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

Answer 1

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)

Z_max = 23 when x = 2 and y = 3.