the constraints of a problem are listed below. what are the vertices of the feasible region?\nx + y ≤ 7\nx…

Explanation:

Step1: Find intersection of $x = 0$ and $y = 0$

The intersection of $x=0$ and $y = 0$ gives the point $(0,0)$.

Step2: Find intersection of $x=0$ and $x + y=7$

Substitute $x = 0$ into $x + y=7$, we get $0 + y=7$, so $y = 7$. The point is $(0,7)$.

Step3: Find intersection of $y = 0$ and $x + y=7$

Substitute $y = 0$ into $x + y=7$, we get $x+0 = 7$, so $x = 7$. The point is $(7,0)$.

Step4: Find intersection of $x + y=7$ and $x-2y=-2$

From $x + y=7$, we have $x=7 - y$. Substitute it into $x-2y=-2$: [ \begin{align*} 7 - y-2y&=-2\ 7-3y&=-2\ -3y&=-9\ y& = 3 \end{align*} ] Substitute $y = 3$ into $x=7 - y$, we get $x=7 - 3=4$. The point is $(4,3)$.

Step5: Find intersection of $x=0$ and $x - 2y=-2$

Substitute $x = 0$ into $x - 2y=-2$, we get $0-2y=-2$, so $y = 1$. The point is $(0,1)$.

We check the non - negativity constraints $x\geq0$ and $y\geq0$ for all these intersection points. The vertices of the feasible region are found by considering the intersection of the boundary lines of the inequalities.

Answer:

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