A company manufacutres two types of sweaters type A and type B. It costs 360 to make type A sweater and 120 to make a type B sweater. The company can make atmost 300 sweater and spent atmost 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 200 for each sweater of type A and 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

Explanation

To formulate the given problem as a Linear Programming Problem (LPP) to maximize the profit for the company, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = Number of type A sweaters manufactured
- \( y \) = Number of type B sweaters manufactured

### Step 2: Formulate the Objective Function
The objective is to maximize the profit \( Z \). The profit from type A sweaters is \( 200 \) per unit, and from type B sweaters, it is \( 120 \) per unit. Therefore, the objective function can be expressed as:
\[ Z = 200x + 120y \]

### Step 3: Identify the Constraints
1. **Cost Constraint**: The total cost of manufacturing cannot exceed \( 72,000 \). The cost of type A sweater is \( 360 \) and type B sweater is \( 120 \). Thus, the constraint is:
\[ 360x + 120y \leq 72000 \]

2. **Production Limit Constraint**: The total number of sweaters produced cannot exceed \( 300 \):
\[ x + y \leq 300 \]

3. **Difference Constraint**: The number of type B sweaters cannot exceed the number of type A sweaters by more than \( 100 \):
\[ y - x \leq 100 \quad \text{or} \quad y \leq x + 100 \]

4. **Non-negativity Constraints**: The number of sweaters produced cannot be negative:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Summarize the LPP
Now, we can summarize the Linear Programming Problem as follows:

**Maximize:**
\[ Z = 200x + 120y \]

**Subject to:**
\[ \begin{align*} 360x + 120y & \leq 72000 \quad \text{(Cost Constraint)} \\ x + y & \leq 300 \quad \text{(Production Limit)} \\ y & \leq x + 100 \quad \text{(Difference Constraint)} \\ x & \geq 0 \\ y & \geq 0 \end{align*} \]