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.
Answer & explanation
Correct answer: option 1
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 $2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0$.