A small firm manufactures neclaces and bracelets.The total number of neclaces and bracelets that it can handle per day is at most 24.It takes one hour to make a bracelet and half an hour to make a neclace.The maxium number of hours available per day is 16. If the profit on a neclace is ₹ 100 and that on a braclet is ₹ 300. Formulate an LLP for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of ecah must be produced.

Answer

To formulate the Linear Programming Problem (LPP) for the given scenario, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of necklaces produced per day
- \( y \) = number of bracelets produced per day

### Step 2: Formulate the Objective Function
The objective is to maximize profit. The profit from each necklace is ₹100 and from each bracelet is ₹300. Therefore, the objective function can be expressed as:
\[ \text{Maximize } Z = 100x + 300y \]

### Step 3: Set Up the Constraints
1. **Total Production Constraint**: The total number of necklaces and bracelets produced cannot exceed 24.
\[ x + y \leq 24 \]

2. **Time Constraint**: It takes 1 hour to make a bracelet and 0.5 hours (or half an hour) to make a necklace. The maximum available time is 16 hours. Therefore, the time constraint can be expressed as:
\[ x + 0.5y \leq 16 \]
To eliminate the decimal, we can multiply through by 2:
\[ 2x + y \leq 32 \]

3. **Non-negativity Constraints**: Since the number of necklaces and bracelets cannot be negative:
\[ x \geq 0, \quad y \geq 0 \]

4. **Minimum Production Constraints**: At least one of each item must be produced:
\[ x \geq 1, \quad y \geq 1 \]

### Step 4: Summary of the LPP
The Linear Programming Problem can be summarized as follows:

**Objective Function**:
\[ \text{Maximize } Z = 100x + 300y \]

**Subject to Constraints**:
1. \( x + y \leq 24 \)
2. \( 2x + y \leq 32 \)
3. \( x \geq 1 \)
4. \( y \geq 1 \)
5. \( x \geq 0 \)
6. \( y \geq 0 \)