The minimum value of \(Z = 4x + 3y\) subject to the constraints \(x + 2y \ge 8\), \(3x + y \ge 9\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 17
Both constraints are of the “≥” type, so the feasible region lies away from the origin and is unbounded. Its corner points are where the boundary lines meet the axes and each other.
Line \(3x + y = 9\) meets the y-axis at (0, 9), and line \(x + 2y = 8\) meets the x-axis at (8, 0). The two lines meet where
\(y = 9 - 3x\)
\(x + 2(9 - 3x) = 8\)
\(18 - 5x = 8\)
\(x = 2,\ y = 3\)
Value of Z at the corner points:
\(Z(0, 9) = 27\)
\(Z(2, 3) = 8 + 9 = 17\)
\(Z(8, 0) = 32\)
The smallest corner value is 17. As the region is unbounded, 17 is the minimum only if the half-plane \(4x + 3y \lt 17\) has no point in common with the feasible region. For every feasible point,
\(4x + 3y = (x + 2y) + (3x + y)\)
\(\ge 8 + 9 = 17\)
So no feasible point gives Z below 17, and the minimum value is 17, at (2, 3).
Why the other options are incorrect
Option (1) 27 — Incorrect. This is Z at the corner (0, 9); Z at (2, 3) is smaller.
Option (2) 32 — Incorrect. This is Z at the corner (8, 0), the largest of the corner values.
Option (4) Z has no minimum value — Incorrect. An unbounded region can still have a minimum; here no feasible point gives Z below 17.