Question

the constraints of a problem are listed below. what are the vertices of the feasible region?
2x + 3y ≤ 6
2x - 6y ≥ -3
x ≥ 0
y ≥ 0

○ (0, 0), (0, 0.5), (0, 2), (3, 0)
○ (0, 0), (0, 2), (1.5, 1), (3, 0)
● (0, 0), (0, 0.5), (1.5, 1), (3, 0)
○ (0, 0), (0, 0.5), (1, 1.5), (3, 0)

Explanation:

Step1: Find intersection with \(x = 0\)

For \(x = 0\), substitute into \(2x - 6y\geq - 3\): \(0-6y\geq - 3\Rightarrow y\leq0.5\). And from \(2x + 3y\leq6\), \(y\leq2\). But with \(x = 0\) and \(y\geq0\), the intersection on \(x = 0\) is \((0,0)\) and \((0,0.5)\) (since \(y\leq0.5\) from the second inequality).

Step2: Find intersection with \(y = 0\)

For \(y = 0\), substitute into \(2x+3y\leq6\): \(2x\leq6\Rightarrow x\leq3\). And from \(2x - 6y\geq - 3\), \(2x\geq - 3\) (always true as \(x\geq0\)). So intersection on \(y = 0\) is \((0,0)\) and \((3,0)\).

Step3: Find intersection of \(2x + 3y=6\) and \(2x - 6y=-3\)

Subtract the second equation from the first: \((2x + 3y)-(2x - 6y)=6-(-3)\Rightarrow9y = 9\Rightarrow y = 1\). Substitute \(y = 1\) into \(2x+3y = 6\): \(2x+3=6\Rightarrow2x = 3\Rightarrow x = 1.5\). So the intersection point is \((1.5,1)\).

Step4: Verify vertices

The vertices are the intersections of the boundary lines: \((0,0)\) (intersection of \(x = 0\) and \(y = 0\)), \((0,0.5)\) (intersection of \(x = 0\) and \(2x - 6y=-3\)), \((1.5,1)\) (intersection of \(2x + 3y = 6\) and \(2x - 6y=-3\)), and \((3,0)\) (intersection of \(y = 0\) and \(2x + 3y = 6\)).

Answer:

(0, 0), (0, 0.5), (1.5, 1), (3, 0)