Applications to Diet Manufacturing and Transport: Exam Tips

Linear programming shines in three classic settings: diet problems (meet nutrition needs at minimum cost), manufacturing problems (respect machine and labour hours while maximising profit), and transport problems (ship goods from sources to destinations at minimum cost). Examiners return to these three templates again and again.

They matter because the real world is full of "best use of limited resources" decisions. Learning these templates gives you a ready-made translation kit: read the words, write the inequalities, and the mathematics takes over from there.

Key Concepts

• Decision variables: the quantities you choose, such as kilograms of food A and food B, or units of product X and product Y.
• Objective: minimise cost or maximise profit, written as Z = ax + by.
• Requirement constraints: "at least" becomes ≥ (nutrition, demand).
• Capacity constraints: "at most" or "no more than" becomes ≤ (machine hours, supply, budget).
• Translation rule: define every variable in words first — marks and clarity both depend on it.

How It Works

1. Diet problem: minimise cost C = c₁x + c₂y subject to nutrient constraints like p₁x + p₂y ≥ required units, with x, y ≥ 0.
2. Manufacturing problem: maximise profit P = p₁x + p₂y subject to resource constraints like h₁x + h₂y ≤ available hours, with x, y ≥ 0.
3. Transport problem: minimise shipping cost subject to supply limits (≤) at sources and demand fulfilment (≥) at destinations.
4. Solve by the graphical or corner-point method, then translate the numbers back into the original story.
5. Exam tip: always state what x and y stand for, list constraints with reasons, and verify the final answer in the original wording.

Worked Example

Diet problem: food A costs ₹20 per kg and gives 2 units of protein per kg; food B costs ₹30 per kg and gives 4 units of protein per kg. You need at least 24 units of protein at minimum cost. Let x = kg of A and y = kg of B. Minimise C = 20x + 30y subject to 2x + 4y ≥ 24, x ≥ 0, y ≥ 0. The boundary line 2x + 4y = 24 simplifies to x + 2y = 12, with intercepts (12, 0) and (0, 6). The feasible region lies above this line and is unbounded upward, so the minimum occurs at a corner of the boundary: at (12, 0), C = 20(12) + 30(0) = 240; at (0, 6), C = 20(0) + 30(6) = 180. Check feasibility of (0, 6): 2(0) + 4(6) = 24 ≥ 24, satisfied. The minimum cost is ₹180, achieved by buying 6 kg of food B and none of food A.

Common Misconceptions

• Mixing up ≥ and ≤. "At least 24 units of protein" means 2x + 4y ≥ 24 — the ≥ side, not ≤.
• Forgetting non-negativity. Negative kilograms of food are meaningless; x ≥ 0 and y ≥ 0 are always part of the model.
• Reporting x = 0 without comment. A zero value is a real decision ("buy no food A") — state it in words, not just as a number.

Real-World Applications

• Hospitals and schools design minimum-cost nutritious meal plans.
• Factories schedule product mixes across limited machines and shifts.
• Logistics companies minimise freight cost from warehouses to retail stores.
• Airlines and railways allocate seats and routes under capacity limits.

Diet, manufacturing and transport problems share one structure: an objective plus resource limits. Translate the words into inequalities carefully, and the corner-point method delivers the optimal plan.