Question

Refer to question 15. Determine the maximum distance that the man can travel.

Answer 1

Referring to solution 15, we have

Maximise Z =x+y, subject to
$2x+3y\le 120,8x+5y\le 400,x\ge 0,y\ge 0$
On solving, we get
8x+5y =400 and 2x+3y =120, we get
$x=\frac{300}{7},y=\frac{80}{7}$
From the shaded feasible region, it is clear that coordinates of corner points are (0,0).(50,0), $(\frac{300}{7},\frac{80}{7})$ and (0,40).

$\begin{array}{cc}\text{Corner points}& \text{Corresponding value of Z=x+y}\\ (0,0)& 0\\ (50,0)& 50\\ \frac{300}{7},\frac{80}{7}& \frac{380}{7}=54\frac{2}{7}km\leftarrow Maximum\\ (0,40)& 40\end{array}$
Hence, the maximum distance that the man can travel is 54 $\frac{2}{7}km.$