To start, you need to represent the cost of the trip as a function of the speed. Let's use where is the speed in km/hr.

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 isA 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?

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:

After simplifying and canceling units, you get

The numerator always gets smaller at a linear rate as 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.