The maximum value of \(Z = kx + 2y\) \((k \gt 0)\), subject to the constraints \(x + 2y \le 10\), \(x + y \le 6\), \(x \ge 0\), \(y \ge 0\), occurs at the point (2, 4). Then the set of all possible values of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1 \le k \le 2\)
The lines x + 2y = 10 and x + y = 6 meet where y = 4 and x = 2. The line x + y = 6 meets the x-axis at (6, 0) and x + 2y = 10 meets the y-axis at (0, 5). So the corner points of the bounded region are (0, 0), (6, 0), (2, 4) and (0, 5).
Values of Z = kx + 2y:
\(Z(0, 0) = 0\)
\(Z(6, 0) = 6k\)
\(Z(2, 4) = 2k + 8\)
\(Z(0, 5) = 10\)
The maximum occurs at (2, 4) exactly when Z(2, 4) is at least every other corner value:
\(2k + 8 \ge 6k\), so \(k \le 2\)
\(2k + 8 \ge 10\), so \(k \ge 1\)
So \(1 \le k \le 2\). At k = 1 the maximum 10 is shared with (0, 5), and at k = 2 the maximum 12 is shared with (6, 0); in both cases it still occurs at (2, 4).
Why the other options are incorrect
Option (1) \(0 \lt k \le 1\) — Incorrect. For k = 1/2, Z(2, 4) = 9 but Z(0, 5) = 10, so the maximum is at (0, 5).
Option (3) \(2 \le k \le 4\) — Incorrect. For k = 3, Z(2, 4) = 14 but Z(6, 0) = 18, so the maximum is at (6, 0).
Option (4) \(k \ge 4\) — Incorrect. For k = 4, Z(2, 4) = 16 but Z(6, 0) = 24.