Let $f(x,y)=2x-3y+180$. For the coordinates of point $P$, substitute $y=0$ into $3x+y=36$, we have
$\begin{array}{rcl}
3x + (0) & = & 36 \\
x & = & 12
\end{array}$
Therefore, $P=(12,0)$. Hence, we have
$\begin{array}{rcl}
f(12,0) & = & 2(12)-3(0) + 180 \\
& = & 204
\end{array}$
For the coordinates of point $Q$,
$\left\{ \begin{array}{ll}
3x+y = 36 & \ldots \unicode{x2460} \\
x+y = 20 & \ldots \unicode{x2461}
\end{array}\right.$
$\unicode{x2460} – \unicode{x2461}$, we have
$\begin{array}{rcl}
2x & = & 16 \\
x & = & 8
\end{array}$
Sub. $x=8$ into $\unicode{x2461}$, we have
$\begin{array}{rcl}
(8) + y & = & 20 \\
y & = & 12
\end{array}$
Therefore, $Q=(8,12)$. Hence, we have
$\begin{array}{rcl}
f(8,12) & = & 2(8) – 3(12) +180 \\
& = & 160
\end{array}$
For the coordinates of $R$, substitute $x=0$ into $x+y=20$, we have
$\begin{array}{rcl}
(0) + y & = & 20 \\
y & = & 20
\end{array}$
Therefore, $R=(0,20)$. Hence, we have
$\begin{array}{rcl}
f(0,20) & = & 2(0) – 3(20) +180 \\
& = & 120
\end{array}$
For the point $O$, we have
$\begin{array}{rcl}
f(0,0) & = & 2(0) -3(0) +180 \\
& = & 180
\end{array}$
Therefore, the least value is $120$.
