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in Mathematics by (7.8k points)

A Furniture dealer deals in only two items, table and chair. He has Rs. 5000 to invest and a space to store at most 60 pieces. A table costs him Rs. 250 and a chair Rs. 50. He can sell a table at a profit of Rs. 50 and a chair at a profit Rs. 15. Assuming that he can sell all the items that he buys, how should he invest his money in order that he may maximize his profit.

1 Answer

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Best answer

Let x and y be the number of tables and chairs. 

cost of x table = Rs. 250 and cost of y chair = Rs. 50 

Since the dealer is maximum invest Rs. 5000 and the maximum number of items.

Also, the dealer want to sell a table and chair at the profit Rs. 50 and Rs. 15 respectively.

So, from the above explanation, we get following mathematical form as follows

250x + 50y ≤ 5000 

5x + y ≤ 100

x + y ≤ 60, x ≥ 0,y ≥ 0

and objective function Z = 50x + 15y

Now, we have to maximize Z = 50x + 15y

Subject to constraints

Now, to solve first of all draw a graph of equation ( l) to (3) corresponds to- in equations. It is clear from the graph that OABC be a feasible region, which is bounded. The co-ordinates of corner points of feasible region are O(0, 0), A(20, 0), B(10,50). The co-ordinate of the point B we get by solving equations (1) and (2). Lastly, applying corner point method to find maximum values of objective function Z as follows :

It is clear from above table the maximum values of Z is 125O at the point (10, 50). Thus the maximum profit to the dealer is Rs. 1250 for buying 10 tables and 50 chairs.

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