A firm manufactures two products A and B on which the profit earned per unit are Rs. 3 and Rs. 4 respectively. Each product is processed on two machines `M_(1)` and `M_(2)`. Product A requires one minute of processing time on `M_(1)` and two minutes on `M_(2)` while B requires one minute on `M_(1)` and one minute on `M_(2)`. Machine `M_(1)` is available for not more than 7 hours 30 minutes, while machine `M_(2)` is available for 10 hrs during any working day. Find the number of units of products A and B to manufactured to get maximum profit.

Answer

To solve the problem of maximizing profit from manufacturing products A and B, we can follow these steps:

### Step 1: Define the 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 earned per unit for product A is Rs. 3, and for product B is Rs. 4. Therefore, the total profit \( Z \) can be expressed as:
\[ Z = 3X + 4Y \]
We want to maximize \( Z \).

### Step 3: Identify the Constraints
From the problem statement, we have the following constraints based on the processing time available on machines \( M_1 \) and \( M_2 \):

1. **Machine M1**:
- Each unit of A requires 1 minute and each unit of B requires 1 minute.
- Total available time for M1 is 7 hours 30 minutes = 450 minutes.
\[ X + Y \leq 450 \]

2. **Machine M2**:
- Each unit of A requires 2 minutes and each unit of B requires 1 minute.
- Total available time for M2 is 10 hours = 600 minutes.
\[ 2X + Y \leq 600 \]

### Step 4: Set Up the System of Inequalities
We can summarize the constraints as:
1. \( X + Y \leq 450 \)
2. \( 2X + Y \leq 600 \)
3. \( X \geq 0 \)
4. \( Y \geq 0 \)

### Step 5: Graph the Constraints
To find the feasible region, we can graph the inequalities on a coordinate system where the x-axis represents product A (X) and the y-axis represents product B (Y).

1. **For \( X + Y = 450 \)**:
- When \( X = 0 \), \( Y = 450 \) (point A).
- When \( Y = 0 \), \( X = 450 \) (point B).

2. **For \( 2X + Y = 600 \)**:
- When \( X = 0 \), \( Y = 600 \) (point C).
- When \( Y = 0 \), \( X = 300 \) (point D).

### Step 6: Identify the Corner Points
The feasible region is bounded by the lines and the axes. The corner points of the feasible region can be found by solving the equations:
1. Intersection of \( X + Y = 450 \) and \( 2X + Y = 600 \):
\[ X + Y = 450 \quad (1) \]
\[ 2X + Y = 600 \quad (2) \]
Subtract (1) from (2):
\[ 2X + Y - (X + Y) = 600 - 450 \]
\[ X = 150 \]
Substitute \( X = 150 \) into (1):
\[ 150 + Y = 450 \Rightarrow Y = 300 \]
So, one corner point is \( (150, 300) \).

2. The other corner points are:
- \( (0, 450) \) from \( X + Y = 450 \)
- \( (300, 0) \) from \( 2X + Y = 600 \)
- \( (0, 600) \) is not feasible since \( Y \) cannot exceed 450.

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 3X + 4Y \) at each corner point:
1. At \( (0, 450) \):
\[ Z = 3(0) + 4(450) = 1800 \]
2. At \( (150, 300) \):
\[ Z = 3(150) + 4(300) = 450 + 1200 = 1650 \]
3. At \( (300, 0) \):
\[ Z = 3(300) + 4(0) = 900 \]

### Step 8: Determine the Maximum Profit
The maximum profit occurs at the point \( (0, 450) \) with a profit of Rs. 1800.

### Conclusion
To maximize profit, the firm should manufacture:
- **0 units of product A**
- **450 units of product B**