Two tailors A and B are paid ₹ 225 and ₹ 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 at least 90 shirts and 48 pants.Solve the problem graphically.

Answer

To solve the problem of minimizing the labor cost for tailors A and B while producing at least 90 shirts and 48 pants, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of days tailor A works
- \( y \) = number of days tailor B works

### Step 2: Formulate the Objective Function
The labor cost per day for each tailor is given as:
- Tailor A: ₹225 per day
- Tailor B: ₹300 per day

Thus, the objective function to minimize the total labor cost \( Z \) is:
\[ Z = 225x + 300y \]

### 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 48 pants, which gives us the following constraints:

1. For shirts:
\[ 9x + 15y \geq 90 \]
Simplifying this constraint by dividing everything by 3:
\[ 3x + 5y \geq 30 \quad \text{(Constraint 1)} \]

2. For pants:
\[ 6x + 6y \geq 48 \]
Simplifying this constraint by dividing everything by 6:
\[ x + y \geq 8 \quad \text{(Constraint 2)} \]

### Step 4: Non-negativity Constraints
Since the number of days cannot be negative, we have:
\[ x \geq 0, \quad y \geq 0 \]

### Step 5: Graph the Constraints
To graph the constraints, we will convert the inequalities into equations:

1. From \( 3x + 5y = 30 \):
- When \( x = 0 \), \( y = 6 \) (y-intercept)
- When \( y = 0 \), \( x = 10 \) (x-intercept)

2. From \( x + y = 8 \):
- When \( x = 0 \), \( y = 8 \) (y-intercept)
- When \( y = 0 \), \( x = 8 \) (x-intercept)

### Step 6: Identify the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies both inequalities. The feasible region will be bounded by the lines and the axes.

### Step 7: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 3x + 5y = 30 \) and \( x + y = 8 \):
- From \( x + y = 8 \), we can express \( y = 8 - x \).
- Substitute into \( 3x + 5(8 - x) = 30 \):
\[ 3x + 40 - 5x = 30 \implies -2x = -10 \implies x = 5 \]
- Substitute \( x = 5 \) back into \( y = 8 - x \):
\[ y = 8 - 5 = 3 \]
- So, one corner point is \( (5, 3) \).

2. The other corner points are:
- \( (0, 6) \) from \( 3x + 5y = 30 \)
- \( (10, 0) \) from \( 3x + 5y = 30 \)
- \( (0, 8) \) from \( x + y = 8 \)
- \( (8, 0) \) from \( x + y = 8 \)

### Step 8: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = 225x + 300y \) at each corner point:
1. At \( (0, 6) \):
\[ Z = 225(0) + 300(6) = 1800 \]
2. At \( (10, 0) \):
\[ Z = 225(10) + 300(0) = 2250 \]
3. At \( (5, 3) \):
\[ Z = 225(5) + 300(3) = 1125 + 900 = 2025 \]
4. At \( (0, 8) \):
\[ Z = 225(0) + 300(8) = 2400 \]
5. At \( (8, 0) \):
\[ Z = 225(8) + 300(0) = 1800 \]

### Step 9: Determine the Minimum Value
The minimum value of \( Z \) occurs at the point \( (5, 3) \) with:
\[ Z = 2025 \]

### Conclusion
The minimum labor cost to produce at least 90 shirts and 48 pants is ₹2025 when tailor A works for 5 days and tailor B works for 3 days.