Q.A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs 400 and each small van is Rs 200. Not more than Rs 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.

Concept: Optimization Word Problem (Linear Programming Problem)

Let $x$ = number of large vans, $y$ = number of small vans.

Step 1 – Objective function

Minimise cost: $Z = 400x + 200y$

Step 2 – Constraints

• Package capacity: $200x + 80y \geq 1200$ (at least 1200 packages)
• Budget: $400x + 200y \leq 3000$
• Large vans ≤ small vans: $x \leq y$ …

The core of any optimisation word problem is translating real-world conditions into mathematical relationships. Here, we have two types of vans — large and small — each with a carrying capacity and a cost. The goal is to move exactly 1200 packages at minimum cost, but we are limited by a budget of Rs 3000 and a rule that the number of large vans cannot exceed the number of small vans.

Let’s break it down step by step.

1. Define the decision variables

   Let $x$ = number of large vans used, and $y$ = number of small vans used.

   These are the quantities we can choose. Since we cannot hire a fraction of a van, $x$ and $y$ are non-negative integers — but in a standard LPP formulation, we first write them as $\geq 0$ and note the integer condition separately if needed.

2. Objective function: minimise cost

   Each large van costs Rs 400, each small van costs Rs 200.

   Total cost $Z = 400x + 200y$.

   We want to minimise $Z$.

3. Constraint 1: Capacity

   Large van carries 200 packages, small van carries 80.

   Total packages carried = $200x + 80y$.

   This must be at least 1200 (we can carry more, but not less):

4. Constraint 2: Budget Total cost cannot exceed Rs 3000:

5. Constraint 3: Large vans ≤ small vans The number of large vans cannot exceed the number of small vans:

6. Non-negativity constraints You cannot hire a negative number of vans: $x \geq 0, \quad y \geq 0$ …

Method: Formulating a Cost-Minimisation LPP

Use this to set up a word problem whose goal is to minimise a cost subject to a requirement to be met, a budget cap, and a comparison condition.

Steps

Step 1: Define the decision variables.

Let $x$ and $y$ be the counts of the two options chosen (e.g. two van types). Note they are non-negative (and often whole numbers).

Step 2: Translate each condition into an inequality — mind the direction.

• A quantity that must be met or exceeded (packages to transport) gives $\ge$: $(\text{cap. of }x)\,x+(\text{cap. of }y)\,y\ge \text{required}$.
• A cap (budget) gives $\le$: $(\text{cost }x)\,x+(\text{cost }y)\,y\le \text{budget}$. …

Common Mistakes

Mistake 1: Writing the capacity condition as $\le$ or $=$.

Why it's wrong: the firm must move at least 1200 packages, so $200x+80y\ge1200$; using $\le$ or $=$ wrongly caps or fixes the load. Correct approach: a minimum requirement is a $\ge$ constraint, a budget cap is $\le$ ($400x+200y\le3000$).

Mistake 2: Reversing the "large ≤ small" condition. …