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anonymous
 3 years ago
A truck is to be driven 200 km along a level highway at x km per hour. Petrol costs 8 cents per litre and is used at the rate of \[25+\frac{ x^2 }{ 112 }\] litres per hour. The driver receives 2 dollars per hour. What is the most economical speed (to the nearest km per hour and the cost of the trip
i) if there is no speed limit,
ii) if the speed must not exceed 60km per hour?
anonymous
 3 years ago
A truck is to be driven 200 km along a level highway at x km per hour. Petrol costs 8 cents per litre and is used at the rate of \[25+\frac{ x^2 }{ 112 }\] litres per hour. The driver receives 2 dollars per hour. What is the most economical speed (to the nearest km per hour and the cost of the trip i) if there is no speed limit, ii) if the speed must not exceed 60km per hour?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Topic: Calculus Sub Topic: Maxima and Minima.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[C(x) = 0.08(25 + \frac{ x^{2} }{ 112 }) * \frac{ 200 }{ x } + 2(\frac{ 200 }{ x })\] This is the equation I came up with for Cost.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How did you do the first part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let me rewrite that , (i used american units and accidentally wrote it in the wrong place)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The first thing we know is that the rate of petrol is $0.08/litre And the second thing we know is that the petrol is used at a rate of 25 + x^2/112 litre/h so \[0.08 \frac{ dollar}{ litre } * (25 + \frac{ x^{2} }{ 112 }) \frac{ litre }{ hour }\] Gives us the units dollars/hour

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The only thing we need to know is how long the trip takes. We know speed = distance / time speed = x distance = 200 km time = t Therefore time, t = 200/x If we multiply what we have so far by (200/x) (which is in hours), then the hours unit cancels leaving an expression in dollars

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So far, I'm up to: \[0.08*(25+\frac{ x^2 }{ 112 })\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0.08 \frac{ dollar }{ litre } * (25 + \frac{ x^{2} }{ 112 }) \frac{ litre }{ hour } * \frac{ 200 }{ x } hours\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you get that, and do you then add: \[\frac{ 400 }{ x }\] for the money he receives per hour?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay. Now I'm getting somewhere.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So do we differentiate that with respect to x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes because C(x) is the cost evaluated at x. So we need to minimize cost by differentiating with respect to x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for the assistance @LogicalApple . Appreciate it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the second part of the question, keep in mind that the point we evaluated in the first part was a minimum. Due to the nature of the hyperbola, you can consider it an absolute minimum whenever x > 0 This should help you answer the second part of the question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I already finished the question.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The hard part was what you did for me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did it on paper and I was like "hm.. I can't think in litres/gallon"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, I did it again lol I mean litres/hour!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well I'm going to close this and put up another question. Thanks again for helping me.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hope you can help me again if you see fit to.
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