If John drives a car at a speed of 60 kms/hour he has to spend Rs 5 per km on petrol. If he drives at a faster speed of 90 kms/hour,

Let John travel xl km at a speed of 60 km/ hour and x₁ km at a speed of 90 km/hour. Therefore, time required to travel a distance of x₁ km is x₁/60 hours and the time required to travel a distance of x₂ km is x₂/90 hours.

∴ total time required to travel is \(\left(\frac{x_1}{60} + \frac{x_2}{90}\right)\) hours.

Since he wishes to travel the maximum distance within an hour,

\(\frac{x_1}{60} + \frac{x_2}{90} \leq 1\)

He has to spend Rs 5 per km on petrol at a speed of 60 km/hour and Rs 8 per km at a speed of 90 km/hour.

∴ the total cost of travelling is Rs (5x + 8x ) Since he has Rs 600 to spend on petrol, 5x₁ + 8x₂ ≤ 600

Since distance is never negative, x₁ ≥ 0, x₂ ≥ 0. 

Total distance travelled by John is z. = (x₁ + x₂) km.

This is the linear function which is to be maximized. 

Hence, it is objective function.

Hence, the given LPP can be formulated as : Maximize z = x₁ + x₂ , subject to

\(\frac{x_1}{60} + \frac{x_2}{90} \leq 1,\) 5x₁ + 8x₂ ≤ 600, x₁ ≥ 0, x₂ ≥ 0.