A company manufactures two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is Rs. 40 on each unit of product of type A and Rs. 50 on each unit of type B. How many units of each type should the company manufacture so as to earn maximum profit ? Use linear programming to find the solution.

Answer

To solve the problem using linear programming, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of units of product A manufactured
- \( y \) = number of units of product B manufactured

### Step 2: Formulate the Objective Function
The profit from product A is Rs. 40 per unit and from product B is Rs. 50 per unit. Therefore, the objective function to maximize profit \( Z \) can be written as:
\[ Z = 40x + 50y \]

### Step 3: Set Up the Constraints
Each unit of product A requires:
- 3 grams of nickel
- 1 gram of chromium

Each unit of product B requires:
- 1 gram of nickel
- 2 grams of chromium

The company has a total of:
- 9 grams of nickel
- 8 grams of chromium

From this information, we can set up the following constraints:

1. Nickel constraint:
\[ 3x + 1y \leq 9 \]

2. Chromium constraint:
\[ 1x + 2y \leq 8 \]

3. Non-negativity constraints:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Graph the Constraints
To graph the constraints, we will convert the inequalities into equations:

1. For \( 3x + y = 9 \):
- If \( x = 0 \), then \( y = 9 \) (point (0, 9))
- If \( y = 0 \), then \( x = 3 \) (point (3, 0))

2. For \( x + 2y = 8 \):
- If \( x = 0 \), then \( y = 4 \) (point (0, 4))
- If \( y = 0 \), then \( x = 8 \) (point (8, 0))

Now, plot these lines on a graph and shade the feasible region that satisfies all constraints.

### Step 5: Identify the Corner Points
The feasible region will be bounded by the lines we have graphed. The corner points (vertices) of the feasible region can be found by solving the equations of the lines:

1. Intersection of \( 3x + y = 9 \) and \( x + 2y = 8 \):
- Solve the system of equations:
\[ 3x + y = 9 \quad (1) \]
\[ x + 2y = 8 \quad (2) \]

From (2), express \( y \):
\[ y = \frac{8 - x}{2} \]
Substitute into (1):
\[ 3x + \frac{8 - x}{2} = 9 \]
Multiply through by 2 to eliminate the fraction:
\[ 6x + 8 - x = 18 \]
\[ 5x = 10 \quad \Rightarrow \quad x = 2 \]
Substitute \( x = 2 \) back into (2):
\[ 2 + 2y = 8 \quad \Rightarrow \quad 2y = 6 \quad \Rightarrow \quad y = 3 \]
Thus, one corner point is \( (2, 3) \).

2. Other corner points can be found by checking intersections with the axes and the constraints:
- \( (0, 0) \)
- \( (3, 0) \)
- \( (0, 4) \)

### Step 6: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 40x + 50y \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 40(0) + 50(0) = 0 \]
2. At \( (3, 0) \):
\[ Z = 40(3) + 50(0) = 120 \]
3. At \( (0, 4) \):
\[ Z = 40(0) + 50(4) = 200 \]
4. At \( (2, 3) \):
\[ Z = 40(2) + 50(3) = 80 + 150 = 230 \]

### Step 7: Determine the Maximum Profit
The maximum profit occurs at the point \( (2, 3) \) with a profit of Rs. 230.

### Conclusion
The company should manufacture:
- 2 units of product A
- 3 units of product B