Sketch the graph of the system of inequalities.
$\left\{\begin{array}{l}
x – 21 = 0 & \ldots \unicode{x2460} \\
x – y – 35 = 0 & \ldots \unicode{x2461} \\
x + 5y – 91 = 0 & \ldots \unicode{x2462} \\
3x + 2y = 0 & \ldots \unicode{x2463}
\end{array}\right.$
$\unicode{x2460} – \unicode{x2461}$, we have
$\begin{array}{rcl}
y + 14 & = & 0 \\
y & = & -14
\end{array}$
Therefore, the intersection point of $\unicode{x2460}$ and $\unicode{x2461}$ is $(21, -14)$.
$\unicode{x2462} – \unicode{x2460}$, we have
$\begin{array}{rcl}
5y -70 & = & 0 \\
y & = & 14
\end{array}$
Therefore, the intersection point of $\unicode{x2462}$ and $\unicode{x2460}$ is $(21, 14)$.
$\unicode{x2463} – 3\times \unicode{x2461}$, we have
$\begin{array}{rcl}
5y + 105 & = & 0 \\
y & = & -21
\end{array}$
Sub. $y = -21$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
x – (-21) -35 & = & 0 \\
x & = & 14
\end{array}$
Therefore, the intersection point of $\unicode{x2461}$ and $\unicode{x2463}$ is $(14, -21)$.
$\unicode{x2463} – 3\times \unicode{x2462}$, we have
$\begin{array}{rcl}
-13y + 273 & = & 0 \\
y & = & 21
\end{array}$
Sub. $y = 21$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
x + 5(21) -91 & = & 0 \\
x & = & -14
\end{array}$
Therefore, the intersection point of $\unicode{x2462}$ and $\unicode{x2463}$ is $(-14, 21)$.
At point $(21, -14)$, the value of $5x + 6y + 234$
$\begin{array}{cl}
= & 5(21) + 6(-14) + 234 \\
= & 255
\end{array}$
At point $(21, 14)$, the value of $5x + 6y + 234$
$\begin{array}{cl}
= & 5(21) + 6(14) + 234 \\
= & 423
\end{array}$
At point $(14, -21)$, the value of $5x + 6y + 234$
$\begin{array}{cl}
= & 5(14) + 6(-21) + 234 \\
= & 178
\end{array}$
At point $(-14, 21)$, the value of $5x + 6y + 234$
$\begin{array}{cl}
= & 5(-14) + 6(21) + 234 \\
= & 290
\end{array}$
Therefore, the least value is $178$.
