Two tailors A and B earns 15 and 20 per dayrespectively. A can stitch 6 shirts and 4 paints while Bcan stitch 10 shirts and 4 paints per day. To minimise the cost to stitch 60 shirts and 32 paints, how manydays should they work?

Explanation

To solve the problem of how many days tailors A and B should work to minimize the cost of stitching 60 shirts and 32 pants, we can follow these steps:

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

### Step 2: Set Up the Production Constraints
From the problem:
- Tailor A can stitch 6 shirts and 4 pants per day.
- Tailor B can stitch 10 shirts and 4 pants per day.

Thus, the total production constraints can be formulated as:
1. For shirts:
\[ 6x + 10y \geq 60 \]
2. For pants:
\[ 4x + 4y \geq 32 \]

### Step 3: Set Up the Cost Function
The cost incurred by each tailor per day is:
- Tailor A earns \( 15 \) per day.
- Tailor B earns \( 20 \) per day.

Thus, the total cost function to minimize is:
\[ Z = 15x + 20y \]

### Step 4: Formulate the Inequalities
We can rewrite the inequalities from the constraints:
1. From \( 6x + 10y \geq 60 \):
\[ 6x + 10y = 60 \]
2. From \( 4x + 4y \geq 32 \):
\[ 4x + 4y = 32 \]

### Step 5: Find Intercepts for Graphing
To graph the equations, we need to find the intercepts.

**For the first equation \( 6x + 10y = 60 \)**:
- When \( x = 0 \):
\[ 10y = 60 \implies y = 6 \quad \text{(Point: (0, 6))} \]
- When \( y = 0 \):
\[ 6x = 60 \implies x = 10 \quad \text{(Point: (10, 0))} \]

**For the second equation \( 4x + 4y = 32 \)**:
- When \( x = 0 \):
\[ 4y = 32 \implies y = 8 \quad \text{(Point: (0, 8))} \]
- When \( y = 0 \):
\[ 4x = 32 \implies x = 8 \quad \text{(Point: (8, 0))} \]

### Step 6: Graph the Inequalities
Plot the points (0, 6), (10, 0), (0, 8), and (8, 0) on a graph. The feasible region will be bounded by these lines in the first quadrant.

### Step 7: Identify Corner Points
The corner points of the feasible region are:
1. (0, 6)
2. (10, 0)
3. (0, 8)
4. (8, 0)
5. Intersection of the two lines.

To find the intersection of the lines \( 6x + 10y = 60 \) and \( 4x + 4y = 32 \), we can solve these equations simultaneously.

### Step 8: Solve the System of Equations
From \( 4x + 4y = 32 \):
\[ x + y = 8 \implies y = 8 - x \]

Substituting \( y \) in the first equation:
\[ 6x + 10(8 - x) = 60 \]
\[ 6x + 80 - 10x = 60 \]
\[ -4x + 80 = 60 \implies -4x = -20 \implies x = 5 \]
Substituting \( x = 5 \) back to find \( y \):
\[ y = 8 - 5 = 3 \]
Thus, the intersection point is (5, 3).

### Step 9: Evaluate the Cost Function at Corner Points
Now evaluate \( Z = 15x + 20y \) at the corner points:
1. At (0, 6):
\[ Z = 15(0) + 20(6) = 120 \]
2. At (10, 0):
\[ Z = 15(10) + 20(0) = 150 \]
3. At (5, 3):
\[ Z = 15(5) + 20(3) = 75 + 60 = 135 \]
4. At (0, 8):
\[ Z = 15(0) + 20(8) = 160 \]
5. At (8, 0):
\[ Z = 15(8) + 20(0) = 120 \]

### Step 10: Determine Minimum Cost
The minimum cost occurs at the point (5, 3) with \( Z = 135 \).

### Conclusion
To minimize the cost of stitching 60 shirts and 32 pants, tailor A should work for **5 days** and tailor B should work for **3 days**.