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Azteck

  • 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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  1. Azteck
    • 3 years ago
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    Topic: Calculus Sub Topic: Maxima and Minima.

  2. LogicalApple
    • 3 years ago
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    Is it 74.83 km/h?

  3. Azteck
    • 3 years ago
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    Yes!!

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

  5. Azteck
    • 3 years ago
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    75km/hour

  6. Azteck
    • 3 years ago
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    How did you do the first part?

  7. LogicalApple
    • 3 years ago
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    sorry

  8. Azteck
    • 3 years ago
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    gallon?

  9. Azteck
    • 3 years ago
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    American?

  10. LogicalApple
    • 3 years ago
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    let me rewrite that , (i used american units and accidentally wrote it in the wrong place)

  11. LogicalApple
    • 3 years ago
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    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

  12. LogicalApple
    • 3 years ago
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    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

  13. Azteck
    • 3 years ago
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    So far, I'm up to: \[0.08*(25+\frac{ x^2 }{ 112 })\]

  14. LogicalApple
    • 3 years ago
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    \[0.08 \frac{ dollar }{ litre } * (25 + \frac{ x^{2} }{ 112 }) \frac{ litre }{ hour } * \frac{ 200 }{ x } hours\]

  15. Azteck
    • 3 years ago
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    So you get that, and do you then add: \[\frac{ 400 }{ x }\] for the money he receives per hour?

  16. LogicalApple
    • 3 years ago
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    yes

  17. Azteck
    • 3 years ago
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    okay. Now I'm getting somewhere.

  18. Azteck
    • 3 years ago
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    So do we differentiate that with respect to x?

  19. LogicalApple
    • 3 years ago
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    Yes because C(x) is the cost evaluated at x. So we need to minimize cost by differentiating with respect to x.

  20. Azteck
    • 3 years ago
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    Thanks for the assistance @LogicalApple . Appreciate it.

  21. LogicalApple
    • 3 years ago
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    Not a problem

  22. LogicalApple
    • 3 years ago
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    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.

  23. Azteck
    • 3 years ago
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    Yeah I already finished the question.

  24. LogicalApple
    • 3 years ago
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    Oh, nice

  25. Azteck
    • 3 years ago
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    The hard part was what you did for me.

  26. LogicalApple
    • 3 years ago
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    I did it on paper and I was like "hm.. I can't think in litres/gallon"

  27. LogicalApple
    • 3 years ago
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    Ah, I did it again lol I mean litres/hour!

  28. Azteck
    • 3 years ago
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    hahah.

  29. Azteck
    • 3 years ago
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    Well I'm going to close this and put up another question. Thanks again for helping me.

  30. Azteck
    • 3 years ago
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    Hope you can help me again if you see fit to.

  31. LogicalApple
    • 3 years ago
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    Good luck!

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