Q.A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

Concept: Linear Programming Constraints — translating real-world limits into mathematical inequalities.

Step 1: Define variables

Let $x$ = number of type A sweaters, $y$ = number of type B sweaters.

Step 2: Constraints from the problem

• Total sweaters: $x + y \leq 300$
• Cost limit: $360x + 120y \leq 72000$ (simplify by dividing by 120: $3x + y \leq 600$)
• Type B not exceed type A by more than 100: $y - x \leq 100$
• Non-negativity: $x \geq 0, y \geq 0$ …

Why This Approach Works

Linear Programming Problems (LPP) are about optimising a linear objective (here, profit) under linear constraints. The key is to translate the real-world conditions into mathematical inequalities. Each constraint represents a limit — on total production, budget, or a relationship between the two products. The non-negativity constraints are always present because you cannot produce a negative number of sweaters.

Let’s define the decision variables first:

• Let $x$ = number of type A sweaters produced per day.
• Let $y$ = number of type B sweaters produced per day.

Now, we convert each condition into an inequality.

Step-by-Step Formulation

1. Objective Function (Profit to maximise)

Profit from type A: Rs 200 per sweater → $200x$

Profit from type B: Rs 120 per sweater → $120y$

Total profit $Z = 200x + 120y$.

We want to maximise $Z$.

2. Constraint 1: Maximum number of sweaters

“At most 300 sweaters” means the total production cannot exceed 300.

3. Constraint 2: Maximum cost (budget)

Cost per type A: Rs 360 → $360x$

Cost per type B: Rs 120 → $120y$

Total cost cannot exceed Rs 72,000.

We can simplify this by dividing through by 120 (the common factor):

4. Constraint 3: Relationship between type B and type A

“The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100.” …

Method: Formulating a Linear Programming Problem

Use this whenever a word problem describes one quantity to make as large or as small as possible (profit, cost, distance) under several stated limits, and asks you only to set it up as an LPP -- not to solve it.

Steps

Step 1: Name the decision variables.

Pick the two quantities you are free to choose and give each a letter, e.g. let $x$ = number of items of the first kind and $y$ = number of the second. Every line that follows must be written using only these variables.

Step 2: Write the objective function.

Identify the single quantity to be optimised and express it as

where $a$ and $b$ are the per-unit contributions read straight from the problem (here, the profit on each type). State explicitly whether you maximise or minimise it.

Step 3: Convert each stated limit into a linear inequality. …

Common Mistakes

Mistake 1: Forgetting the non-negativity constraints.

Why it's wrong: dropping $x \ge 0,\ y \ge 0$ leaves the model allowing a negative number of sweaters, which is meaningless. Correct approach: always append $x \ge 0,\ y \ge 0$ -- they are part of the LPP, not optional.

Mistake 2: Mistranslating "type B cannot exceed type A by more than 100."

Why it's wrong: writing $y \le 100$ or $y - x \ge 100$ changes the condition entirely. The phrase caps the difference $y - x$. Correct approach: write $y - x \le 100$ (equivalently $y \le x + 100$). …