The maximum value of the objective function Z = 3x + 5y, subject to the constraints: \(x+y \ge 6\); \(3x+y \ge 8\); \(x \le 6\); \(x, y \ge 0\) occurs at:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Not a finite point
The constraints \(x+y \ge 6\) and \(3x+y \ge 8\) only push the region away from the origin, and \(x \le 6\) limits x but not y. So nothing limits y from above: the feasible region is unbounded.
For example, every point (0, y) with \(y \ge 8\) is feasible:
\(0+y \ge 6,\quad 0+y \ge 8,\quad 0 \le 6\)
At such a point
\(Z = 3(0)+5y = 5y\)
which becomes as large as we like when y grows. So Z has no maximum value, and the maximum does not occur at any finite point.
(For comparison, Z(1, 5) = 28 and Z(0, 8) = 40 at two of the corner points, and Z(2, 4) = 26; all of these are exceeded, for example Z(0, 10) = 50.)