For the LPP:
Maximise z = 4x + y
subject to constraints
x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0 the maximum value of z is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 120
$z = 4x + y$
$x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0$
finding intersection
$x + y = 50$ ...(1)
$3x + y = 90$ ...(2)
eq. (2) - eq. (1)
$2x=40⇒x=20$
from (1) $y=30$
| corner points | value of Z |
| $O(0,0)$ | $Z_O=0$ |
| $A(0,50)$ | $Z_A=50$ |
| $B(20,30)$ | $Z_B=110$ |
| $C(30,0)$ | $Z_C=120$ |
Maximum value of Z → 120