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Question
If a young man drives his vehicle at 25 km/hr, he has to spend ₹2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to ₹5 per km. He has ₹100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
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Solution
Let us assume that the man travels x km when the speed is 25 km/hour and y km when the speed is 40 km/hour.
Thus, the total distance travelled is (x + y) km.
Now, it is given that the man has Rs 100 to spend on petrol.
Total cost of petrol = 2x + 5y ≤ 100
Now, time taken to travel x km = \[\frac{x}{25}\] h Time taken to travel y km = \[\frac{y}{40}\] h Now, it is given that the maximum time is 1 hour. So,
\[ \Rightarrow 8x + 5y \leq 200\]
Maximise Z = x + y
subject to the constraints
2x + 5y ≤ 100
8x + 5y ≤ 200
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are O(0, 0), A(25, 0), B \[\left( \frac{50}{3}, \frac{40}{3} \right)\] and C(0, 20).
The value of the objective function at these points are given in the following table.
| Corner Points | Z = x + y |
| (0, 0) | 0 + 0 = 0 |
| (25, 0) | 25 + 0 = 25 |
|
\[\left( \frac{50}{3}, \frac{40}{3} \right)\]
|
\[\frac{50}{3} + \frac{40}{3} = 30\]
|
| (0, 20) | 0 + 20 = 20 |
So, the maximum value of Z is 30 at \[x = \frac{50}{3}, y = \frac{40}{3}\]
Thus, the maximum distance that the man can travel in one hour is 30 km.
Hence, the distance travelled by the man at the speed of 25 km/hour is \[\frac{50}{3}\] km, and the distance travelled by him at the speed of 40 km/hour is \[\frac{40}{3}\] Km.
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