The minimum value of \(Z = x + 2y\) subject to the constraints \(x + y \le 4\), \(y - x \le 2\) and \(y \ge 0\) only (x may also take negative values) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-2\)
Here there is no condition \(x \ge 0\), so the region can extend to the left of the y-axis. It lies above \(y = 0\), below \(x + y = 4\) and below \(y = x + 2\). Its corner points are:
\(y = 0\) and \(y = x + 2\): \((-2, 0)\)
\(y = 0\) and \(x + y = 4\): \((4, 0)\)
\(x + y = 4\) and \(y = x + 2\): \(2x + 2 = 4\), so \((1, 3)\)
The region is this bounded triangle. Values of Z:
\(Z(-2, 0) = -2\)
\(Z(4, 0) = 4\)
\(Z(1, 3) = 7\)
The minimum value is \(-2\), at \((-2, 0)\).
Why the other options are incorrect
Option (1) 0 — Incorrect. This is Z at the origin, the answer only if \(x \ge 0\) were given; the feasible point \((-2, 0)\) gives the smaller value \(-2\).
Option (2) 4 — Incorrect. This is Z at (4, 0), which is larger than \(-2\).
Option (3) 2 — Incorrect. The feasible point \((-2, 0)\) gives \(-2\), which is smaller.