A man has ₹1,500 for the purchase of Rice and Wheat.A bag of rice and a bag of wheat costs ₹ 180 and ₹ 120 respectively.He has storage capacity of 10 bags only.He earns a profit of ₹ 11 and ₹ 9 per bag of Rice and Wheat respectively.Formulate the problem as an LPP to find the number of bags of each type he should buy to maximize the profit and solve it graphically.

Answer

To solve this linear programming problem step-by-step, we will follow the formulation of the objective function and constraints, and then graphically solve it to find the optimal solution.

### Step 1: Define the Variables
Let:
- \( x \) = number of bags of rice
- \( y \) = number of bags of wheat

### Step 2: Formulate the Objective Function
The profit earned from rice and wheat is given as:
- Profit from rice = ₹11 per bag
- Profit from wheat = ₹9 per bag

Thus, the objective function to maximize the profit \( Z \) can be expressed as:
\[ Z = 11x + 9y \]

### Step 3: Formulate the Constraints
1. **Budget Constraint**: The total cost of rice and wheat should not exceed ₹1500.
- Cost of rice = ₹180 per bag
- Cost of wheat = ₹120 per bag
\[ 180x + 120y \leq 1500 \]

2. **Storage Capacity Constraint**: The total number of bags cannot exceed 10.
\[ x + y \leq 10 \]

3. **Non-negativity Constraints**: The number of bags cannot be negative.
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Rewrite the Constraints
To make it easier to graph, we can simplify the budget constraint:
\[ 180x + 120y \leq 1500 \implies 3x + 2y \leq 25 \]

Now, our constraints are:
1. \( 3x + 2y \leq 25 \)
2. \( x + y \leq 10 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 5: Graph the Constraints
To graph the constraints, we find the intercepts for each equation:

1. For \( 3x + 2y = 25 \):
- If \( x = 0 \): \( 2y = 25 \) → \( y = 12.5 \)
- If \( y = 0 \): \( 3x = 25 \) → \( x = 8.33 \)

2. For \( x + y = 10 \):
- If \( x = 0 \): \( y = 10 \)
- If \( y = 0 \): \( x = 10 \)

### Step 6: Identify the Feasible Region
Plot these lines on a graph and shade the feasible region that satisfies all constraints. The feasible region is bounded by the axes and the lines \( 3x + 2y = 25 \) and \( x + y = 10 \).

### Step 7: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 3x + 2y = 25 \) and \( x + y = 10 \):
- From \( x + y = 10 \), we can express \( y = 10 - x \).
- Substituting into \( 3x + 2(10 - x) = 25 \):
\[ 3x + 20 - 2x = 25 \implies x = 5 \quad \text{and} \quad y = 5 \]
- Corner point: \( (5, 5) \)

2. Other corner points are:
- \( (0, 10) \) from \( x + y = 10 \)
- \( (8.33, 0) \) from \( 3x + 2y = 25 \)

### Step 8: Evaluate the Objective Function at Each Corner Point
1. At \( (0, 10) \):
\[ Z = 11(0) + 9(10) = 90 \]
2. At \( (8.33, 0) \):
\[ Z = 11(8.33) + 9(0) \approx 91.63 \]
3. At \( (5, 5) \):
\[ Z = 11(5) + 9(5) = 55 + 45 = 100 \]

### Step 9: Determine the Maximum Profit
The maximum profit occurs at the corner point \( (5, 5) \) with a profit of ₹100.

### Conclusion
To maximize profit, the man should buy:
- 5 bags of rice
- 5 bags of wheat