Question

A company manufactures two types of ladies dresses C and D. The raw material and labour available per day is given in the table.

Resources	Dress C(x)	Dress D(y)	Max. availability
Raw material	5	4	60
Labour	5	3	50

P is the profit, if P = 50x + 100y, solve this LPP to find x and y to get the maximum profit

Answer 1

Resources	Dress C(x)	Dress D(y)	Max. availability
Raw material	5	4	60
Labour	5	3	50

From the table, the constraints are

5x + 4y ≤ 60

5x + 3y ≤ 50

∴ Given problem can be formulated as follows:

Maximize P = 50x + 100y

Subject to 5x + 4y ≤ 60

5x + 3y ≤ 50, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality	5x + 4y ≤ 60	5x + 3y ≤ 50
Corresponding equation (of line)	5x + 4y = 60	5x + 3y = 50
Intersection of line with X-axis	(12, 0)	(10, 0)
Intersection of line with Y-axis	(0, 15)	`(0, 50/3)`
Region	Origin side	Origin side

Shaded portion OABC is the feasible region whose vertices are O(0, 0), A(10, 0), B and C(0, 15).

B is the point of intersection of 5x + 3y = 50 and 5x + 4y = 60.

Solving the above equations, we get

x = 4, y = 10

∴ B ≡ (4, 10)

Here, the objective function is

P = 50x + 100y

∴ P at O(0, 0) = 50(0) + 100(0) = 0

P at A(10, 0) = 50(10) + 100(0) = 500

P at B(4, 10) = 50(4) + 100(10)

= 200 + 1000

= 1200

P at C(0, 15) = 50(0) + 100(15) = 1500

∴ Maximum value of P is 1500 at x = 0 and y = 15.