For an LPP: Maximize z = 3x+9y;\ x ≥ 0, y ≥ 0 , the feasible region O…

The correct answer is Option (4) → \(x + 3y \le 60, x \le y\)

From the figure, the sides of the triangle OAB lie on three lines:

    OB: the y-axis, x = 0
    OA: the line x = y, through O(0, 0) and A(15, 15)
    AB: the line x + 3y = 60, through B(0, 20), A(15, 15) and (60, 0)

The direction of each inequality follows from a test point inside the region, such as (5, 10):

    \(x + 3y = 5 + 30 = 35 \le 60\), so \(x + 3y \le 60\)
    \(x = 5 \le 10 = y\), so \(x \le y\)

So the region lies on the origin side of x + 3y = 60 and above the line y = x, and the other constraints are \(x + 3y \le 60\) and \(x \le y\).