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 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.

Max $Z = x + y$ subject to $2x + 3y \leq 120, 8x + 5y \leq 400, x, y \geq 0$

The correct answer is Option (1) → Max $Z = x + y$ subject to $2x + 3y \leq 120, 8x + 5y \leq 400, x, y \geq 0$

Let the man covers x km on his motorcycle 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 ₹120 atmost on petrol

$∴ 2x + 3y ≤ 120$ ...(i)

Now, the man has only 1 hr time

$∴ \frac{x}{50} + \frac{y}{80} ≤ 1  ⇒ 8x + 5y ≤ 400$

$x ≥ 0, y ≥ 0$

To have maximum distance $Z = x + y$.

Hence, the required LPP to travel maximum distance is maximise $Z = x + y$, subject to the constraints $2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0$.