The maximum value of \(Z = 4x + 3y\) subject to the constraints \(3x + 2y \le 18\), \(x + 2y \le 10\), \(x \le 5\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 25
The region is bounded, so the maximum of Z is at a corner point.
On the y-axis, x + 2y = 10 gives (0, 5). On the x-axis, 3x + 2y = 18 would give (6, 0), which breaks \(x \le 5\), so the region stops at (5, 0).
The line x = 5 meets 3x + 2y = 18 where
\(2y = 18 - 15 = 3\)
\(\displaystyle y = \frac{3}{2}\)
The lines 3x + 2y = 18 and x + 2y = 10 meet where
\(2x = 18 - 10 = 8\)
\(x = 4,\ y = 3\)
The corner points are (0, 0), (5, 0), \(\left(5, \frac{3}{2}\right)\), (4, 3) and (0, 5).
\(Z(0, 0) = 0\)
\(Z(5, 0) = 20\)
\(\displaystyle Z\left(5, \frac{3}{2}\right) = 20 + \frac{9}{2} = \frac{49}{2}\)
\(Z(4, 3) = 16 + 9 = 25\)
\(Z(0, 5) = 15\)
The maximum is 25, at (4, 3).
Why the other options are incorrect
Option (2) \(\displaystyle\frac{49}{2}\) — Incorrect. This is Z at the corner \(\left(5, \frac{3}{2}\right)\), just below 25.
Option (3) 24 — Incorrect. This is Z at (6, 0), which breaks \(x \le 5\).
Option (4) 20 — Incorrect. This is Z at the corner (5, 0).