For a linear programming problem with a bounded feasible region, the maximum value of \(Z = ax + by\) \((a, b \gt 0)\) occurs at the two corner points \((2, 3)\) and \((5, 1)\). Which of the following points is also an optimal solution?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\left(\displaystyle\frac{7}{2}, 2\right)\)
Since Z has the same value at (2, 3) and (5, 1):
\(2a + 3b = 5a + b\)
\(2b = 3a\)
So \(Z = \frac{a}{2}(2x + 3y)\), and the maximum value corresponds to \(2x + 3y = 13\). Every point of the segment joining the two optimal corners lies in the region and has \(2x + 3y = 13\), so it is optimal. The midpoint of the segment is
\( \displaystyle \left(\frac{2 + 5}{2}, \frac{3 + 1}{2}\right) = \left(\frac{7}{2}, 2\right) \)
\( \displaystyle 2 \times \frac{7}{2} + 3 \times 2 = 13 \)
Why the other options are incorrect
Option (2) \((4, 2)\) — Incorrect. \(2x + 3y = 14 \gt 13\), so Z would exceed its maximum; the point cannot be feasible.
Option (3) \((3, 2)\) — Incorrect. \(2x + 3y = 12 \lt 13\), so Z is below its maximum there.
Option (4) \((7, 4)\) — Incorrect. \(2x + 3y = 26 \gt 13\), so Z would exceed its maximum; the point cannot be feasible.