To solve the manufacturing problem, we will follow a systematic approach using linear programming. Here’s the step-by-step solution:
### Step 1: Define Variables
Let:
- \( x \) = number of pieces of Model A produced
- \( y \) = number of pieces of Model B produced
### Step 2: Set Up the Objective Function
The objective is to maximize profit. The profit from Model A is Rs 8000 per piece, and from Model B is Rs 12000 per piece. Thus, the objective function can be expressed as:
\[
Z = 8000x + 12000y
\]
### Step 3: Set Up the Constraints
From the problem, we have the following constraints based on labor hours for fabricating and finishing:
1. **Fabricating Constraint**:
Each piece of Model A requires 9 hours and each piece of Model B requires 12 hours. The total available hours for fabricating is 180 hours:
\[
9x + 12y \leq 180
\]
2. **Finishing Constraint**:
Each piece of Model A requires 1 hour and each piece of Model 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: Formulate the System of Inequalities
We can summarize the constraints as:
1. \( 9x + 12y \leq 180 \)
2. \( x + 3y \leq 30 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 5: Find the Corner Points
To find the feasible region, we need to determine the intersection points of the constraints.
1. **For \( 9x + 12y = 180 \)**:
- When \( x = 0 \): \( 12y = 180 \) → \( y = 15 \) → Point (0, 15)
- When \( y = 0 \): \( 9x = 180 \) → \( x = 20 \) → Point (20, 0)
2. **For \( x + 3y = 30 \)**:
- When \( x = 0 \): \( 3y = 30 \) → \( y = 10 \) → Point (0, 10)
- When \( y = 0 \): \( x = 30 \) → Point (30, 0)
3. **Finding Intersection of the two lines**:
To find the intersection of \( 9x + 12y = 180 \) and \( x + 3y = 30 \):
- From \( x + 3y = 30 \), express \( x \) in terms of \( y \):
\[
x = 30 - 3y
\]
- Substitute into \( 9(30 - 3y) + 12y = 180 \):
\[
270 - 27y + 12y = 180 \\
-15y = -90 \\
y = 6
\]
- Substitute \( y = 6 \) back to find \( x \):
\[
x = 30 - 3(6) = 12
\]
- Intersection point is (12, 6).
### Step 6: Identify the Feasible Region
The feasible region is bounded by the points:
- (0, 15)
- (20, 0)
- (0, 10)
- (12, 6)
### Step 7: Evaluate the Objective Function at Each Corner Point
1. At (0, 15):
\[
Z = 8000(0) + 12000(15) = 180000
\]
2. At (20, 0):
\[
Z = 8000(20) + 12000(0) = 160000
\]
3. At (0, 10):
\[
Z = 8000(0) + 12000(10) = 120000
\]
4. At (12, 6):
\[
Z = 8000(12) + 12000(6) = 96000 + 72000 = 168000
\]
### Step 8: Determine Maximum Profit
The maximum profit occurs at the point (12, 6):
- Number of pieces of Model A = 12
- Number of pieces of Model B = 6
- Maximum Profit = Rs 168000
### Final Answer
The company should manufacture 12 pieces of Model A and 6 pieces of Model B per week to realize a maximum profit of Rs 168000.
---
