Sketch the system of inequalities, we have
Note that $A=(2,0)$.
For the point $B$, substitute $y=0$ into $4x-y=20$, we have
$\begin{array}{rcl}
4x-(0) & = & 20 \\
x & = & 5
\end{array}$
Therefore, $B=(5,0)$.
For the point $C$,
$\left\{\begin{array}{ll}
x+4y=22 & \ldots \unicode{x2460} \\
4x-y=20 & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2460}$, we have
$\begin{array}{rcl}
x+4y & = & 22 \\
x & = & 22- 4y\ldots \unicode{x2462}
\end{array}$
Sub. $\unicode{x2462}$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
4(22-4y)-y & = & 20 \\
88 – 17y & = & 20 \\
17y & = & 68 \\
y & = & 4
\end{array}$
Sub. $y=4$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
x & = & 22-4(4) \\
& = & 6
\end{array}$
Therefore, $C=(6,4)$.
For the point $D$, substitute $x=2$ into $x+4y=22$, we have
$\begin{array}{rcl}
(2)+4y & = & 22 \\
y & = & 5
\end{array}$
Therefore, $D=(2,5)$.
Let $f(x,y)=3y-4x+15$. At point $A(2,0)$,
$\begin{array}{rcl}
f(2,0) & = & 3(0)-4(2)+15 \\
& = & 7
\end{array}$
At point $B(5,0)$,
$\begin{array}{rcl}
f(5,0) & = & 3(0)-4(5)+15 \\
& = & -5
\end{array}$
At point $C(6,4)$,
$\begin{array}{rcl}
f(6,4) & = & 3(4)-4(6)+15 \\
& = & 3
\end{array}$
At point $D(2,5)$,
$\begin{array}{rcl}
f(2,5) & = & 3(5)-4(2) + 15 \\
& = & 22
\end{array}$
Therefore, the greatest value of $3y-4x+15$ is $22$.
