The minimum value of Z = x + 2y subject to the constraints x + y ≥ 6…

The correct answer is Option (1) → \(\displaystyle\frac{15}{2}\)

The region lies above all three lines and is unbounded. On the y-axis it starts at \(y = 8\) (the largest of 6, 8, 3), and on the x-axis at \(x = 9\) (the largest of 6, 4, 9). The other corners are where the lines meet:

    \(2x + y = 8\) and \(x + y = 6\): \(x = 2, \; y = 4\)

    \(x + y = 6\) and \(x + 3y = 9\): \(2y = 3\), so \( \displaystyle \left(\frac{9}{2}, \frac{3}{2}\right) \)

Both points satisfy the third constraint (\(2 + 12 = 14 \ge 9\) and \(9 + \frac{3}{2} \ge 8\)), so the corner points are (0, 8), (2, 4), \(\left(\frac{9}{2}, \frac{3}{2}\right)\) and (9, 0).

    \(Z(0, 8) = 16\)
    \(Z(2, 4) = 10\)

    \( \displaystyle Z\left(\frac{9}{2}, \frac{3}{2}\right) = \frac{9}{2} + 3 = \frac{15}{2} \)

    \(Z(9, 0) = 9\)

The smallest corner value is \(\frac{15}{2}\). Since the region is unbounded, check the half-plane \(x + 2y \lt \frac{15}{2}\): every feasible point has

    \( \displaystyle x + 2y = \frac{1}{2}(x + y) + \frac{1}{2}(x + 3y) \ge 3 + \frac{9}{2} = \frac{15}{2} \)

So no feasible point gives a smaller value, and the minimum is \(\frac{15}{2}\).

Why the other options are incorrect

• Option (2) 9 — Incorrect. This is Z at the corner (9, 0), larger than \(\frac{15}{2}\).

• Option (3) 6 — Incorrect. This is Z at (6, 0), but there \(x + 3y = 6 \lt 9\), so (6, 0) is not feasible.

• Option (4) 10 — Incorrect. This is Z at the corner (2, 4), the point where two of the lines meet, but it is not the smallest corner value.