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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.
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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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