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Question
Two tailors, A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP
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Solution 1
| Tailors/Product | A(x) | B(y) | Avl |
| Shirts | 6 | 10 | 60 |
| Trousers | 4 | 4 | 32 |
Let A work x days and B work y days.
`∴ x >= 0, y >= 0`
So L.P.P., objective function Min. Z = 300 x + 400 y
Subject to.
`6x + 10y >= 60`
`4x + 4y >= 32`
:. LPP
Min. Z = 300 x + 400 y
Subject to.
`3x + 5y >= 30`
`x + y >= 8`
`x >= 0, y >= 0`
Solution 2
Let tailor A work for x days and tailor B work for y days.
In one day, A can stitch 6 shirts and 4 pairs of trousers whereas B can stitch 10 shirts and 4 pairs of trousers.
Thus, in x days A can stitch 6x shirts and 4x pairs of trousers. Similarly, in y days B can stitch 10y shirts and 4y pairs of trousers.
It is given that the minimum requirement of the shirts and pairs of trousers are respectively 60 and 32 respectively.
Thus,
6x + 10y ≥ 60
4x + 4y ≥ 32
Further it is given that A and B earn Rs 300 and Rs 400 per day respectively. Thus, in x days and y days, A and B earn Rs 300x and Rs 400y respectively.
Let Z denotes the total cost
∴ Z =Rs (300x + 400y)
Number of days cannot be negative.
Therefore, x, y ≥ 0
Hence, the required LPP is as follows:
Minimize Z = 300x + 400y
subject to
6x + 10y ≥ 60
4x + 4y ≥ 32
x ≥ 0, y ≥ 0
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