Question

A manufacturer has two machines X and Y that may run at the most 360 minutes in a day to produce two types of toys A and B. To produce each Toy A, machines X and Y need to run at the most 12 minutes and 6 minutes respectively. To produce each Toy B, machines X and Y need to run at the most 6 minutes and 9 minutes respectively. By selling the toys A and B, the manufacturer makes the profits ₹ 30/- and ₹ 20/- respectively. Formulate a Linear Programming Problem and find the number of toys A and B that should be manufactured in a day to get maximum profit.

Answer 1

Let x and y be the number of toys A and B, respectively.

Z = 30x + 20y (Maximise)

Machine x total time ≤ 360 min:

12x + 6y ≤ 360

Machine y total time ≤ 360  min:

6x + 9y ≤ 360

Also,

x ≥ 0, y ≥ 0

From 12x + 6y = 360 ⇒ 2x + y = 60, intercepts (0, 60), (30, 0)    ...[i]

From 6x + 9y = 360 ⇒ 2x + 3y = 120 intercepts (0, 40) (60, 0)    ...[ii]

Intersection of

2x + y = 60 and 2x + 3y = 120

Subtract equation (1) from (2):

(2x + 3y) − (2x + y) = 120 − 60
2y = 60
y = 30

Substitute y = 30 in (1):

2x + 30 = 60
2x = 30
x = 15

Point: (15, 30)

Hence corner points are:

O(0, 0), P(0, 40), Q(15, 30), R(30, 0)

Value of Z at corner points:

Corner point	Z = 30x + 20y
O(0, 0)	0
P(0, 40)	800
Q(15, 30)	450 + 600 = 1050 → max.
R(30, 0)	900

Maximum profit is ₹ 1050 at (15, 30).