A man rides his motorcycle at the speed of 50 km/h. He has to spend Rs.2 per km on petrol. If he rides it at a faster speed of 80 km/h, the petrol cost increases to Rs.3 per km. He has atmost Rs.120 to spend on petrol and one hour's time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

Let the man rides to his motorcycle to a distance $x\;{\rm{km}}$ at the speed of

$50\;{\rm{km}}/{\rm{h}}$ and to a distance $y\;{\rm{km}}$ at the speed of $80\;{\rm{km}}/{\rm{h}}$.

Therefore, cost on petrol is $2x + 3y$.

Since, he has to spend Rs.120 atmost on petrol.

$\therefore$ $2x + 3y \le 120$ …….(i)

Also, he has atmost one hour's time.
$\therefore \frac{x}{{50}} + \frac{y}{{80}} \le 1$

$\Rightarrow$ $8x + 5y \le 400$ …….(ii)

Also, We have the following conditions as per the question, $x \ge 0,y \ge 0$

[non-negative constraints]

Thus, required LPP to travel maximum distance by him is

Maximise $Z = x + y$, subject to $2x + 3y \le 120,$ $8x + 5y \le 400,$ $x \ge 0,$ $y \ge 0$