Solution:

According to the solution of exercise 15, we have

Maximize $Z=x+y$, subject to the constraints

$2 x+3 y \leq 120 \ldots$ (i)

$8 x+5 y \leq 400 \ldots$ (ii)

$x \geq 0, y \geq 0$

Let’s construct a constrain table for the above

Table for (i)
$$\begin{tabular}{|l|l|l|}
\hline $\mathrm{x}$ & 0 & 60 \\
\hline $\mathrm{y}$ & 40 & 0 \\
\hline
\end{tabular}$$
Table for (ii)
$$\begin{tabular}{|l|l|l|}
\hline $\mathrm{x}$ & 0 & 50 \\
\hline $\mathrm{y}$ & 80 & 0 \\
\hline
\end{tabular}$$
Next, solving (i) and (iii) we obtain
$x=300 / 7$ and $y=80 / 7$
It can be seen that the feasible region whose corner points are $O(0,0), A(50,0), B(300 / 7,80 / 7)$ and $C(0,40)$.
Let’s evaluate the value of $Z$
$$\begin{tabular}{|l|l|}
\hline Corner points & Value of $Z=x+y$ \\
\hline$O(0,0)$ & $Z=0+0=0$ \\
\hline$A(50,0)$ & $Z=50+0=50$ \\
\hline$B(300 / 7,80 / 7)$ & $Z=300 / 7+80 / 7=380 / 7=54.3$ \\
\hline$C(0,40)$ & $Z=0-40=40$ \\
\hline
\end{tabular}$$
From above table the maximum value of $Z$ is $54.3$
So, the maximum distance that the man can travel is $54.3 \mathrm{~km}$ at $(300 / 7,80 / 7)$.