Question

A company uses three machines to manufacture two types of shirts, half sleeves and full sleeves. The number of hours required per week on machine M₁, M₂ and M₃ for one shirt of each type is given in the following table: 

	M1	M2	M3
Half sleeves	1	2	8/5
Full sleeves	2	1	8/5

None of the machines can be in operation for more than 40 hours per week . The profit on each half sleeve shirt is ₹ 1 and the profit on each full sleeve shirt is ₹ 1.50. How many of each type of shirts should be made per week to maximixe the company’s profit?

Answer 1

Let the number of half sleeve shirts made per week be x and the number of full sleeve shirts made per week be y.

So, the objective function is,

Maximize Z = x + 1.50y

Subject to constraints,

x + 2y ≤ 40   ...(i)

2x + y ≤ 40   ...(ii)

`8/5x + 8/5y ≤ 40`

or x + y ≤ 25   ...(iii)

And x ≥ 0 and y ≥ 0

From inequation (i), we have

x + 2y = 40

x	0	40
y	20	0

From inequation (ii), we have

2x + y = 40

x	0	20
y	40	0

From inequation (iii), we have

x + y = 25

x	0	25
y	25	0

Points	Z = x + 1.50y
A(0, 20)	0 + 1.50 × 20 = ₹ 30
B(10, 15)	10 + 1.50 × 15 = ₹ 32.50
C(15, 10)	15 + 1.50 × 10 = ₹ 30
D(20, 0)	20 + 1.50 × 0 = ₹ 20
O(0, 0)	0

So, Z is maximum at B(10, 15).

Hence, 10 half sleeves shirts and 15 full sleeve shirts should be made per week for maximum profit.