A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at re - Mathematics and Statistics

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A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at respective profits of ₹ 300 and ₹ 90. Construct the LPP and find the number of units of each product to be purchased to maximize its profit

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Solution

Let the dealer purchase x units of X and y units of Y.

The dealer can sell items X and Y at respective profits of ₹ 300 and ₹ 90

∴ Total profit = Z = 300x + 90y

∴ Given problem can be formulated as follows:

Maximize Z = 300x + 90y

Subject to x + y ≤ 80

2500x + 1000y ≤ 100000, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality x + y ≤ 80 2500x + 1000y ≤ 100000
Corresponding equation (of line) x + y = 80 2500x + 1000y = 100000
Intersection of line with X-axis (80, 0) (40, 0)
Intersection of line with Y-axis (0, 80) (0, 100)
Region Origin side Origin side

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(40, 0), B and C(0, 80).

B is the point of intersection of the lines 2500x + 1000y = 100000 and x + y = 80.

Solving the above equations, we get

x = `40/3`, y = `200/3`

∴ B ≡ `(40/3, 200/3)`

Here, the objective function is

Z = 300x + 90y

∴ Z at O(0, 0) = 300(0) + 90(0) = 0

Z at A(40, 0) = 300 (40) + 90(0)

= 12000

Z at B`(40/3, 200/3) = 300(40/3) + 90(200/3)`

= 4000 + 6000

= 10000

Z at C(0, 80) = 300 (0) + 90(80)

= 7200

∴ Z has maximum value 12000 at x = 40 and y = 0.

∴ The dealer should purchase 40 units of X and 0 units of Y to maximize its profit.

  Is there an error in this question or solution?
Chapter 2.6: Linear Programming - Q.4 (D)

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