Question

A dealer in rural area wishes to purchase a number of sewing machines. He has only Rs. 5,760 to invest and has space for at most 20 items for storage. An electronic sewing machine cost him Rs. 360 and a manually operated sewing machine Rs. 240. He can sell an electronic sewing machine at a profit of Rs. 22 and a manually operated sewing machine at a profit of Rs. 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximise his profit? Make it as a LPP and solve it graphically.

Answer 1

Suppose dealer purchase x electronic sewing machines and y manually operated sewing machines. If Z denotes the total profit. Then according to question (Objective function) Z = 22x + 18 y 

Also, x + y ≤ 20 

360x + 240y ≤ 5760 ⇒ 9x + 6y ≤ 144 

x ≥ 0, y ≥ 0.

We have to maximise Z subject to above constraint. 

To solve graphically, at first we draw the graph of line corresponding to given inequations and shade the feasible region OABC. 

The corner points of the feasible region OABC are O(0, 0), A(16, 0), B(8, 12) and C(0, 20). 

Now the value of objective function Z at corner points are obtained in table as

Corner points	Z = 22x + 18y
O(0, 0)	Z = 0
A(16, 0)	Z = 22 x 16 + 18 x 0 = 352
B(8, 12)	Z = 22 x 8 + 18 x 12 = 392 maximum
C(0, 20)	Z = 22 x 0 + 18 x 20 = 360

From table, it is obvious that Z is maximum when x = 8 and y = 12.

Hence, dealer should purchase 8 electronic sewing machines and 12 manually operated sewing machines to obtain the maximum profit ` 392 under given condition.