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Question
If a young man drives his vehicle at 25 km/hr, he has to spend Rs 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs 5/per km. He has Rs 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
Solution
Let young man drives x km at a speed of 25 km/hr and y km at a speed of \[40 km/hr\] Clearly, \[x, y \geq 0\]
It is given that, he spends Rs 2 per km if he drives at a speed of \[25 km/hr\] and Rs 5 per km if he drives at a speed of \[40 km/hr\] . Therefore, money spent by him when he travelled x km and y km is Rs 2x and Rs 5y respectively.
It is given that he has a maximum of Rs 100 to spend.
Thus, \[2x + 5y \leq 100\]
\[\text{ Time spent by him when travelling with a speed of 25 km/hr } = \frac{x}{25}hr\]
\[\text{ Time spent by him when travelling with a speed of 40 km/hr} = \frac{x}{40}hr\]
Also, the available time is of 1 hour.
\[\frac{x}{25} + \frac{y}{40} \leq 1\]
\[ \Rightarrow 40x + 25y \leq 1000\]
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[x + y\]
\[2x + 5y \leq 100\]
\[40x + 25y \leq 1000\]
2x + 5y = 100, 40x + 25y = 1000, x = 0 and y = 0
Region represented by 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets the coordinate axes at \[A\left( 50, 0 \right)\] and \[B\left( 0, 20 \right)\] respectively. By joining these points we obtain the line 2x + 5y = 100. Clearly (0,0) satisfies the 2x + 5y = 100. So,the region which contains the origin represents the solution set of the inequation 2x + 5y ≤ 100.
Region represented by 40x + 25y ≤ 1000:
The line 40x + 25y = 1000 meets the coordinate axes at
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 5y ≤ 100, 40x + 25y ≤ 1000, x ≥ 0, and y ≥ 0 are as follows

The values of Z at these corner points are as follows
Corner point | Z = x + y |
O | 0 |
B | 20 |
E | 30 |
C | 25 |
The maximum value of Z is 30 which is attained at E.
Thus, the maximum distance travelled by the young man is 30 kms, if he drives
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