The maximum value of \(Z = 4x + 3y\) subject to the constraints \(x + y \le 7\), \(2x + y \le 10\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 24
The feasible region is bounded, so the maximum of Z is at a corner point.
On the axes, 2x + y = 10 meets the x-axis at (5, 0) and x + y = 7 meets the y-axis at (0, 7); the other intercepts, (7, 0) and (0, 10), each break one of the constraints.
The lines x + y = 7 and 2x + y = 10 meet where
\(x = 10 - 7 = 3\)
\(y = 7 - 3 = 4\)
The corner points are (0, 0), (5, 0), (3, 4) and (0, 7).
\(Z(0, 0) = 0\)
\(Z(5, 0) = 20\)
\(Z(3, 4) = 12 + 12 = 24\)
\(Z(0, 7) = 21\)
The maximum is 24, at (3, 4).
Why the other options are incorrect
Option (2) 21 — Incorrect. This is Z at the corner (0, 7), which is less than 24.
Option (3) 20 — Incorrect. This is Z at the corner (5, 0).
Option (4) 28 — Incorrect. This is Z at (7, 0), which is not feasible: there 2x + y = 14 > 10.