The minimum value of \(Z = 4x + y\) subject to the constraints \(x + y \ge 5\), \(y \le 3\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 11
The line \(x + y = 5\) meets \(y = 3\) at (2, 3) and the x-axis at (5, 0). The region lies above \(x + y = 5\) and below \(y = 3\), and extends without limit to the right, so it is unbounded.
The corner points are (2, 3) and (5, 0). Values of Z:
\(Z(2, 3) = 8 + 3 = 11\)
\(Z(5, 0) = 20\)
For every feasible point, \(x \ge 5 - y\) and \(y \le 3\), so
\(4x + y \ge 4(5 - y) + y = 20 - 3y \ge 20 - 9 = 11\)
So the minimum value is 11, at (2, 3).
Why the other options are incorrect
Option (1) 5 — Incorrect. This is Z at (0, 5), which breaks \(y \le 3\).
Option (2) 20 — Incorrect. This is Z at (5, 0); Z is smaller at (2, 3).
Option (4) Z has no minimum value — Incorrect. The region is unbounded, but \(Z \ge 11\) at every feasible point and \(Z = 11\) at (2, 3).