A man rides his motorcycle at the speed of 50 \mathrm{~km} / hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 \mathrm{~km} / hour, the petrol cost increases to Rs 3 per km. He has at most 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.
A man rides his motorcycle at the speed of 50 \mathrm{~km} / hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 \mathrm{~km} / hour, the petrol cost increases to Rs 3 per km. He has at most 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.

Solution:

Suppose the man covers \mathrm{x} \mathrm{km} on his motorcycle at the speed of 50 \mathrm{~km} / \mathrm{hr} and covers \mathrm{y} \mathrm{km} at the speed of 50 \mathrm{~km} / \mathrm{hr} and covers y \mathrm{km} at the speed of 80 \mathrm{~km} / \mathrm{hr}.

Therefore, cost of petrol =2 x+3 y

The man has to spend Rs 120 atmost on petrol

\Rightarrow 2 \mathrm{x}+3 \mathrm{y} \leq 120\dots \dots(i)

Now, the man has only 1 \mathrm{hr} time

Therefore, x / 50+y / 80 \leq 1 \Rightarrow 8 x+5 y \leq 400 \ldots(ii)

And, \quad x \geq 0, y \geq 0

To have maximum distance \mathrm{Z}=\mathrm{x}+\mathrm{y}.

So, the required LPP to travel maximum distance is maximize \mathrm{Z}=\mathrm{x}+\mathrm{y}, subject to the constraints

2 x+3 y \leq 120,8 x+5 y \leq 400, x \geq 0, y \geq 0