A furniture dealer deals in only items - tables and chairs. He has `₹ 15,000` to invest and a space to store atmost `60` pieces. A table costs him `₹ 750` and chair `₹ 150` . Suppose he makes x tables any y chairs
The graphical solution of the inequations representing given data is

Options

A.
B.
C.
D. None of these

Answer

To solve the problem step by step, we need to set up the inequalities based on the given information about the furniture dealer's constraints.

### Step 1: Define Variables
Let:
- \( x \) = number of tables
- \( y \) = number of chairs

### Step 2: Set Up the Inequalities
1. **Investment Constraint**: The dealer has ₹ 15,000 to invest. The cost of each table is ₹ 750 and each chair is ₹ 150. Therefore, the inequality representing the investment constraint is:
\[ 750x + 150y \leq 15000 \]
Dividing the entire inequality by 150 to simplify:
\[ 5x + y \leq 100 \]

2. **Storage Constraint**: The dealer can store at most 60 pieces of furniture (tables and chairs combined). Therefore, the inequality representing the storage constraint is:
\[ x + y \leq 60 \]

3. **Non-negativity Constraints**: Since the dealer cannot produce a negative number of tables or chairs, we have:
\[ x \geq 0 \quad \text{and} \quad y \geq 0 \]

### Step 3: Graph the Inequalities
To graph the inequalities, we need to find the intercepts and plot the lines.

1. **For the first inequality \( 5x + y = 100 \)**:
- When \( x = 0 \):
\[ y = 100 \quad \text{(Point: (0, 100))} \]
- When \( y = 0 \):
\[ 5x = 100 \implies x = 20 \quad \text{(Point: (20, 0))} \]

2. **For the second inequality \( x + y = 60 \)**:
- When \( x = 0 \):
\[ y = 60 \quad \text{(Point: (0, 60))} \]
- When \( y = 0 \):
\[ x = 60 \quad \text{(Point: (60, 0))} \]

### Step 4: Identify the Feasible Region
The feasible region is where all the inequalities overlap. This region is bounded by the lines \( 5x + y = 100 \) and \( x + y = 60 \), and it must also satisfy \( x \geq 0 \) and \( y \geq 0 \).

### Step 5: Find the Intersection Point
To find the intersection of the lines \( 5x + y = 100 \) and \( x + y = 60 \), we can solve these equations simultaneously.

1. From \( x + y = 60 \), we can express \( y \):
\[ y = 60 - x \]

2. Substitute \( y \) into the first equation:
\[ 5x + (60 - x) = 100 \]
Simplifying:
\[ 5x + 60 - x = 100 \implies 4x + 60 = 100 \implies 4x = 40 \implies x = 10 \]

3. Substitute \( x = 10 \) back into \( y = 60 - x \):
\[ y = 60 - 10 = 50 \]

### Step 6: Conclusion
Thus, the solution to the problem is:
- Number of tables \( x = 10 \)
- Number of chairs \( y = 50 \)

### Summary
The graphical solution shows that the dealer can purchase a maximum of 10 tables and 50 chairs while adhering to the constraints of investment and storage.