A factory manufactures two types of screws, A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on hand-operated machines to manufacture a packet of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machines to manufacture a packet of screws 'B'. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of 70 paise and screws 'B' at a profit of Rs. 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximise his profit ? Formulate the above L.P.P. and solve it graphically and find the maximum profit.

Answer

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

### Step 1: Define Variables
Let:
- \( x \) = number of packets of screw A produced
- \( y \) = number of packets of screw B produced

### Step 2: Formulate the Objective Function
The profit from each type of screw is given as:
- Profit from screw A = 0.7 rupees per packet
- Profit from screw B = 1 rupee per packet

Thus, the total profit \( Z \) can be expressed as:
\[ Z = 0.7x + 1y \]

### Step 3: Set Up the Constraints
Each machine is available for 4 hours a day, which is equivalent to 240 minutes. The time taken for each type of screw is as follows:
- For screw A:
- Automatic machine: 4 minutes
- Hand-operated machine: 6 minutes
- For screw B:
- Automatic machine: 6 minutes
- Hand-operated machine: 3 minutes

From this, we can derive the following constraints:

1. Time on the automatic machine:
\[ 4x + 6y \leq 240 \quad \text{(1)} \]

2. Time on the hand-operated machine:
\[ 6x + 3y \leq 240 \quad \text{(2)} \]

3. Non-negativity constraints:
\[ x \geq 0, \quad y \geq 0 \quad \text{(3)} \]

### Step 4: Convert Inequalities to Equations
To graphically solve the inequalities, we convert them to equations:

1. \( 4x + 6y = 240 \)
2. \( 6x + 3y = 240 \)

### Step 5: Find Intercepts for Graphing
**For the first equation \( 4x + 6y = 240 \):**
- If \( x = 0 \):
\[ 6y = 240 \Rightarrow y = 40 \quad (0, 40) \]
- If \( y = 0 \):
\[ 4x = 240 \Rightarrow x = 60 \quad (60, 0) \]

**For the second equation \( 6x + 3y = 240 \):**
- If \( x = 0 \):
\[ 3y = 240 \Rightarrow y = 80 \quad (0, 80) \]
- If \( y = 0 \):
\[ 6x = 240 \Rightarrow x = 40 \quad (40, 0) \]

### Step 6: Plot the Lines and Identify Feasible Region
Plot the lines on a graph:
- Line from (0, 40) to (60, 0) for \( 4x + 6y = 240 \)
- Line from (0, 80) to (40, 0) for \( 6x + 3y = 240 \)

The feasible region is where the two lines intersect and is bounded by the axes.

### Step 7: Find Intersection Point
To find the intersection of the two lines:
1. From \( 4x + 6y = 240 \) and \( 6x + 3y = 240 \):
- Multiply the second equation by 2:
\[ 12x + 6y = 480 \]
- Subtract the first equation from this:
\[ 12x + 6y - (4x + 6y) = 480 - 240 \Rightarrow 8x = 240 \Rightarrow x = 30 \]
2. Substitute \( x = 30 \) back into the first equation:
\[ 4(30) + 6y = 240 \Rightarrow 120 + 6y = 240 \Rightarrow 6y = 120 \Rightarrow y = 20 \]
Thus, the intersection point is \( (30, 20) \).

### Step 8: Evaluate Profit at Corner Points
The corner points of the feasible region are:
1. \( (0, 0) \): \( Z = 0.7(0) + 1(0) = 0 \)
2. \( (40, 0) \): \( Z = 0.7(40) + 1(0) = 28 \)
3. \( (30, 20) \): \( Z = 0.7(30) + 1(20) = 21 + 20 = 41 \)
4. \( (0, 40) \): \( Z = 0.7(0) + 1(40) = 40 \)

### Step 9: Determine Maximum Profit
The maximum profit occurs at the point \( (30, 20) \):
- Number of packets of screw A = 30
- Number of packets of screw B = 20
- Maximum Profit = 41 rupees

### Final Answer
The factory owner should produce 30 packets of screw A and 20 packets of screw B to maximize profit, resulting in a maximum profit of 41 rupees.