The minimum value of Z = 3x + 2y subjected to the constraints 2x + y…

The correct answer is Option (3) → 12

Both main constraints are of the "\(\ge\)" type, so the feasible region is unbounded. Its corner points are found from the boundary lines.

On the y-axis, \(2x + y \ge 7\) needs \(y \ge 7\) and \(x + 2y \ge 8\) needs \(y \ge 4\), giving (0, 7).
On the x-axis, the two need \(x \ge 3.5\) and \(x \ge 8\), giving (8, 0).

Solving \(2x + y = 7\) and \(x + 2y = 8\) together:

    \(y = 7 - 2x\)
\(x + 2(7 - 2x) = 8\)
\(-3x = -6,\quad x = 2,\quad y = 3\)

Values of Z at the corner points:

    \(Z(0, 7) = 14\)
\(Z(2, 3) = 6 + 6 = 12\)
\(Z(8, 0) = 24\)

The smallest corner value is 12. Since the region is unbounded, check the open half plane \(3x + 2y \lt 12\): it has no point in common with the feasible region, because for every feasible point

    \( \displaystyle 3x + 2y = \frac{4}{3}(2x + y) + \frac{1}{3}(x + 2y) \ge \frac{4}{3}(7) + \frac{1}{3}(8) = 12 \)

So the minimum value of Z is 12, at (2, 3).