A dealer deals in two products X & Y. He has ₹1,00,000/- to invest & space to store 80 pieces. Product X costs ₹ 2500/- & product Y costs ₹ 1000/- per unit. Construct the LPP and find the number of units of each product to be purchased.

Answer

To solve the problem, we need to construct a Linear Programming Problem (LPP) based on the given constraints and then find the optimal solution. Let's break it down step by step.

### Step 1: Define the Variables
Let:
- \( x \) = number of units of product X
- \( y \) = number of units of product Y

### Step 2: Formulate the Constraints
From the problem statement, we have the following constraints:

1. **Investment Constraint**: The total cost of products X and Y should not exceed ₹1,00,000.
\[ 2500x + 1000y \leq 100000 \]

2. **Storage Constraint**: The total number of units of products X and Y should not exceed 80.
\[ x + y \leq 80 \]

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

### Step 3: Rewrite the Constraints
We can simplify the investment constraint:
\[ 2500x + 1000y \leq 100000 \]
Dividing the entire inequality by 500 gives:
\[ 5x + 2y \leq 200 \]

So, our constraints are now:
1. \( 5x + 2y \leq 200 \)
2. \( x + y \leq 80 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 4: Graph the Constraints
To find the feasible region, we will graph the inequalities.

1. **For \( x + y = 80 \)**:
- If \( x = 0 \), then \( y = 80 \) (Point A: (0, 80))
- If \( y = 0 \), then \( x = 80 \) (Point B: (80, 0))

2. **For \( 5x + 2y = 200 \)**:
- If \( y = 0 \), then \( x = 40 \) (Point C: (40, 0))
- If \( x = 0 \), then \( y = 100 \) (Point D: (0, 100))

### Step 5: Identify the Feasible Region
The feasible region is the area where all constraints overlap, bounded by the lines we have drawn.

### Step 6: Find the Corner Points
The corner points of the feasible region are:
1. Point A: (0, 80)
2. Point B: (80, 0)
3. Point C: (40, 0)
4. Point D: (0, 100)

To find the intersection of the lines \( x + y = 80 \) and \( 5x + 2y = 200 \):
- From \( x + y = 80 \), we can express \( y = 80 - x \).
- Substitute into \( 5x + 2(80 - x) = 200 \):
\[ 5x + 160 - 2x = 200 \implies 3x = 40 \implies x = \frac{40}{3} \]
- Substitute \( x \) back to find \( y \):
\[ y = 80 - \frac{40}{3} = \frac{240 - 40}{3} = \frac{200}{3} \]
Thus, the intersection point is:
\[ \left(\frac{40}{3}, \frac{200}{3}\right) \]

### Step 7: Evaluate the Objective Function
Since we are not provided with an objective function to maximize or minimize, we can evaluate the feasible points:
1. (0, 80)
2. (80, 0)
3. \(\left(\frac{40}{3}, \frac{200}{3}\right)\)

### Conclusion
The possible solutions are:
- Purchase 0 units of X and 80 units of Y.
- Purchase 40 units of X and 0 units of Y.