Optimizing speed and cost.

  • #1
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Homework Statement


The cost of fuel per kilometre for a truck travelling [tex]v[/tex] kilometres per hour is given by the equation [tex] C(v) = \frac{v}{100}+\frac{25}{v}[/tex]. Assume the driver is paid $40/h. What speed would give the lowest cost, including fuel and wagesm for a 1000-km trip?


Homework Equations



[tex]C(v) = \frac{v}{100}+\frac{25}{v}[/tex]

The Attempt at a Solution



[tex]C(v) = \frac{v^2+2500}{100v}[/tex]

I simplified it into one expression. From here, I differentiate and find the minimum speed. I divide $40/h by the speed to get $/km which I can then use to solve again?

I do not think this is right.
 

Answers and Replies

  • #2
35,115
6,856

Homework Statement


The cost of fuel per kilometre for a truck travelling [tex]v[/tex] kilometres per hour is given by the equation [tex] C(v) = \frac{v}{100}+\frac{25}{v}[/tex]. Assume the driver is paid $40/h. What speed would give the lowest cost, including fuel and wagesm for a 1000-km trip?


Homework Equations



[tex]C(v) = \frac{v}{100}+\frac{25}{v}[/tex]

The Attempt at a Solution



[tex]C(v) = \frac{v^2+2500}{100v}[/tex]

I simplified it into one expression. From here, I differentiate and find the minimum speed.
You don't want the minimum speed, you want the minimum cost.
I divide $40/h by the speed to get $/km which I can then use to solve again?

I do not think this is right.
I agree.
The total cost for the trip is the cost for fuel plus the driver's wages. You are given the cost per km as a function of the speed, but you also need to add in the driver's wages, which are also a function of the speed (the faster he drives, the fewer hours, so the less he gets).
 
  • #3
145
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That is the part I am stuck at. If I solve the derivative, I would find out the speed which would give the lowest fuel cost per kilometre.
 
  • #4
145
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While I tried to do this, I got a local max at x=-50 and local min at x=50. Something does not seem right.
 
  • #5
145
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While I tried to do this, I got a local max at v=-50 and local min at v=50. Something does not seem right.

If I am trying to find the minimum cost, I should be using maximum speed which occurs at v=-50.
 
  • #6
35,115
6,856
That is the part I am stuck at. If I solve the derivative, I would find out the speed which would give the lowest fuel cost per kilometre.
What is the function you have for the total cost of the trip? What is it that you are taking the derivative of? You DO NOT want to take the derivative of C(v).
 
  • #7
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What is the function you have for the total cost of the trip? What is it that you are taking the derivative of? You DO NOT want to take the derivative of C(v).

Would I have to multiply the $40/h in to C(v) to get the total cost?

Edit: I know I have to do something with 40.
 
Last edited:
  • #8
35,115
6,856
Yes, you have to do something with the $40/hr.

Look at what I wrote at the bottom of post #2. If what I have isn't clear, ask questions about it.
 
  • #9
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Yes, you have to do something with the $40/hr.

Look at what I wrote at the bottom of post #2. If what I have isn't clear, ask questions about it.

I think I understand this now.

You know that speed is equal to distance/time. Time = distance/speed.

[tex] t = \frac{1000}{v}[/tex]

Now, to make the total cost function you would write: [tex]C(t) = (\frac{v}{100}+\frac{25}{v})(1000) + 40t[/tex]

Substitute d/t into the equation, differentiate and solve, you would get how many hours (minimum). Then you can find speed.
 
  • #10
35,115
6,856
That's a little different from what I did, but it seems reasonable, so I think it will work. Notice that your cost function represents fuel cost for the trip + cost of driver's wages, which is what I've been saying you need to work with.
 

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