1. ## optimization problem

I'm having trouble coming up with an equation to differentiate for this problem

A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/L of gas. Gas costs 0.68$/L. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. Drivers are paid 35$/h in wages and benefits. Fixed costs for running the truck are $15.50/h. If a trip of 450km is planned, what speed will minimize the operating expenses? Anybody know? 2. Originally Posted by Hasan1 A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/L of gas. Gas costs 0.68$/L. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. Drivers are paid 35$/h in wages and benefits. Fixed costs for running the truck are$15.50/h. If a trip of 450km is planned, what speed will minimize the operating expenses?
Let the speed of the truck in km/h be $\displaystyle v$. The fuel efficiency can be modeled as

$\displaystyle FE = 8 - 0.1(v - 110) = 118 - 0.1v$

This gives us km/L, so $\displaystyle \frac1{FE}$ gives us the liters used per kilometer. Then we can find the cost of gas as a function of speed:

$\displaystyle \text{cost of gas} = \frac{1\text{ L}}{\left[118 - 0.1v\right]\text{ km}}\cdot\frac{\$0.68}{1\text{ L}}\cdot450\text{ km}\displaystyle =\$\frac{0.68\cdot450}{118 - 0.1v}=\$\frac{306}{118 - 0.1v}$Now we will need to know how long the trip will take in order to calculate the costs of driver wages and truck maintenance. Again, using$\displaystyle v$as speed:$\displaystyle \text{time} = \frac{\text{distance}}{\text{speed}}\displaystyle \Rightarrow\text{trip time} = \frac{450}v$So, the total trip will take$\displaystyle \frac{450}v$hours. The total cost, in dollars, is therefore$\displaystyle C =\frac{306}{118 - 0.1v} + 35\left(\frac{450}v\right) + 15.5\left(\frac{450}v\right)$Now just minimize$\displaystyle C$. 3. A truck crossing the prairies at a constant speed of 110 km/h gets 8 km/L of gas. Gas costs 0.68$/L. The truck loses 0.10 km/L in fuel efficiency for each km/h increase in speed. Drivers are paid 35$/h in wages and benefits. Fixed costs for running the truck are$15.50/h. If a trip of 450km is planned, what speed will minimize the operating expenses?
Perhaps you are so worried about the derivative that you aren't able to think on the more fundamental concepts involved. Remember, calculus should make life easier for certain types of problems. If it confuses you and causes you to forget your prerequisites, something is wrong.

We need operating expenses.

Rule #1 - Things should make sense.

We're thinking from the point of view of the truck owner. Paying the driver is part of the expense.

Rule #2 - Break it down to manageable pieces

Voyage Expenses = Driver + Gas + Fixed Costs

Rule #3 - What do we need in order to talk about these things in a useful way?

Everything is in Hours and Km, so it would be good to define things in these terms. Let's define a few things.

H = Hours to complete the trip.

We have variable gas costs, so it probably is helpful to define a function that will express this.

Speed of Truck = 110 kmh + t kmh
t kmh = number of kmh over 100 kmh for which we know the mileage.

I'm not really getting a feel for the problem, so let's just play with it a little.

At 110 kmh, we have 450 km / 110 kmh = 4.0909 hours and
450 km / 8 km/L = 56.25 L and
56.25 L * $0.68/L =$38.25

Now we have the flavor of the calculation.

At (110+t) kmh, we have 450 km / (110+t kmh) = H and
450 km / (8-0.10t km/L) and
450 km / (8-0.10t km/L) * $0.68/L Gas Expense = [450 km / (8-0.10t km/L)] *$0.68/L

Voyage Expenses = [450 km / (110+t kmh)]*($35.00/h) + [450 km / (8-0.10t km/L)] *$0.68/L + [450 km / (110+t kmh)]*(\$15.50/h)

After a very, very careful surgery consisting of algebra and unit checks, this gives:

Voyage Expenses(t) = 22725/(t+110) + 3060/(t-80)

Interestingly, maximizing that seems to be a problem of finding the endpoints.

Note: That was very complicated. Just think it through, one step at a time.