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Homework Help: Optimization - help needed setting up a system of equations?

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data
    A truck driving over a flat interstate at a constant rate of 50 mph gets 4 miles to the gallon. Fuel costs $0.89 per gallon. For each mile per hour increase in speed, the truck loses a tenth of a mile per gallon of its mileage. Drivers get $27.50 per hour in wages, and the fixed costs for running the truck amount to $11.33 per hour. What constant speed, between 50 mph and 65 mph, should the dispatcher require on a straight run through 260 miles of Kansas interstate to minimize the total cost of operating the truck?

    2. Relevant equations
    50 mph for 4 miles/gallon $0.89/gallon cost
    1 mph increase in speed --> -(1/10) decrease in mile/gallon mileage
    $27.50/hr drivers' wages + $11.33/hr cost to operate = $38.83/hr cost (cost goes up with time)
    38.83 = dc/dt ?

    3. The attempt at a solution
    We're looking for optimum cost (C), I think.
    I am really stumped with a system of equations for this one.

    I could probably go a lot further with this one on my own if I just had a place to start. Does anyone have any enlightening thoughts?
  2. jcsd
  3. Jan 10, 2010 #2


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    Science Advisor

    At 50 mph, the truck gets 4 m/g but loses .1 m/g for each 10 mph above that: v- 50 is "mph above 50" and (v- 50)/10 measures how many 10 mph above 50. At speed v, the truck gets 4- .1(v-50)/10= 4- .01v+ .5= 4.5- .01v mpg. That is the same as 1/(4.5- .01v) gallons per mile. Since each gallon of gas cost $0.89 (this is an old problem!) the cost would be .89/(4.5- .01v) dollars per mile. Sice the total trip is 260 miles, the cost of gas would be 260(.89)/(4.5- .01v)= 231.4/(4.5- .01v) dollars.

    At v mph, 260 miles would take 260/v hours and so the driver would be payed 27.50(260)/v= 7150/v dollars.

    Since the truck costs an additional 11.33 per hour to run, that is 11.33(260)/v= 2945.8 dollars.

    Add those to get the total cost of the trip, the function to be optimized.
  4. Jan 10, 2010 #3
    Wow, thank you! That makes complete sense. I think.
    Last edited: Jan 10, 2010
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