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Question
If John drives a car at a speed of 60 km/hour, he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 km/hour, the cost of petrol increases ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as L.P.P.
Solution
Let John travel x1 km at speed of 60 km/hr and x2 km at a speed of 90 km/hr.
∴ Total distance = (x1 + x2) km
Time = `"Distance"/"Speed"`
Time to travel x1 km = `(x_1/60)` hours and time to travel x2 km = `(x_2/90)` hours.
∴ Total time = `(x_1/60 + x_2/90)"hours"`
But John wishes to travel maximum distance within an hour.
∴ `x_1/(60) + x_2/(90) ≤ 1`
John has to spend ₹ 5 per km at 60 km/hr and ₹ 8 per km at 90 km/hr.
∴ Total cost = ₹ (5x1 + 8x2)
But John has ₹ 600 to spend on petrol
∴ 5x1 + 8x2 ≤ 600
Since x1 and x2 cannot be negative, we have x1 ≥ 0, x2 ≥ 0
∴ Given problem can be formulated as follows:
Maximize Z = x1 + x2,
Subject to `x_1/(60) + x_2/(90) ≤1, 5x_1 + 8x_2 ≤ 600`, x1 ≥ 0, x2 ≥ 0.
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