To solve the problem, we will formulate it as a Linear Programming Problem (LPP) and then find the optimal solution to maximize profit.
### Step 1: Define the Variables
Let:
- \( x \) = number of Type A circuits produced
- \( y \) = number of Type B circuits produced
### Step 2: Formulate the Objective Function
The profit from Type A circuits is 50 per unit and from Type B circuits is 60 per unit. Therefore, the objective function to maximize profit \( Z \) is:
\[
Z = 50x + 60y
\]
### Step 3: Formulate the Constraints
We have the following constraints based on the availability of resistors, transistors, and capacitors:
1. **Resistors Constraint**:
Each Type A circuit requires 20 resistors and each Type B circuit requires 10 resistors. The total available resistors are 200. Thus, the constraint is:
\[
20x + 10y \leq 200
\]
Simplifying this gives:
\[
2x + y \leq 20 \quad \text{(Constraint 1)}
\]
2. **Transistors Constraint**:
Each Type A circuit requires 10 transistors and each Type B circuit requires 20 transistors. The total available transistors are 120. Thus, the constraint is:
\[
10x + 20y \leq 120
\]
Simplifying this gives:
\[
x + 2y \leq 12 \quad \text{(Constraint 2)}
\]
3. **Capacitors Constraint**:
Each Type A circuit requires 10 capacitors and each Type B circuit requires 30 capacitors. The total available capacitors are 150. Thus, the constraint is:
\[
10x + 30y \leq 150
\]
Simplifying this gives:
\[
x + 3y \leq 15 \quad \text{(Constraint 3)}
\]
### Step 4: Non-negativity Constraints
Since the number of circuits produced cannot be negative, we have:
\[
x \geq 0, \quad y \geq 0
\]
### Step 5: Summary of the LPP
Maximize:
\[
Z = 50x + 60y
\]
Subject to:
\[
\begin{align*}
2x + y & \leq 20 \quad \text{(Constraint 1)} \\
x + 2y & \leq 12 \quad \text{(Constraint 2)} \\
x + 3y & \leq 15 \quad \text{(Constraint 3)} \\
x & \geq 0 \\
y & \geq 0
\end{align*}
\]
### Step 6: Finding the Corner Points
To find the corner points of the feasible region, we will solve the equations formed by the constraints:
1. **Intersection of Constraint 1 and Constraint 2**:
\[
2x + y = 20 \quad (1)
\]
\[
x + 2y = 12 \quad (2)
\]
Solving these simultaneously gives \( (8, 4) \).
2. **Intersection of Constraint 2 and Constraint 3**:
\[
x + 2y = 12 \quad (2)
\]
\[
x + 3y = 15 \quad (3)
\]
Solving these simultaneously gives \( (10.5, 1.5) \).
3. **Intersection of Constraint 1 and Constraint 3**:
\[
2x + y = 20 \quad (1)
\]
\[
x + 3y = 15 \quad (3)
\]
Solving these simultaneously gives \( (0, 20) \).
4. **Intercepts**:
- For \( x = 0 \) in Constraint 1: \( y = 20 \)
- For \( y = 0 \) in Constraint 1: \( x = 10 \)
- For \( x = 0 \) in Constraint 2: \( y = 6 \)
- For \( y = 0 \) in Constraint 2: \( x = 12 \)
- For \( x = 0 \) in Constraint 3: \( y = 5 \)
- For \( y = 0 \) in Constraint 3: \( x = 15 \)
### Step 7: Evaluate the Objective Function at Corner Points
Now we will evaluate \( Z = 50x + 60y \) at the corner points:
1. \( (0, 20) \): \( Z = 50(0) + 60(20) = 1200 \)
2. \( (8, 4) \): \( Z = 50(8) + 60(4) = 400 + 240 = 640 \)
3. \( (10.5, 1.5) \): \( Z = 50(10.5) + 60(1.5) = 525 + 90 = 615 \)
4. \( (15, 0) \): \( Z = 50(15) + 60(0) = 750 \)
### Step 8: Determine the Maximum Profit
The maximum profit occurs at the point \( (15, 0) \) with:
\[
\text{Maximum Profit} = 750
\]
### Conclusion
To maximize profit, the manufacturer should produce:
- **15 circuits of Type A** and **0 circuits of Type B**.
- The maximum profit will be **750**.
