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

- katieb

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- anonymous

Topic: Calculus
Sub Topic: Maxima and Minima.

- anonymous

Is it 74.83 km/h?

- anonymous

Yes!!

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## More answers

- anonymous

\[C(x) = 0.08(25 + \frac{ x^{2} }{ 112 }) * \frac{ 200 }{ x } + 2(\frac{ 200 }{ x })\]
This is the equation I came up with for Cost.

- anonymous

75km/hour

- anonymous

How did you do the first part?

- anonymous

sorry

- anonymous

gallon?

- anonymous

American?

- anonymous

let me rewrite that , (i used american units and accidentally wrote it in the wrong place)

- anonymous

The 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

The 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

So far, I'm up to:
\[0.08*(25+\frac{ x^2 }{ 112 })\]

- anonymous

\[0.08 \frac{ dollar }{ litre } * (25 + \frac{ x^{2} }{ 112 }) \frac{ litre }{ hour } * \frac{ 200 }{ x } hours\]

- anonymous

So you get that, and do you then add:
\[\frac{ 400 }{ x }\] for the money he receives per hour?

- anonymous

yes

- anonymous

okay. Now I'm getting somewhere.

- anonymous

So do we differentiate that with respect to x?

- anonymous

Yes because C(x) is the cost evaluated at x. So we need to minimize cost by differentiating with respect to x.

- anonymous

Thanks for the assistance @LogicalApple . Appreciate it.

- anonymous

Not a problem

- anonymous

For 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

Yeah I already finished the question.

- anonymous

Oh, nice

- anonymous

The hard part was what you did for me.

- anonymous

I did it on paper and I was like "hm.. I can't think in litres/gallon"

- anonymous

Ah, I did it again lol
I mean litres/hour!

- anonymous

hahah.

- anonymous

Well I'm going to close this and put up another question. Thanks again for helping me.

- anonymous

Hope you can help me again if you see fit to.

- anonymous

Good luck!

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