Question

unique car ltd.manufactures three type of cars A,b,$ c.these cars are manufactured at two different plants of the company having different manufacturing capacities.the following details pertaining to the manufacturing process are provided:
manufacturing maximum production(of cars) operating cost of plants
plants A B C
1 50 100 100 2500
2 60 60 200 3500
demand of cars 2500 3000 7000 _
Based on the given information,find the least number of days of operations per month so as to minimize the total cost of operations at the two plants (use graphical method)?

Answer 1

To minimize the total cost of operations for manufacturing cars of types A, B, and C at two plants, we can formulate this as a linear programming problem. Below are the steps to solve it graphically:

### Given Data

- **Plants**:
- Plant 1: Operating Cost = \$2500
- Plant 2: Operating Cost = \$3500

- **Maximum Production Capacity**:
- **Plant 1**:
- A: 50 cars
- B: 100 cars
- C: 100 cars
- **Plant 2**:
- A: 60 cars
- B: 60 cars
- C: 200 cars

- **Demand for Cars**:
- A: 2500 cars
- B: 3000 cars
- C: 7000 cars

### Step 1: Define Decision Variables
Let:
- \( x_1 \) = number of days Plant 1 operates
- \( x_2 \) = number of days Plant 2 operates

### Step 2: Formulate the Objective Function
The total operating cost for the plants can be represented as:
\[
\text{Total Cost} = 2500x_1 + 3500x_2
\]
Our goal is to minimize this cost function.

### Step 3: Formulate Constraints
We need to ensure that the production meets the demand for each car type:
1. For Car A:
\[
50x_1 + 60x_2 \geq 2500
\]

2. For Car B:
\[
100x_1 + 60x_2 \geq 3000
\]

3. For Car C:
\[
100x_1 + 200x_2 \geq 7000
\]

Additionally, we have non-negativity constraints:
\[
x_1 \geq 0, \quad x_2 \geq 0
\]

### Step 4: Solve the Constraints
We can rearrange the inequalities to express them in terms of \( x_2 \):
1. From \( 50x_1 + 60x_2 \geq 2500 \):
\[
x_2 \geq \frac{2500 - 50x_1}{60}
\]

2. From \( 100x_1 + 60x_2 \geq 3000 \):
\[
x_2 \geq \frac{3000 - 100x_1}{60}
\]

3. From \( 100x_1 + 200x_2 \geq 7000 \):
\[
x_2 \geq \frac{7000 - 100x_1}{200}
\]

### Step 5: Graphical Method
To find the feasible region:

1. Plot each of these equations on a graph where the x-axis is \( x_1 \) (days for Plant 1) and the y-axis is \( x_2 \) (days for Plant 2).
2. Identify the intersection points of these lines and the feasible area defined by the inequalities.
3. The vertices of this feasible region will give potential solutions for minimizing the cost.

### Step 6: Calculate Costs at Vertices
Once you have the vertices of the feasible region:
- Evaluate the objective function \( 2500x_1 + 3500x_2 \) at each vertex.
- The minimum value will give you the optimal number of days of operations per month for each plant.

### Conclusion
This approach will allow you to not only find the least number of days of operations needed but also help minimize the total costs incurred by Unique Car Ltd. If you have graphing tools or software, you might find it easier to visualize and solve this linear programming problem.