Linear Programming — Class 12 Maths Solution

Referring to Solution 15, We have the following conditions as per the question,

Maximise $Z = x + y$, subject to

$2x + 3y \le 120,$ $8x + 5y \le 400,$ $x \ge 0,$ $y \ge 0$

On solving,
$8x + 5y = 400$ 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) ,

$(50,0),\left( {\frac{{300}}{7},\frac{{80}}{7}} \right)$

and (0,40).

Hence, the maximum distance that the man can travel is $54\frac{2}{7}\;{\rm{km}}$.