For the LPP: minimize \(z = 6x + 3y\) subject to the constraints
\(4x + y \ge 80\)
\(x + 5y \ge 115\)
\(3x + 2y \le 150\)
\(x \ge 0, y \ge 0\) then the minimum value of z is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 150
The minimum of z over a bounded feasible region occurs at a corner point. The corners are where pairs of boundary lines meet.
\(4x + y = 80\) and \(x + 5y = 115\): from the first, \(y = 80 - 4x\), so
\(x + 5(80 - 4x) = 115\)
\(-19x = -285,\quad x = 15,\quad y = 20\)
\(4x + y = 80\) and \(3x + 2y = 150\):
\(3x + 2(80 - 4x) = 150\)
\(-5x = -10,\quad x = 2,\quad y = 72\)
\(x + 5y = 115\) and \(3x + 2y = 150\): from the first, \(x = 115 - 5y\), so
\(3(115 - 5y) + 2y = 150\)
\(-13y = -195,\quad y = 15,\quad x = 40\)
Each of these points satisfies the third constraint, so all three are corners. No corner lies on an axis: on \(x = 0\), \(4x + y \ge 80\) needs \(y \ge 80\) while \(3x + 2y \le 150\) needs \(y \le 75\); on \(y = 0\), \(x + 5y \ge 115\) needs \(x \ge 115\) while \(3x + 2y \le 150\) needs \(x \le 50\). So the feasible region is the triangle with these three corners.
\(z(15, 20) = 90 + 60 = 150\)
\(z(2, 72) = 12 + 216 = 228\)
\(z(40, 15) = 240 + 45 = 285\)
The minimum value of z is 150, at (15, 20).
Why the other options are incorrect
Option (2) 228 — Incorrect. This is z at the corner (2, 72), which is not the smallest corner value.
Option (3) 120 — Incorrect. It is below the smallest corner value 150, so no feasible point gives it.
Option (4) 100 — Incorrect. It is below the smallest corner value 150, so no feasible point gives it.