The minimum value of \(Z = 6x + 4y\) subject to the constraints \(x \ge 2\), \(y \ge 3\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 24
The region is the part of the plane to the right of x = 2 and above y = 3. It is unbounded, with only one corner point, (2, 3).
\(Z(2, 3) = 12 + 12 = 24\)
For every feasible point,
\(6x + 4y \ge 6(2) + 4(3) = 24\)
So Z never falls below 24, and the minimum value is 24, at (2, 3).
Why the other options are incorrect
Option (1) 0 — Incorrect. This is Z at (0, 0), which breaks both \(x \ge 2\) and \(y \ge 3\).
Option (3) 12 — Incorrect. This counts only 6x at x = 2 and leaves out 4y = 12.
Option (4) Z has no minimum value — Incorrect. The region is unbounded, but Z can only grow as x and y grow; its least value is 24.