Let $f(x,y)=2x-3y+35$.
Note that the coordinates of $S$ is $(0,3)$. Then we have
$\begin{array}{rcl}
f(0,3) & = & 2(0)-3(3)+35 \\
& = & 29
\end{array}$
Consider the coordinates of $R$.
$\left\{ \begin{array}{l}
RS:~y = 3 \\
QR:~x + y = 5
\end{array} \right.$
After solving, we have $R=(2,3)$. Hence,
$\begin{array}{rcl}
f(2,3) & = & 2(2)-3(3)+35 \\
& = & 30
\end{array}$
Consider the coordinates of $Q$.
$\left\{ \begin{array}{l}
QR:~x+y=5 \\
PQ:~x-y=7
\end{array} \right.$
After solving, we have $Q=(6,-1)$. Hence,
$\begin{array}{rcl}
f(6,-1) & = & 2(6)-3(-1)+35 \\
& = & 50
\end{array}$
Consider the coordinates of $P$. Substitute $x=0$ into the equation of $PQ$, we have
$\begin{array}{rcl}
(0)-y & = & 7 \\
y & = & -7
\end{array}$
Therefore, $P=(0,-7)$. Hence
$\begin{array}{rcl}
f(0,-7) & = & 2(0)-3(-7)+35 \\
& = & 56
\end{array}$
Therefore, $2x-3y+35$ attain its greatest value at $P$.
