A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 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.
Let’s assume the man covers x km on his motorcycle at the speed of 50km/hr and covers y km at the speed of 50 km/hr and covers y km at the speed of 80 km/hr.
So, cost of petrol = 2x + 3y
The man has to spend Rs 120 atmost on petrol
⇒ 2x + 3y ≤ 120 …. (i)
Now, the man has only 1 hr time
So, x/50 + y/80 ≤ 1 ⇒ 8x + 5y ≤ 400 … (ii)
And, x ≥ 0, y ≥ 0
To have maximum distance Z = x + y.
Therefore, the required LPP to travel maximum distance is maximize Z = x + y, subject to the constraints
2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0.
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