For the L.P.P. Maximize \(z = 10x + 6y\) subjected to
\(3x + y \le 12\),
\(2x + 5y \le 34\),
\(x, y \ge 0\).
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
\(x, y \ge 0\) keeps the region in the first quadrant. Both other constraints are of the form \(\le\), and the origin satisfies them, so each one keeps the region on the origin side of its line.
Intercepts of the boundary lines:
\(3x + y = 12\): (4, 0) and (0, 12)
\(2x + 5y = 34\): (17, 0) and (0, 6.8)
The lines meet where \(y = 12 - 3x\):
\(2x + 5(12 - 3x) = 34\)
\(-13x = -26,\quad x = 2,\quad y = 6\)
So the feasible region is the quadrilateral with corners (0, 0), (4, 0), (2, 6) and (0, 6.8). It is non-empty and lies inside a finite polygon, so it is bounded.