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Math Help - optimization problem - urgent !

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    optimization problem - urgent !

    Tomorrow I have a test on optimization and unfortunately this problem has stumped me. Please help me!

    A truck crossing the prairies at a constant speed of 110km/h gets gas mileage of 8km/L. Gas costs $1.15 per litre. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. The driver is paid $35 per hour in wages and benefits. Fixed costs for running the truck are $15.50 per hour. If a trip of 450km is planned, what speed will minimize operating expenses?

    Thanks in advance !
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    Quote Originally Posted by checkmarks View Post
    Tomorrow I have a test on optimization and unfortunately this problem has stumped me. Please help me!
    To start, you need to represent the cost of the trip as a function of the speed. Let's use C(s), where s is the speed in km/hr.

    A truck crossing the prairies at a constant speed of 110km/h gets gas mileage of 8km/L. Gas costs $1.15 per litre. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. The driver is paid $35 per hour in wages and benefits. Fixed costs for running the truck are $15.50 per hour. If a trip of 450km is planned, what speed will minimize operating expenses?
    Treat this like a unit conversion problem. To get the fuel costs, convert km to L, and then L to $, and so on. At 110 km/hr, the cost is

    C(110)=\text{fuel}+\text{wages}+\text{fixed costs}

    =(450\text{ km})\left(\frac{1 \text{ L}}{8\text{ km}}\right)\left(\frac{\$1.15}{1\text{ L}}\right)+(450\text{ km})\left(\frac{1\text{ hr.}}{110\text{ km}}\right)\left(\frac{\$35+\$15.50}{1\text{ hr.}}\right).

    Since we lose 0.10 km/L in efficiency with every km/hr. increase in speed, we can modify the appropriate conversion factor to get the general function:

    C(s)=(450\text{ km})\left(\frac{1 \text{ L}}{\left[8-0.1(s-110)\right]\text{ km}}\right)\left(\frac{\$1.15}{1\text{ L}}\right)+(450\text{ km})\left(\frac{1\text{ hr.}}{s\text{ km}}\right)\left(\frac{\$50.50}{1\text{ hr.}}\right).

    After simplifying and canceling units, you get

    C(s)=\frac{\$517.5}{19-0.1s}+\frac{\$22725}s

    =\frac{\$431775-\$1755s}{19s-0.1s^2}.

    The numerator always gets smaller at a linear rate as s increases. But the denominator is a quadratic representing a parabola which opens downward. The larger the denominator, the smaller the fraction. So what is the maximum point of this parabola? This will give the smallest cost.
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