Subject to constraints $2x+4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4, x, y ≥ 0$; the maximum value of $Z=3x+15y$ is:

48

The correct answer is Option (3) → 48

Equations to find intersection points,

$2x+4y ≤ 8$   ...(1)

$3x + y ≤ 6$   ...(2)

$x + y ≤ 4$   ...(3)

(1) Intersection point of (1) and (2)

$⇒2x+4y ≤ 8$

$x+2y ≤ 4$

$x=4-2y$

$⇒3x + y ≤ 6$

$3(4-2y)+y=6$

$12-6y+y=6$

$5y=6$

$y=1.2$

$∴x=4-2(1.2)=1.6$

(2) Intersection of (2) & (3)

$⇒x + y = 4$

$y=4-x$

$⇒3x+y=6$

$3x+4-x=6$

$2x+4=6$

$2x=2$

$x=1$

$∴y=4-(1)=3$

(3) Intersection of (1) & (3)

$⇒x+y=4$

$x=4-y$

$⇒2x+4y=8$

$2(4-y)+4y=8$

$8+2y=8$

$2y=0$

$y=0$

$∴x=4-0=4$

Evaluate $Z=3x+15y$

∴ Maximum value of Z occurs at (1, 3) is equal to 48.