The tailors A and B are per Rs.225 and Rs.300 per day respectively. A can stitch 9 shirts and 6 pants while B can stitch 15 shirts and 6 pants per day. Form a linear programming problem to minimize the labour cost to produce atleast 90 shirts and 45 pants. Solve the problem graphically.

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let:
- \( X \) = Number of days tailor A works
- \( Y \) = Number of days tailor B works

### Step 2: Formulate the Objective Function
We need to minimize the labor cost, which can be expressed as:
\[ Z = 225X + 300Y \]

### Step 3: Formulate the Constraints
From the problem, we know:
- Tailor A can stitch 9 shirts and 6 pants per day.
- Tailor B can stitch 15 shirts and 6 pants per day.

We need to produce at least 90 shirts and 45 pants. Thus, we can set up the following inequalities:

1. For shirts:
\[ 9X + 15Y \geq 90 \]

2. For pants:
\[ 6X + 6Y \geq 45 \]

### Step 4: Simplify the Constraints
We can simplify the inequalities:
1. Dividing the first constraint by 3:
\[ 3X + 5Y \geq 30 \]

2. Dividing the second constraint by 6:
\[ X + Y \geq 7.5 \]

Since \( X \) and \( Y \) must be whole numbers (days), we can round up the second constraint:
\[ X + Y \geq 8 \]

### Step 5: Non-negativity Constraints
We also have the non-negativity constraints:
\[ X \geq 0 \]
\[ Y \geq 0 \]

### Step 6: Summary of the Linear Programming Problem
We now have the following linear programming problem:
- Minimize:
\[ Z = 225X + 300Y \]
- Subject to:
1. \( 3X + 5Y \geq 30 \)
2. \( X + Y \geq 8 \)
3. \( X \geq 0 \)
4. \( Y \geq 0 \)

### Step 7: Graphical Solution
1. **Plot the constraints** on a graph.
- For \( 3X + 5Y = 30 \):
- When \( X = 0 \), \( Y = 6 \) (point (0, 6))
- When \( Y = 0 \), \( X = 10 \) (point (10, 0))
- For \( X + Y = 8 \):
- When \( X = 0 \), \( Y = 8 \) (point (0, 8))
- When \( Y = 0 \), \( X = 8 \) (point (8, 0))

2. **Identify the feasible region** by checking which side of the lines satisfies the inequalities.

3. **Find the corner points** of the feasible region:
- Intersection of \( 3X + 5Y = 30 \) and \( X + Y = 8 \):
- Substitute \( Y = 8 - X \) into \( 3X + 5(8 - X) = 30 \)
- Solve for \( X \):
\[ 3X + 40 - 5X = 30 \]
\[ -2X = -10 \]
\[ X = 5 \]
\[ Y = 8 - 5 = 3 \]
- Intersection point is (5, 3).

4. **Evaluate the objective function** at each corner point:
- At (0, 8): \( Z = 225(0) + 300(8) = 2400 \)
- At (5, 3): \( Z = 225(5) + 300(3) = 1125 + 900 = 2025 \)
- At (10, 0): \( Z = 225(10) + 300(0) = 2250 \)

### Step 8: Conclusion
The minimum cost occurs at the point (5, 3):
- Tailor A works for 5 days and Tailor B works for 3 days.
- Minimum labor cost is \( Z = 2025 \).