For the L.P.P. Maximize $z = 10x + 6y$ subjected to
$3x + y ≤ 12$,
$2x+5y ≤ 34$,
$x, y ≥ 0$.
Then the feasible region represented by system of inequalities is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Bounded in first quadrant
Given: Maximize $z = 10x + 6y$
Subject to the constraints:
$3x + y \leq 12$
$2x + 5y \leq 34$
$x \geq 0,\ y \geq 0$
The feasible region lies in the first quadrant, bounded by the lines:
1. $3x + y = 12$ ⟶ $y = 12 - 3x$
2. $2x + 5y = 34$ ⟶ $y = \frac{34 - 2x}{5}$
Both lines intersect within the first quadrant and the constraints form a closed polygonal region.
Also, $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant only.
Thus, the feasible region is closed and bounded.
