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.
The profit on a bat and on a racket is Rs. 5 and Rs. 3.50 respectively. Find the maximum profit to the factory on a day when it produces (i) only bats, (ii) only rackets and (iii) works at full capacity.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Define Variables
Let:
- \( X \) = Number of cricket bats produced
- \( Y \) = Number of tennis rackets produced

### Step 2: Formulate the Objective Function
We want to maximize the profit, which can be expressed as:
\[ Z = 5X + 3.5Y \]
where \( Z \) is the total profit.

### Step 3: Set Up the Constraints
From the problem, we have the following constraints based on machine time and craftsman's time:

1. **Machine Time Constraint**:
- A cricket bat takes 1.5 hours and a tennis racket takes 2.5 hours.
- Total available machine time is 80 hours.
\[ 1.5X + 2.5Y \leq 80 \]
To simplify, we can multiply through by 2 to eliminate the decimals:
\[ 3X + 5Y \leq 160 \]

2. **Craftsman's Time Constraint**:
- A cricket bat takes 2 hours and a tennis racket takes 1.5 hours.
- Total available craftsman's time is 70 hours.
\[ 2X + 1.5Y \leq 70 \]
Again, we can multiply through by 2 to simplify:
\[ 4X + 3Y \leq 140 \]

3. **Non-negativity Constraints**:
\[ X \geq 0, \quad Y \geq 0 \]

### Step 4: Graph the Constraints
Now we will graph the inequalities:
1. \( 3X + 5Y \leq 160 \)
2. \( 4X + 3Y \leq 140 \)

To graph these, we find the intercepts for each line.

**For \( 3X + 5Y = 160 \)**:
- When \( X = 0 \): \( Y = 32 \) (intercept on Y-axis)
- When \( Y = 0 \): \( X = \frac{160}{3} \approx 53.33 \) (intercept on X-axis)

**For \( 4X + 3Y = 140 \)**:
- When \( X = 0 \): \( Y = \frac{140}{3} \approx 46.67 \) (intercept on Y-axis)
- When \( Y = 0 \): \( X = 35 \) (intercept on X-axis)

### Step 5: Identify the Feasible Region
The feasible region is the area where all constraints overlap. This region will be bounded by the lines we graphed and the axes.

### Step 6: Find Corner Points
The optimal solution will lie at one of the corner points of the feasible region. We can find the corner points by solving the equations of the lines:

1. **Intersection of \( 3X + 5Y = 160 \) and \( 4X + 3Y = 140 \)**:
- Solve these equations simultaneously to find the intersection point.

2. **Other corner points**:
- The intercepts we calculated earlier (0, 32), (35, 0), and any other points where the lines intersect the axes.

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = 5X + 3.5Y \) at each corner point to find the maximum profit.

1. At \( (0, 32) \):
\[ Z = 5(0) + 3.5(32) = 112 \]

2. At \( (35, 0) \):
\[ Z = 5(35) + 3.5(0) = 175 \]

3. At the intersection point (let's say it is \( (20, 20) \)):
\[ Z = 5(20) + 3.5(20) = 100 + 70 = 170 \]

### Step 8: Determine Maximum Profit
From the evaluations:
- Profit at \( (0, 32) \) = 112
- Profit at \( (35, 0) \) = 175
- Profit at \( (20, 20) \) = 170

The maximum profit is **Rs. 175** when the factory produces **only tennis rackets**.