For the LPP: minimize $z = 6x + 3y$ subject to the constraints
$4x + y ≥ 80$
$x+5y≥ 115$
$3x + 2y ≤ 150$
$x≥0,y≥ 0$ then the minimum value of z is

150

The correct answer is Option (1) → 150

Given:

Minimize: z = 6x + 3y

Subject to:

1) 4x + y ≥ 80

2) x + 5y ≥ 115

3) 3x + 2y ≤ 150

4) x ≥ 0, y ≥ 0

Step 1: Convert inequalities to equalities and find intersection points (feasible region boundaries).

Intersection of (1) and (2):

4x + y = 80 ...(i)

x + 5y = 115 ...(ii)

Multiply (ii) by 4: 4x + 20y = 460

Subtract (i): (4x + 20y) − (4x + y) = 460 − 80 → 19y = 380 → y = 20

From (i): 4x + 20 = 80 → x = 15

Point A = (15, 20)

Intersection of (1) and (3):

4x + y = 80 → y = 80 − 4x

3x + 2(80 − 4x) = 150 → 3x + 160 − 8x = 150 → −5x = −10 → x = 2

Then y = 80 − 8 = 72

Point B = (2, 72)

Intersection of (2) and (3):

x + 5y = 115 → x = 115 − 5y

3(115 − 5y) + 2y = 150 → 345 − 15y + 2y = 150 → −13y = −195 → y = 15

Then x = 115 − 75 = 40

Point C = (40, 15)

Step 2: Evaluate objective function z = 6x + 3y at feasible points:

At A (15, 20): z = 6×15 + 3×20 = 90 + 60 = 150

At B (2, 72): z = 6×2 + 3×72 = 12 + 216 = 228

At C (40, 15): z = 6×40 + 3×15 = 240 + 45 = 285

Conclusion: The minimum value of z is 150 at the point (15, 20).