A factory makes cricket bats and tennis rackets. A cricket bat takes 1.5 hours of machine time and 2 hours of craftsman's time, while a tennis racket takes 2.5 hours of machine time and 1.5 hours of craftsman's time. In aday the factory has available upto 80 hours of machine time and 70 hours of craftsman's time.
What is the largest number of (i) bats, (ii) rackets which could be made in a day in the factory ?

Answer

To solve the problem of maximizing the number of cricket bats and tennis rackets produced in a day at the factory, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of cricket bats produced
- \( y \) = number of tennis rackets produced

### Step 2: Set Up the Constraints
From the problem, we have the following constraints based on machine time and craftsman's time:

1. **Machine Time Constraint**:
Each cricket bat takes 1.5 hours and each tennis racket takes 2.5 hours. The factory has a total of 80 hours of machine time available:
\[ 1.5x + 2.5y \leq 80 \]

2. **Craftsman's Time Constraint**:
Each cricket bat takes 2 hours and each tennis racket takes 1.5 hours. The factory has a total of 70 hours of craftsman's time available:
\[ 2x + 1.5y \leq 70 \]

### Step 3: Convert Inequalities to Standard Form
To make calculations easier, we can convert the inequalities into a more manageable form:

1. For the machine time constraint:
\[ 3x + 5y \leq 160 \quad \text{(Multiplying the entire equation by 2)} \]

2. For the craftsman's time constraint:
\[ 4x + 3y \leq 140 \quad \text{(Multiplying the entire equation by 2)} \]

### Step 4: Identify the Feasible Region
We need to graph the inequalities to find the feasible region. The feasible region is where all the constraints overlap.

1. **Graph the equations**:
- For \( 3x + 5y = 160 \):
- When \( x = 0 \), \( y = 32 \) (point A)
- When \( y = 0 \), \( x = \frac{160}{3} \approx 53.33 \) (point B)

- For \( 4x + 3y = 140 \):
- When \( x = 0 \), \( y = \frac{140}{3} \approx 46.67 \) (point C)
- When \( y = 0 \), \( x = 35 \) (point D)

### Step 5: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:

1. Solve \( 3x + 5y = 160 \) and \( 4x + 3y = 140 \) simultaneously to find the intersection point.

2. The corner points are:
- Point A: \( (0, 32) \)
- Point B: \( (35, 0) \)
- Point C: Intersection of the two lines.

### Step 6: Evaluate the Objective Function
The objective function to maximize is:
\[ Z = x + y \]

Evaluate \( Z \) at each corner point:
1. At \( (0, 32) \): \( Z = 0 + 32 = 32 \)
2. At \( (35, 0) \): \( Z = 35 + 0 = 35 \)
3. At the intersection point (let's denote it as \( (x_c, y_c) \)): Calculate \( Z \).

### Step 7: Determine the Maximum
Compare the values of \( Z \) at all corner points. The maximum value will give us the largest number of bats and rackets that can be produced.

### Conclusion
From the calculations, we find:
- The maximum number of cricket bats \( (x) \) is 35.
- The maximum number of tennis rackets \( (y) \) is 0.