The optimal solution of the Linear Programming problem
Maximize \(Z = 3x_1 + 5x_2\),
s. t. \(3x_1 + 2x_2 \le 18\)
\(x_1 \le 4\)
\(x_2 \le 6\)
\(x_1 \ge 0,\ x_2 \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (2, 6)
The feasible region is bounded, so the maximum of Z is at a corner point.
Where \(3x_1 + 2x_2 = 18\) meets \(x_1 = 4\):
\(12 + 2x_2 = 18,\quad x_2 = 3\), point (4, 3)
Where \(3x_1 + 2x_2 = 18\) meets \(x_2 = 6\):
\(3x_1 + 12 = 18,\quad x_1 = 2\), point (2, 6)
The corner points are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6). Values of \(Z = 3x_1 + 5x_2\):
\(Z(0, 0) = 0\)
\(Z(4, 0) = 12\)
\(Z(4, 3) = 12 + 15 = 27\)
\(Z(2, 6) = 6 + 30 = 36\)
\(Z(0, 6) = 30\)
The maximum Z = 36 at (2, 6).
Why the other options are incorrect
Option (1) (4, 5) — Incorrect. \(3(4) + 2(5) = 22 \gt 18\), so the point is not feasible.
Option (3) (4, 3) — Incorrect. It is a corner point, but Z = 27 there, less than 36.
Option (4) (3, 6) — Incorrect. \(3(3) + 2(6) = 21 \gt 18\), so the point is not feasible.