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Question
Solve the following L.P.P.:
Maximize Z = 4x + 5y
subject to 2x + y ≥ 4
x + y ≤ 5,
0 ≤ x ≤ 3,
0 ≤ y ≤ 3
Solution
The constraints x + y ≤ 5 and 2x + y ≥ 4 becomes x + y = 5 and 2x + y = 4
Points on the above lines are
(5,0); (0,5) and (2,0); (0,4) respectively by
The feasible region is polygon AEPQR, where
(i) P is the intersection of x + y = 5 and x = 3
`therefore` P ≡ (3,2)
(ii) Q isthe intersection of x + y = 5 and y = 3
`therefore` Q ≡ (2,3)
(iii) R is the intersection of 2x + y = 4 and y = 3
`therefore R ≡ (1/2 . 3)`
| Corner Point | Value of Z = 4x + 5y |
| A(2,0) | Za = 4(2) + 5(0) = 8 |
| E(3,0) | Ze = 4(3) + 5(0) = 12 |
| P(3,2) | ZP = 4(3) + 5(2) = 22 |
| Q(2,3) | ZQ = 4(2) + 5(3) = 23 |
| R`(1/2 . 3)` | ZA = 4 `(1/2)` + 5(3) = 17 |
Z attains its maximum at Q(2,3)
Zmax = 23 when x = 2 and y = 3.
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