1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook