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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. - Mathematics and Statistics

Sum

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.

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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.

Concept: Linear Programming Problem (L.P.P.)
  Is there an error in this question or solution?
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