A furniture dealer deals in tables and chairs. He has ₹ 1,50,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 1500 and a chair ₹ 750. Construct the inequations and find the feasible solution.

#### Solution

Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0.

The dealer has a space to store at most 60 pieces. ∴ x + y ≤ 60

Since, the cost of each table is ₹ 1500 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 1500x + 750y. Since the dealer has ₹ 1,50,000 to invest, 1500x + 750y ≤ 1,50,000 = 150x + 750y ≤ 15000 = 2x + y ≤ 200

Hence the system of inequations are

x + y ≤ 60, 2x + y ≤ 200

First we draw the lines AB and CD whose equations are

x + y = 60 and 2x + y ≤ 200, x ≥ 0, y ≥ 0 respectively.

Line |
Equation |
Points on the X-axis |
Points on the Y-axis |
Sign |
Region |

AB | x + y = 60 | A(60,0) | B(0,60) | ≤ | origin side of line AB |

CD | 2x + y = 200 | C(100,0) | D(0,200) | ≤ | origin side of line CD |

The feasible solution is OAPDO which is shaded in the graph.