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:

120

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) $y=30$

Maximum value of Z → 120