Consider the following L.P.P Max. z = 5x+2y , subject to -2x-3y ≤ -6,…

The correct answer is Option (1) → Its feasible region is bounded and has 4 corner points

The first constraint is \(2x + 3y \ge 6\). The boundary lines are

    \(L_1: 2x + 3y = 6,\quad L_2: x - 2y = 2\)
    \(L_3: 3x + 2y = 12,\quad L_4: -3x + 2y = 3\)

The axes do not touch the region:

    on y = 0: \(2x \ge 6\) gives \(x \ge 3\), but \(x \le 2\) — impossible
    on x = 0: \(3y \ge 6\) gives \(y \ge 2\), but \(2y \le 3\) gives \(y \le 1.5\) — impossible

So the region is bounded by the four lines alone. Their intersections:

    \(\displaystyle L_1 \cap L_2: \left(\frac{18}{7}, \frac{2}{7}\right),\quad L_2 \cap L_3: \left(\frac{7}{2}, \frac{3}{4}\right)\)

    \(\displaystyle L_3 \cap L_4: \left(\frac{3}{2}, \frac{15}{4}\right),\quad L_4 \cap L_1: \left(\frac{3}{13}, \frac{24}{13}\right)\)

Each of these satisfies the other two constraints and \(x, y \ge 0\) (for example, at \(\left(\frac{18}{7}, \frac{2}{7}\right)\): \(3x + 2y = \frac{58}{7} \le 12\) and \(-3x + 2y = -\frac{50}{7} \le 3\)). The remaining pairs meet outside the first quadrant:

    \(L_1 \cap L_3: (4.8, -1.2),\quad L_2 \cap L_4: (-2.5, -2.25)\)

So the feasible region is a quadrilateral: bounded, with 4 corner points.