The corner points of the bounded feasible region of a linear programming problem are \((1, 1)\), \((6, 1)\), \((4, 5)\) and \((0, 3)\). For the objective function \(Z = x + 2y\), which of the following is NOT a value of Z at any point of the feasible region?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15
On a bounded feasible region, the minimum and the maximum of Z occur at corner points, and Z takes every value between them, because the region is in one piece and Z changes continuously.
\(Z(1, 1) = 3\)
\(Z(6, 1) = 8\)
\(Z(4, 5) = 14\)
\(Z(0, 3) = 6\)
So the values of Z on the region are exactly the numbers from 3 to 14. The maximum value is 14, so Z never equals 15.
Why the other options are incorrect
Option (1) 3 — Incorrect. Z = 3 at the corner (1, 1).
Option (2) \(\displaystyle\frac{17}{2}\) — Incorrect. It lies between 3 and 14; for example, at the midpoint (2.5, 3) of the corners (1, 1) and (4, 5), Z = 2.5 + 6 = 8.5.
Option (4) 14 — Incorrect. Z = 14 at the corner (4, 5).