To solve the problem of maximizing the profit from manufacturing teaching aids A and B, we will formulate it as a Linear Programming Problem (LPP) and solve it step by step.
### Step 1: Define the Variables
Let:
- \( X \) = number of pieces of type A manufactured
- \( Y \) = number of pieces of type B manufactured
### Step 2: Formulate the Objective Function
The profit from each piece of type A is 80 and from type B is 120. Therefore, the objective function to maximize profit \( Z \) is given by:
\[
Z = 80X + 120Y
\]
### Step 3: Formulate the Constraints
From the problem, we have the following constraints based on labor hours for fabricating and finishing:
1. **Fabricating Constraint**:
Each type A requires 9 hours and each type B requires 12 hours for fabricating. The total available hours for fabricating is 180.
\[
9X + 12Y \leq 180
\]
Simplifying this gives:
\[
3X + 4Y \leq 60 \quad \text{(Constraint 1)}
\]
2. **Finishing Constraint**:
Each type A requires 1 hour and each type B requires 3 hours for finishing. The total available hours for finishing is 30.
\[
1X + 3Y \leq 30
\]
This remains as:
\[
X + 3Y \leq 30 \quad \text{(Constraint 2)}
\]
3. **Non-negativity Constraints**:
Since the number of pieces cannot be negative, we have:
\[
X \geq 0, \quad Y \geq 0
\]
### Step 4: Graph the Constraints
To find the feasible region, we need to graph the constraints.
1. **For Constraint 1**: \( 3X + 4Y = 60 \)
- If \( X = 0 \): \( 4Y = 60 \) → \( Y = 15 \) (Point: (0, 15))
- If \( Y = 0 \): \( 3X = 60 \) → \( X = 20 \) (Point: (20, 0))
2. **For Constraint 2**: \( X + 3Y = 30 \)
- If \( X = 0 \): \( 3Y = 30 \) → \( Y = 10 \) (Point: (0, 10))
- If \( Y = 0 \): \( X = 30 \) (Point: (30, 0))
### Step 5: Identify the Feasible Region
Plotting these points on a graph will show the feasible region bounded by the lines of the constraints. The intersection points of the lines will also be calculated.
### Step 6: Find Intersection Points
To find the intersection of the two lines:
1. From \( 3X + 4Y = 60 \)
2. From \( X + 3Y = 30 \)
Substituting \( X = 30 - 3Y \) into the first equation:
\[
3(30 - 3Y) + 4Y = 60
\]
\[
90 - 9Y + 4Y = 60
\]
\[
-5Y = -30 \Rightarrow Y = 6
\]
Substituting \( Y = 6 \) back into \( X + 3Y = 30 \):
\[
X + 3(6) = 30 \Rightarrow X + 18 = 30 \Rightarrow X = 12
\]
Thus, the intersection point is \( (12, 6) \).
### Step 7: Evaluate the Objective Function at Corner Points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 10) \)
3. \( (12, 6) \)
4. \( (20, 0) \)
Calculating \( Z \) at each corner:
1. \( Z(0, 0) = 80(0) + 120(0) = 0 \)
2. \( Z(0, 10) = 80(0) + 120(10) = 1200 \)
3. \( Z(12, 6) = 80(12) + 120(6) = 960 + 720 = 1680 \)
4. \( Z(20, 0) = 80(20) + 120(0) = 1600 \)
### Step 8: Conclusion
The maximum profit occurs at the point \( (12, 6) \), meaning the company should manufacture:
- 12 pieces of type A
- 6 pieces of type B
