Here's the question you clicked on:
Azteck
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?
Topic: Calculus Sub Topic: Maxima and Minima.
\[C(x) = 0.08(25 + \frac{ x^{2} }{ 112 }) * \frac{ 200 }{ x } + 2(\frac{ 200 }{ x })\] This is the equation I came up with for Cost.
How did you do the first part?
let me rewrite that , (i used american units and accidentally wrote it in the wrong place)
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
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
So far, I'm up to: \[0.08*(25+\frac{ x^2 }{ 112 })\]
\[0.08 \frac{ dollar }{ litre } * (25 + \frac{ x^{2} }{ 112 }) \frac{ litre }{ hour } * \frac{ 200 }{ x } hours\]
So you get that, and do you then add: \[\frac{ 400 }{ x }\] for the money he receives per hour?
okay. Now I'm getting somewhere.
So do we differentiate that with respect to x?
Yes because C(x) is the cost evaluated at x. So we need to minimize cost by differentiating with respect to x.
Thanks for the assistance @LogicalApple . Appreciate it.
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.
Yeah I already finished the question.
The hard part was what you did for me.
I did it on paper and I was like "hm.. I can't think in litres/gallon"
Ah, I did it again lol I mean litres/hour!
Well I'm going to close this and put up another question. Thanks again for helping me.
Hope you can help me again if you see fit to.