A furniture dealer deals in tables and chairs. He has ₹ 15,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 150 and a chair ₹ 750. Construct the inequations and find the feasible 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 ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15000
Hence the system of inequations are
x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are
x + y = 60 and 150x + 750y = 15000 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||150x + 750y = 15000||C(100,0)||D(0,20)||≤||origin side of line CD|
The feasible solution is OAPDO which is shaded in the graph.