 SAR. Beyond this point dates and flour are both exhausted, so more butter adds nothing.
Step 2 - value of the extra butter. Moving from 10 kg to 12 kg of butter raises profit from 1,380 to 1,560, i.e. 90 SAR per kilogram. At a cost of 40 SAR per kilogram, each of those 2 kg nets 50 SAR.
Answer: buy 2 kg only, pay 80 SAR, gain 180 SAR, net +100 SAR. Buying the full 4 kg would waste 80 SAR on butter the bakery cannot convert into boxes.