Food Truck Menu Mix Under Time and Item Limits
A two-person food truck sells only tacos and burritos at a weekday office-park spot. Fixed daily costs - permit, fuel, and commissary rent - total $180 regardless of how much is sold. Each taco uses $2.10 of ingredients and sells for $4.50; each burrito uses $3.40 and sells for $6.25. The service window and grill allow at most 220 items to be sold in a day, and total prep-and-serve labour is capped at 420 minutes: 1.5 minutes per taco and 2.5 minutes per burrito, including taking payment. The operators can prepare any mix of the two items and expect to sell out of whatever they make. Before tomorrow's shift they want to know the minimum number of items that covers the day's fixed costs if they sell only one product, whether a plan of 140 tacos and 80 burritos is workable, and which feasible mix maximizes daily profit.
1. Write the constraint system and daily profit function, then test the plan of 140 tacos and 80 burritos for feasibility and profit.
Read the worked answer
Define variables: t = tacos made, b = burritos made, both non-negative integers.
Constraints:
- Item cap: t + b ≤ 220
- Labour: 1.5t + 2.5b ≤ 420
- Non-negativity/integrality: t, b ≥ 0
Profit function (revenue minus ingredient cost minus fixed cost): P = (4.50 − 2.10)t + (6.25 − 3.40)b − 180 = 2.40t + 2.85b − 180.
Test (140, 80):
- Items: 140 + 80 = 220 ≤ 220 ✓ (cap is exactly binding)
- Labour: 1.5(140) + 2.5(80) = 210 + 200 = 410 ≤ 420 ✓ with 10 minutes to spare
- Profit: 2.40(140) + 2.85(80) − 180 = 336 + 228 − 180 = $384.00
The plan is feasible and clears fixed costs comfortably.
2. Find the break-even volume if the truck sells only tacos, and separately if it sells only burritos. Explain why the two numbers differ.
Read the worked answer
Tacos only: 2.40t = 180 → t = 75 tacos. Labour used: 75 × 1.5 = 112.5 min, well under 420, so the time constraint is irrelevant.
Burritos only: 2.85b = 180 → b = 63.16. Since items are whole, check both sides:
- 63 burritos: 2.85 × 63 = 179.55 → profit = −$0.45 (still a loss)
- 64 burritos: 2.85 × 64 = 182.40 → profit = $2.40 ✓
- Labour for 64: 64 × 2.5 = 160 min, also under 420. So break-even is 75 tacos or 64 burritos.
Why they differ: break-even on a single product depends only on contribution margin per item ($2.40 vs $2.85), so the higher-margin burrito needs fewer units. That ranking reverses once labour is scarce - see the next task - which is exactly the trap in reading break-even as a menu recommendation.
3. Which feasible mix maximizes daily profit? Justify using the corner points of the feasible region, and comment on what drives the answer.
Read the worked answer
Corner points of the region bounded by t + b ≤ 220, 1.5t + 2.5b ≤ 420, t, b ≥ 0:
- (0, 0): P = −$180
- (220, 0): labour 330 ✓; P = 2.40(220) − 180 = 528 − 180 = $348.00
- (0, 168): labour-limited, 420 ÷ 2.5 = 168; P = 2.85(168) − 180 = 478.80 − 180 = $298.80
- Intersection of the two lines: substitute t = 220 − b into 1.5t + 2.5b = 420 → 330 + b = 420 → b = 90, t = 130. P = 2.40(130) + 2.85(90) − 180 = 312 + 256.50 − 180 = $388.50
Maximum is $388.50 at 130 tacos and 90 burritos.
What drives it: the burrito wins on margin per item ($2.85 vs $2.40) but loses badly on margin per minute of labour ($1.14 vs $1.60). Because 420 minutes is the binding resource at the optimum, the taco-heavy mix wins. If the operators hired a third person and labour rose above roughly 550 minutes, the item cap would bind instead and the answer would flip toward burritos.