Thread: optimization problem - urgent !

1. 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 ! 2. Originally Posted by checkmarks 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.