The maximum value of \(Z = 5x + 3y\) subject to the constraints \(3x + 5y \le 15\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 25
The feasible region is the triangle cut off from the first quadrant by the line 3x + 5y = 15. It is bounded, so the maximum is at a corner point.
The line meets the x-axis at (5, 0) and the y-axis at (0, 3). The corner points are (0, 0), (5, 0) and (0, 3).
\(Z(0, 0) = 0\)
\(Z(5, 0) = 25\)
\(Z(0, 3) = 9\)
The maximum is 25, at (5, 0).
Why the other options are incorrect
Option (1) 9 — Incorrect. This is Z at the corner (0, 3), which is less than 25.
Option (2) 15 — Incorrect. This is the right-hand side of the constraint, not a value of Z at a corner.
Option (3) Z has no maximum value — Incorrect. The region is a bounded triangle, so Z has a maximum.