Subject to constraints $2x+4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4, x, y ≥ 0$; the maximum value of $Z=3x+15y$ is:
Answer & explanation
Correct answer: option 3
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 $Z=3x+15y$
| Intersection point | $Z=3x+15y$ |
| $(1.6,1.2)$ | $22.8$ |
| $(1,3)$ | $48$ |
| $(4,0)$ | $12$ |
∴ Maximum value of Z occurs at (1, 3) is equal to 48.