To solve the problem of maximizing the profit from producing belts A and B, we will follow these steps:
### Step 1: Define the Variables
Let:
- \( x \) = number of belts of type A produced
- \( y \) = number of belts of type B produced
### Step 2: Formulate the Objective Function
The profit from each type of belt is given. Therefore, the objective function to maximize profit \( Z \) is:
\[
Z = 2x + 1.5y
\]
### Step 3: Identify the Constraints
From the problem, we have the following constraints:
1. **Material Constraint**: The total number of belts produced cannot exceed the material available for 800 belts.
\[
x + y \leq 800
\]
2. **Time Constraint**: Since a belt of type A takes twice as much time as a belt of type B, if we assume that producing 1000 belts of type B takes 1 day, then the time taken for \( x \) belts of type A and \( y \) belts of type B is:
\[
2x + y \leq 1000
\]
3. **Buckle Constraints**:
- For type A, the maximum number of belts is limited by the number of buckles available:
\[
x \leq 400
\]
- For type B, the maximum number of belts is limited by the number of buckles available:
\[
y \leq 700
\]
### Step 4: List the Constraints
Thus, we have the following system of inequalities:
1. \( x + y \leq 800 \)
2. \( 2x + y \leq 1000 \)
3. \( x \leq 400 \)
4. \( y \leq 700 \)
### Step 5: Graph the Constraints
To find the feasible region, we will graph the inequalities on the coordinate plane.
1. **Graph \( x + y = 800 \)**: This line intersects the axes at (800, 0) and (0, 800).
2. **Graph \( 2x + y = 1000 \)**: This line intersects the axes at (500, 0) and (0, 1000).
3. **Graph \( x = 400 \)**: A vertical line at \( x = 400 \).
4. **Graph \( y = 700 \)**: A horizontal line at \( y = 700 \).
### Step 6: Identify the Feasible Region
The feasible region is the area where all the constraints overlap. This region will be bounded by the lines we have graphed.
### Step 7: Find the Corner Points
To find the maximum profit, we need to evaluate the objective function at the corner points of the feasible region. The corner points can be found by solving the equations of the lines where they intersect.
1. **Point A (0, 700)**: Intersection of \( y = 700 \) and \( x = 0 \).
2. **Point B (400, 700)**: Intersection of \( x = 400 \) and \( y = 700 \).
3. **Point C (200, 600)**: Intersection of \( x + y = 800 \) and \( 2x + y = 1000 \).
4. **Point D (400, 200)**: Intersection of \( x = 400 \) and \( 2x + y = 1000 \).
5. **Point E (0, 0)**: Origin.
### Step 8: Evaluate the Objective Function at Each Corner Point
Now we will calculate \( Z \) at each of these points:
1. **At Point A (0, 700)**:
\[
Z = 2(0) + 1.5(700) = 1050
\]
2. **At Point B (400, 700)**:
\[
Z = 2(400) + 1.5(700) = 800 + 1050 = 1850 \quad \text{(not feasible since } y > 700\text{)}
\]
3. **At Point C (200, 600)**:
\[
Z = 2(200) + 1.5(600) = 400 + 900 = 1300
\]
4. **At Point D (400, 200)**:
\[
Z = 2(400) + 1.5(200) = 800 + 300 = 1100
\]
5. **At Point E (0, 0)**:
\[
Z = 2(0) + 1.5(0) = 0
\]
### Step 9: Determine the Maximum Profit
The maximum profit occurs at Point C (200, 600) with a profit of Rs. 1300.
### Conclusion
To maximize profit, the company should produce:
- **200 belts of type A**
- **600 belts of type B**
