Question

the constraints of a problem are listed below. what are the vertices of the feasible region?
x + y ≤ 7
x - 2y ≤ -2
x ≥ 0
y ≥ 0
○ (0, 0), (0, 1), (4, 3), (7, 0)
○ (0, 1), (4, 3), (7, 0)
○ (0, 1), (0, 7), (2, 5)
○ (0, 1), (0, 7), (4, 3)

Explanation:

Step1: Identify axis intercepts

First, find vertices from intersections with axes:

• Intersection of $x=0$ and $y=0$: $(0,0)$
• Intersection of $x+y=7$ and $x=0$: $(0,7)$
• Intersection of $x+y=7$ and $y=0$: $(7,0)$
• Intersection of $x-2y=-2$ and $x=0$: $0-2y=-2 \implies y=1$, so $(0,1)$

Step2: Find intersection of two lines

Solve the system of $x+y=7$ and $x-2y=-2$:
Subtract the second equation from the first:
$$(x+y)-(x-2y)=7-(-2)$$
$$3y=9 \implies y=3$$
Substitute $y=3$ into $x+y=7$:
$$x=7-3=4$$
So the intersection is $(4,3)$

Step3: Validate feasible region points

Check which points satisfy all constraints $x+y\leq7$, $x-2y\leq-2$, $x\geq0$, $y\geq0$:

• $(0,0)$: $0-0=0 > -2$, violates $x-2y\leq-2$, so not in feasible region.
• $(0,1)$: Satisfies all constraints.
• $(4,3)$: Satisfies all constraints.
• $(7,0)$: $7-0=7 > -2$, violates $x-2y\leq-2$, so not in feasible region.
• $(0,7)$: $0-14=-14 \leq -2$, $0+7=7\leq7$, satisfies all constraints.

Answer:

(0, 1), (0, 7), (4, 3)