the constraints of a problem are listed below. what are the vertices of the feasible region?\nx + 3y ≤ 6\n4x…

Explanation:

Step1: Find intersection with x - axis

Set (y = 0) in the inequalities. For (x+3y\leq6), when (y = 0), we get (x\leq6). For (4x + 6y\geq9), when (y=0), we have (4x\geq9) or (x\geq\frac{9}{4}). The intersection of (x\geq0), (y = 0) with (4x+6y\geq9) gives the point ((\frac{9}{4},0)), and with (x + 3y\leq6) gives ((6,0)) (since (\frac{9}{4}\leq6)).

Step2: Find intersection with y - axis

Set (x = 0) in the inequalities. For (x+3y\leq6), when (x = 0), we get (y\leq2). For (4x+6y\geq9), when (x = 0), we have (6y\geq9) or (y\geq\frac{3}{2}). The intersection of (x = 0), (y\geq0) with (4x + 6y\geq9) gives the point ((0,\frac{3}{2})), and with (x+3y\leq6) gives ((0,2)) (since (\frac{3}{2}\leq2)).

Step3: Consider the origin

The origin ((0,0)) does not satisfy (4x + 6y\geq9), so it is not a vertex of the feasible - region.

Answer:

((0,\frac{3}{2}),(0,2),(6,0),(\frac{9}{4},0))