The minimum value of \(Z = 5x + 7y\) subject to the constraints \(2x + y \ge 8\), \(x + 2y \ge 10\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 38
Both constraints are of the “≥” type, so the feasible region lies away from the origin and is unbounded.
Line 2x + y = 8 meets the y-axis at (0, 8), and line x + 2y = 10 meets the x-axis at (10, 0). The two lines meet where
\(y = 8 - 2x\)
\(x + 2(8 - 2x) = 10\)
\(16 - 3x = 10\)
\(x = 2,\ y = 4\)
Value of Z at the corner points:
\(Z(0, 8) = 56\)
\(Z(2, 4) = 10 + 28 = 38\)
\(Z(10, 0) = 50\)
The smallest corner value is 38. As the region is unbounded, it is the minimum only if no feasible point gives Z below 38. For every feasible point,
\(5x + 7y = (2x + y) + 3(x + 2y)\)
\(\ge 8 + 30 = 38\)
So the minimum value is 38, at (2, 4).
Why the other options are incorrect
Option (1) 50 — Incorrect. This is Z at the corner (10, 0); Z at (2, 4) is smaller.
Option (2) 56 — Incorrect. This is Z at the corner (0, 8), 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 38.