Consider the LPP, Max Z= 2x + y, subject to the conditions

$3x+2y ≤ 6$

$4x+y ≤ 4 $

$x ≥ 0, y ≥0,$ then the maximum value of the objective function is :

$\frac{16}{5}$

The correct answer is Option (2) → $\frac{16}{5}$

$3x+2y ≤ 6⇒\frac{x}{2}+\frac{y}{3}≤ 6$

$4x+y ≤ 4⇒\frac{x}{1}+\frac{y}{4}≤1$

$x, y ≥0$ ⇒ first quadrant

finding intersection 

$3x+2y=6$   ...(1)

$4x+y=4$   ...(2)

Eq. (2) × 2 - Eq. (1)

$8x+6y-3x-2y=8-6$

$5x=2$

$⇒x=\frac{2}{5}$

from (2) $\frac{8}{5}+y=4$

so $y=\frac{12}{5}$

Maximum value → $Z_B=\frac{16}{5}$