The maximum value of \(Z = 3x + 2y\) subject to the constraints \(x + 2y \le 10\), \(x \le 2y\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20
The constraint \(x \le 2y\) keeps the region on the side of the line \(x = 2y\) that contains the y-axis. That line meets \(x + 2y = 10\) where
\(2y + 2y = 10\)
\(y = 2.5, \; x = 5\)
So the feasible region is the triangle with corner points (0, 0), (5, 2.5) and (0, 5). Values of Z:
\(Z(0, 0) = 0\)
\(Z(5, 2.5) = 15 + 5 = 20\)
\(Z(0, 5) = 10\)
The bounded region gives the maximum value 20, at (5, 2.5).
Why the other options are incorrect
Option (1) 10 — Incorrect. This is Z at (0, 5), which is smaller than Z at (5, 2.5).
Option (3) 30 — Incorrect. This is Z at (10, 0), but there \(x = 10 \gt 2y = 0\), so (10, 0) is not feasible.
Option (4) 15 — Incorrect. Z takes the value 20 at the feasible point (5, 2.5), so 15 is not the maximum.