Determine the maximum distance that the man can travel.

Question

A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2per 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.

Determine the maximum distance that the man can travel.

Answer 1

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.

Maximize Z = x + y, subject to the constraints

2x + 3y ≤ 120 … (i)

8x + 5y ≤ 400 … (ii)

x ≥ 0, y ≥ 0

Now, let’s construct a constrain table for the above

Table for (i)

Next, solving equation (i) and (iii) we get

x = 300/7 and y = 80/7

It’s seen that the feasible region whose corner points are O(0, 0), A(50, 0), B(300/7, 80/7) and C(0, 40).

Let’s evaluate the value of Z

From above table the maximum value of Z is 54.3

Therefore, the maximum distance that the man can travel is 54.3 km at (300/7, 80/7).