Sub. $y=0$ into the equation of $AB$, we have
$\begin{array}{rcl}
2x + 0 – 8 & = & 0 \\
x & = & 4
\end{array}$
Therefore, the coordinates of $A$ are $(4,0)$.
Hence, the value of $x+3y+4$ at $A(4,0)$
$\begin{array}{cl}
= & (4) + 3(0) + 4 \\
= & 8
\end{array}$
Sub. $x=0$ into the equation of $BC$, we have
$\begin{array}{rcl}
2(0) + 3y – 12 & = & 0 \\
y & = & 4
\end{array}$
Therefore, the coordinates of $C$ are $(0,4)$.
Hence, the value of $x+3y +4$ at $C(0,4)$
$\begin{array}{cl}
= & (0) + 3(4) + 4 \\
= & 16
\end{array}$
Consider the coordinates of $B$,
$\left\{ \begin{array}{ll}
2x+y-8=0 & \ldots \unicode{x2460} \\
2x+3y-12=0 & \ldots \unicode{x2461}
\end{array} \right.$
$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
2y – 4 & = & 0 \\
y & = & 2
\end{array}$
Sub. $y=2$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
2x + (2) -8 & = & 0 \\
x & = & 3
\end{array}$
Therefore, the coordinates of $B$ are $(3,2)$.
Hence, the value of $x+3y+4$ at $B(3,2)$
$\begin{array}{cl}
= & (3) + 3(2) + 4 \\
= & 13
\end{array}$
Therefore, the greatest value of $x+3y+4$ is $16$ at $C(0,4)$.
