For the objective function \(Z = 3x - y\) subject to the constraints \(x + y \ge 4\), \(x - y \le 2\), \(y - x \le 2\), \(x \ge 0\), \(y \ge 0\), which of the following is true?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Z has a minimum value 0 but no maximum value
The constraints \(x - y \le 2\) and \(y - x \le 2\) keep every feasible point between the parallel lines y = x − 2 and y = x + 2. With \(x + y \ge 4\), the region is a strip running without end in the direction of the line y = x, so it is unbounded.
The line x + y = 4 meets the two parallel lines at
with x − y = 2: \(2x = 6\), giving (3, 1)
with y − x = 2: \(2y = 6\), giving (1, 3)
These are the only corner points.
\(Z(3, 1) = 9 - 1 = 8\)
\(Z(1, 3) = 3 - 3 = 0\)
Maximum. Every point (t, t) with \(t \ge 2\) is feasible, and there
\(Z = 3t - t = 2t\)
For t = 10 this gives Z = 20 > 8, and Z grows without limit. So Z has no maximum value.
Minimum. At every feasible point, \(4 - x \le y \le x + 2\), so \(4 - x \le x + 2\), that is, \(x \ge 1\). Then
\(3x - y \ge 3x - (x + 2)\)
\(= 2x - 2 \ge 0\)
So Z never falls below 0, and Z(1, 3) = 0. The minimum value is 0.
Why the other options are incorrect
Option (2) Z has a maximum value 8 and a minimum value 0 — Incorrect. 8 is only the larger corner value; the feasible point (10, 10) gives Z = 20.
Option (3) Z has a maximum value 8 but no minimum value — Incorrect. Z has no maximum, and it does have a minimum, 0.
Option (4) Z has neither a maximum nor a minimum value — Incorrect. Z never falls below 0, and it equals 0 at (1, 3), so the minimum exists.