2022-II-36

Ans: B

$\left\{ \begin{array}{ll}
2x+y = 8 & \ldots \unicode{x2460} \\
2x+3y = 16 & \ldots \unicode{x2461} \\
4x+3y = 22 & \ldots \unicode{x2462}
\end{array}\right.$

$\unicode{x2461} -\unicode{x2460}$, we have

$\begin{array}{rcl}
2y & = & 8 \\
y & = & 4
\end{array}$

Sub. $y=4$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
2x+4 & = & 8 \\
2x & = & 4
x & = & 2
\end{array}$

Therefore, $(2,4)$ is one of the vertices of $R$.

$\unicode{x2462} -\unicode{x2461}$, we have

$\begin{array}{rcl}
2x & = & 6 \\
x & = & 3
\end{array}$

Sub. $x=3$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
2(3) +3y & = & 16 \\
3y & = & 10 \\
y & = & \dfrac{10}{3}
\end{array}$

Therefore, $(3, \dfrac{10}{3})$ is one of the vertices of $R$.

$\unicode{x2462} -2\times \unicode{x2460}$, we have $y=6$.

Sub. $y=6$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
2x +6 & = & 8 \\
2x & = & 2\\
x & = & 1
\end{array}$

Therefore, $(1,6)$ is one of the vertices of $R$.

At the vertex $(2,4)$,

$\begin{array}{cl}
& 7x+6y \\
= & 7(2)+6(4) \\
= & 38
\end{array}$

At the vertex $(3,\dfrac{10}{3})$,

$\begin{array}{cl}
& 7x+6y \\
= & 7(3) +6(\dfrac{10}{3}) \\
= & 41
\end{array}$