Bunuel wrote:
The cost of the fuel for running the engine of an army tank is proportional to the square of the speed and is equal to $64 per hour at a speed of 16 kmph. The total cost for the tank per hour equals to the cost of the fuel per hour plus $400 per hour. The tank has to make a journey of 400 km at a constant speed. What is the most cost-efficient speed for this journey ?
A. 10 kmph
B. 20 kmph
C. 35 kmph
D. 40 kmph
E. 50 kmph
Solution:Let’s let c = the cost of the fuel per hour and s = the speed of the tank. We are told that cost is proportional to the square of the speed, so we have:
c = k x s^2,
where k = the constant of proportionality.
Substituting the given information, we have:
64 = k x 16^2
64/256 = k
1/4 = k
We must calculate the per-hour cost, the length of the trip (in hours), and the total cost for the trip for each of the answer choices.
Choice A: 10 kmph: c = ¼ x 10^2 = $25 per hour fuel cost + $400 = $425 per hour.
Length of trip: 400/10 = 40 hours, so the total trip cost is 425 x 40 = $17,000.
Choice B: 20 kmph: c = ¼ x 20^2 = $100 per hour fuel cost + $400 = $500 per hour.
Length of trip: 400/20 = 20 hours, so the total trip cost is 500 x 20 = $10,000.
Choice C: 35 kmph: c = ¼ x 35^2 = $306.25 per hour fuel cost + $400 = $706.25 per hour.
Length of trip: 400/35 = 11.43 hours, so the total trip cost is 706.25 x 11.43 ≈ $8,072.
Choice D: 40 kmph: c = ¼ x 40^2 = $400 per hour fuel cost + $400 = $800 per hour.
Length of trip: 400/40 = 10 hours, so the total trip cost is 800 x 10 = $8,000.
Choice E: 50 kmph: c = ¼ x 50^2 = $625 per hour fuel cost + $400 = $1025 per hour.
Length of trip: 400/50 = 8 hours, so the total trip cost is 1025 x 8 = $8,200.
We see that a speed of 40 kmph is the most cost-efficient speed of the given choices.
Answer: D _________________
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