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प्रश्न
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.
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उत्तर
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).
