For the linear programming problem: Maximise \(Z = 3x - 4y\) subject to \(x - y \le 2\), \(x + 2y \ge 8\), \(x \ge 0\), \(y \ge 0\), which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The maximum value of Z is 4, at \((4, 2)\).
The region lies above \(x + 2y = 8\) and above the line \(x - y = 2\) (that is, \(y \ge x - 2\)). No point of the x-axis is feasible: \(y = 0\) needs \(x \ge 8\) and \(x \le 2\) together. The region extends upwards without limit, so it is unbounded.
Its corner points are (0, 4), where \(x + 2y = 8\) meets \(x = 0\), and the meeting point of the two lines:
\(x - y = 2\) and \(x + 2y = 8\)
\(3y = 6\)
\(y = 2, \; x = 4\)
Values of Z:
\(Z(0, 4) = -16\)
\(Z(4, 2) = 12 - 8 = 4\)
Since the region is unbounded, 4 is the maximum only if the half-plane \(3x - 4y \gt 4\) has no point in common with the region. Every feasible point satisfies
\( \displaystyle 3x - 4y = \frac{10}{3}(x - y) - \frac{1}{3}(x + 2y) \)
\( \displaystyle 3x - 4y \le \frac{10}{3}(2) - \frac{1}{3}(8) = 4 \)
So the maximum value is 4, at (4, 2).
Why the other options are incorrect
Option (1) Z has no maximum value, since the feasible region is unbounded. — Incorrect. An unbounded region can still give a maximum; here \(3x - 4y \le 4\) at every feasible point.
Option (2) The maximum value of Z is \(-16\), at \((0, 4)\). — Incorrect. \(Z(4, 2) = 4\) is larger than \(-16\).
Option (3) The maximum value of Z is 2, at \((6, 4)\). — Incorrect. (6, 4) is feasible and gives \(Z = 2\), but this is less than 4, and (6, 4) is not a corner point.