A manufacturing company makes two types of teaching aids A and B . Each type of A requires 9 labour hours for fabricating and 1 labour hours for finishing.Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing.For fabricating and finishing, the maximum labour hours abailable per week are 180 and 30 respectively.The company makes a profit of ₹ 80 on each type A and ₹ 120 on each type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically.What is the maximum profit per week?

Answer

To solve the problem step-by-step, we will formulate the linear programming problem (LPP) and then solve it graphically.

### Step 1: Define the Variables
Let:
- \( x \) = number of type A teaching aids produced
- \( y \) = number of type B teaching aids produced

### Step 2: Formulate the Objective Function
The profit from type A is ₹80 and from type B is ₹120. Therefore, the objective function to maximize profit \( Z \) is:
\[ Z = 80x + 120y \]

### Step 3: Formulate the Constraints
Based on the problem statement, we have the following constraints:

1. **Fabrication Constraint**: Each type A requires 9 hours and each type B requires 12 hours. The total available hours for fabrication is 180 hours.
\[ 9x + 12y \leq 180 \]

2. **Finishing Constraint**: Each type A requires 1 hour and each type B requires 3 hours. The total available hours for finishing is 30 hours.
\[ x + 3y \leq 30 \]

3. **Non-negativity Constraints**:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Graph the Constraints
To graph the constraints, we will convert the inequalities into equations and find the intercepts.

1. **For the Fabrication Constraint**:
\[ 9x + 12y = 180 \]
- When \( x = 0 \): \( 12y = 180 \) → \( y = 15 \) (y-intercept)
- When \( y = 0 \): \( 9x = 180 \) → \( x = 20 \) (x-intercept)

2. **For the Finishing Constraint**:
\[ x + 3y = 30 \]
- When \( x = 0 \): \( 3y = 30 \) → \( y = 10 \) (y-intercept)
- When \( y = 0 \): \( x = 30 \) (x-intercept)

### Step 5: Identify the Feasible Region
Plot the lines on a graph:
- The line from the fabrication constraint intersects at (20, 0) and (0, 15).
- The line from the finishing constraint intersects at (30, 0) and (0, 10).

The feasible region is the area where both constraints overlap, and it is bounded by the axes.

### Step 6: Find the Corner Points
The corner points of the feasible region can be identified as follows:
1. Intersection of the axes: (0, 0)
2. Intersection of the fabrication line and the finishing line.

To find the intersection of the lines:
From \( 9x + 12y = 180 \) and \( x + 3y = 30 \):
- Multiply the second equation by 4:
\[ 4x + 12y = 120 \]
- Subtract this from the first equation:
\[ 9x + 12y - (4x + 12y) = 180 - 120 \]
\[ 5x = 60 \quad \Rightarrow \quad x = 12 \]
- Substitute \( x = 12 \) into \( x + 3y = 30 \):
\[ 12 + 3y = 30 \quad \Rightarrow \quad 3y = 18 \quad \Rightarrow \quad y = 6 \]

Thus, the corner points are:
- (0, 10)
- (20, 0)
- (12, 6)

### Step 7: Evaluate the Objective Function at Each Corner Point
1. At (0, 10):
\[ Z = 80(0) + 120(10) = 1200 \]
2. At (20, 0):
\[ Z = 80(20) + 120(0) = 1600 \]
3. At (12, 6):
\[ Z = 80(12) + 120(6) = 960 + 720 = 1680 \]

### Step 8: Determine the Maximum Profit
The maximum profit occurs at the point (12, 6):
\[ \text{Maximum Profit} = ₹1680 \]

### Conclusion
To maximize profit, the company should produce:
- 12 pieces of type A
- 6 pieces of type B