To solve the problem of maximizing profit for the manufacturer producing products A and B, we will follow these steps:
### Step 1: Define Variables
Let:
- \( x \) = number of units of product A produced
- \( y \) = number of units of product B produced
### Step 2: Formulate the Objective Function
The profit from product A is ₹5 per unit and from product B is ₹6 per unit. Therefore, the objective function to maximize profit \( Z \) is:
\[
Z = 5x + 6y
\]
### Step 3: Set Up Constraints
We need to consider the constraints based on the machine hours available:
1. **For Machine 1**: Each unit of product A requires 3 hours and each unit of product B requires 2 hours. The total available hours for Machine 1 is 12 hours. Thus, the constraint is:
\[
3x + 2y \leq 12
\]
2. **For Machine 2**: Each unit of product A requires 3 hours and each unit of product B requires 1 hour. The total available hours for Machine 2 is 9 hours. Thus, the constraint is:
\[
3x + y \leq 9
\]
3. **Non-negativity Constraints**:
\[
x \geq 0, \quad y \geq 0
\]
### Step 4: Graph the Constraints
To graph the constraints, we will first find the intercepts for each constraint.
1. **For the first constraint \( 3x + 2y = 12 \)**:
- When \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) (Intercept at (0, 6))
- When \( y = 0 \): \( 3x = 12 \) → \( x = 4 \) (Intercept at (4, 0))
2. **For the second constraint \( 3x + y = 9 \)**:
- When \( x = 0 \): \( y = 9 \) (Intercept at (0, 9))
- When \( y = 0 \): \( 3x = 9 \) → \( x = 3 \) (Intercept at (3, 0))
### Step 5: Identify the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies all constraints. The feasible region will be bounded by the lines and the axes.
### Step 6: Find Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. **Intersection of \( 3x + 2y = 12 \) and \( 3x + y = 9 \)**:
- From \( 3x + y = 9 \), we can express \( y \) as \( y = 9 - 3x \).
- Substitute into the first equation:
\[
3x + 2(9 - 3x) = 12 \\
3x + 18 - 6x = 12 \\
-3x = -6 \\
x = 2
\]
- Substitute \( x = 2 \) back to find \( y \):
\[
y = 9 - 3(2) = 3
\]
- So, one corner point is \( (2, 3) \).
2. **Other corner points are**:
- \( (0, 0) \)
- \( (0, 6) \)
- \( (4, 0) \)
- \( (3, 0) \)
### Step 7: Evaluate the Objective Function at Each Corner Point
Now, we calculate the profit \( Z \) at each corner point:
1. At \( (0, 0) \): \( Z = 5(0) + 6(0) = 0 \)
2. At \( (0, 6) \): \( Z = 5(0) + 6(6) = 36 \)
3. At \( (4, 0) \): \( Z = 5(4) + 6(0) = 20 \)
4. At \( (2, 3) \): \( Z = 5(2) + 6(3) = 10 + 18 = 28 \)
### Step 8: Determine the Maximum Profit
The maximum profit occurs at the point \( (0, 6) \) where \( Z = 36 \).
### Final Answer
The manufacturer should produce 0 units of product A and 6 units of product B to achieve the maximum profit of ₹36.
