Question
Determine the maximum distance that the man can travel.

Answer


We have to maximize z = x + y subject to constraints.
$2\text{x}+3\text{y}\leq120,8\text{x}+5\text{y}\leq400,\text{x}\geq0,\text{y}\geq0$
These inequalities are plotted as shown in the following figure.
From the figure shaded region is bounded with the corner points O(0, 0), A(50, 0), $\text{B}\Big(\frac{300}{7},\frac{80}{7}\Big),$ C(0, 0)
Corner points
Corresponding value of Z = x + y
(0, 0)
0
(50, 2)
50
$\Big(\frac{300}{7},\frac{80}{7}\Big)$
$\frac{380}{7}=54\frac{2}{7}\text{km (maximum)}$ 
(0, 40)
40
 Hence, the maximum distance that the man can travel is $54\frac{2}{7}\text{km}.$

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