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    Aah difficult problem

    A driver on a desert road discovers a hole in the gas tank leaking gas at the constant rate of 4 gallons per hour. This driver, having no way to plug the hole, decides to drive for as long as the gas supply allows. The gauge reading indicates the tank is three-fourths full, which means that the tank contains 14 gallons. The car consumes gas at the rate of 18 miles per gallon at 40 mph. For each 5 mpg below 40 mph add one-half mile per gallon to this rate; for each 5 mpg above 40 mpg, subtract one mile per gallon from this rate. If the driver chooses the best constant speed in order to get the maximum driving distance, find the maximum distance that the 14 gallons will allow. Assume that gas consumption is a continuous functional speed.

    How do i solve this?
    Last edited by mr fantastic; October 15th 2009 at 08:47 PM. Reason: Restored deleted question.
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  2. #2
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    Quote Originally Posted by algaljal View Post
    A driver on a desert road discovers a hole in the gas tank leaking gas at the constant rate of 4 gallons per hour. This driver, having no way to plug the hole, decides to drive for as long as the gas supply allows. The gauge reading indicates the tank is three-fourths full, which means that the tank contains 14 gallons. The car consumes gas at the rate of 18 miles per gallon at 40 mph. For each 5 mpg below 40 mph add one-half mile per gallon to this rate; for each 5 mpg above 40 mpg, subtract one mile per gallon from this rate. If the driver chooses the best constant speed in order to get the maximum driving distance, find the maximum distance that the 14 gallons will allow. Assume that gas consumption is a continuous functional speed.

    How do i solve this?
    let's model it at 40 mph to see how to calculate the vehicle's range in miles.

    (40 mi/hr)/(18 mi/gal) = (20/9) gal/hr

    also lose 4 gal/hr ... so total usage is (56/9) gal/hr

    (14 gal)/(56/9 gal/hr) = (9/4) hrs

    (9/4 hrs)(40 mi/hr) = 90 mi

    to recap ...

    (speed)/(mileage) + 4 gets you usage in gal/hr

    (14 gal)/[(speed)/(mileage) + 4] = hrs

    finally ...

    (14 gal)/[(speed)/(mileage) + 4] * (speed) = range

    (14 gal)(speed)(mileage)/[(speed) + 4(mileage)] = range


    let x = number of 5 mph increments

    start with speeds < 40 mph

    (14)(40 - 5x)(18 + x/2)/[(40 - 5x) + 4(18 + x/2)] = range


    for speeds > 40 mph

    (14)(40 + 5x)(18 - x)/[(40 + 5x) + 4(18 - x)] = range

    you should be able to clean these equations up, find a derivative, and determine a maximum.

    using a calculator, I found the maximum range to be about 101.4 mi at a speed of about 61.4 mph
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