Optimizing Truck Speed and Freight Load for Cost Efficiency on Interstate Travel

[Ph4cm]

A truck traveling interstate, driving at a constant speed of 110km/h, gets 7km/L efficiency and loses 0.1km/L in fuel efficiency for each km/h increase in speed. Costs include diesel ($1.49/L), truck drivers’ wage ($35/hour), and truck maintenance and repairs ($9.50/hour). This truck is mainly used for carrying freight between Adelaide and Sydney (1375km). Explain what would happen if the truck were able to maintain a constant speed over 250km/h?
 Is it reasonable to assume a constant speed for such a trip?
 (Find 2 sources)
 What factors affect the reasonableness of the model the most?
 Suppose that, for every extra freight container the truck carries, the fuel efficiency drops by 0.05km/L. Suggest what will happen to the optimum traveling speed as the load of the truck’s freight increases.
 Find a mathematical relationship between freight size and speed for which cost is a minimum. Discuss your findings.

I've figured out the cost model C= 1375*(801-2.96x)(x(18-0.1x)) (x is speed, C is total cost)

 

[skeeter]

A truck traveling interstate, driving at a constant speed of 110km/h, gets 7km/L efficiency and loses 0.1km/L in fuel efficiency for each km/h increase in speed. Costs include diesel (\$1.49/L), truck drivers’ wage (\$35/hour), and truck maintenance and repairs (\$9.50/hour). This truck is mainly used for carrying freight between Adelaide and Sydney (1375km). Explain what would happen if the truck were able to maintain a constant speed over 250km/h?
 Is it reasonable to assume a constant speed for such a trip?
 (Find 2 sources)
 What factors affect the reasonableness of the model the most?
 Suppose that, for every extra freight container the truck carries, the fuel efficiency drops by 0.05km/L. Suggest what will happen to the optimum traveling speed as the load of the truck’s freight increases.
 Find a mathematical relationship between freight size and speed for which cost is a minimum. Discuss your findings.

I've figured out the cost model C= 1375*(801-2.96x)(x(18-0.1x)) (x is speed, C is total cost)

What are your units for speed, the $x$ in your cost function?

I get cost as a function of speed, $v$, in km/hr ...

$C(v) = 1375 \bigg[\dfrac{44.50}{v} + \dfrac{1.49}{7-0.1(v-110)} \bigg]$

unit analysis of my cost equation ...

$ \$ = km\bigg[\dfrac{\$ /hr}{km/hr} + \dfrac{\$/L}{km/L - \frac{km/L}{km/hr}(km/hr)}\bigg]$

 

[Ph4cm]

Yes the x in the model represents in km/h.
Actually i really needed help in the last two questions. I've solved the first 3 questions. And don't mind the "find 2 sources" one it was my annotation. Would you mind to give me the direction to solve the last 2 questions ?

 

[skeeter]

simplified the initial cost function ...

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freightcost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{29.8}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{29.8}{(180-v-0.5n)^2}\right)$

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$

attached is a calculated table of values for $v$ as a function of $n$ (in the table, $x$ represents $n$ and $Y_1$ is speed, $v$) ...

 

[Ph4cm]

Could you explain more about how you've got the cost function C(v,n) ?
Cause I've got a different answer:
C= 1375∗(1.4918.05−0.1v−0.05n+44.5v)

 

[skeeter]

Could you explain more about how you've got the cost function C(v,n) ?
Cause I've got a different answer:
C= 1375∗(1.4918.05−0.1v−0.05n+44.5v)

I have a better idea ... explain how you derived your function.

 

[Ph4cm]

I have a better idea ... explain how you derived your function.

So the new fuel efficiency function is y(v,n)= 7-0.1(v-110)-0.05(n-1) = 18.05-0.1x-0.05n
Total cost for fuel: (1375*1.49)/(18.05-0.1x-0.05n)
Total cost for wages and repairs: 44.5*1375/x
Total cost for the trip: C(v,n) = 1375∗(1.49/18.05−0.1v−0.05n + 44.5/v)
Could you check if mine is correct ?

 

[skeeter]

So the new fuel efficiency function is y(v,n)= 7-0.1(v-110)-0.05(n-1) = 18.05-0.1x-0.05n

what does n-1 represent?

in the equation I wrote in post #4, $n$ represents the number of additional freight loads

 

[Ph4cm]

what does n-1 represent?

in the equation I wrote in post #4, $n$ represents the number of additional freight loads

Sorry it was my mistake, it should only be n like in your equation at #4. But i can't understand how did you get the 29.8 in the equation at #4. Would you explain for me a little bit more please ? That will really helps me

 

[skeeter]

Sorry it was my mistake, it should only be n like in your equation at #4. But i can't understand how did you get the 29.8 in the equation at #4. Would you explain for me a little bit more please ? That will really helps me

that's my mistake ... should be 14.9. I doubled the numerator and denominator to get the coefficient of $n$ to be 1 instead of 0.5. When I decided against it, I forgot to change the numerator back after I typed it.

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freight

corrected ...

cost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{{\color{red}14.9}}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{{\color{red}14.9}}{(180-v-0.5n)^2}\right)$...this last equation was correct (see the 149 in the radical?)

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$

 

[Ph4cm]

that's my mistake ... should be 14.9. I doubled the numerator and denominator to get the coefficient of $n$ to be 1 instead of 0.5. When I decided against it, I forgot to change the numerator back after I typed it.

$C(v) = 1375\left(\dfrac{44.5}{v} + \dfrac{14.9}{180-v}\right)$

minimum cost occurs at $v \approx 114 \, km/hr$ with no additional freight

corrected ...

cost as a function of speed and freight, where $n$ is the additional number of freight units

$C(v,n) = 1375 \left(\dfrac{44.5}{v} + \dfrac{{\color{red}14.9}}{180-v-0.5n} \right)$

$\dfrac{\partial C}{\partial v} = 1375\left(-\dfrac{44.5}{v^2} + \dfrac{{\color{red}14.9}}{(180-v-0.5n)^2}\right)$...this last equation was correct (see the 149 in the radical?)

$\dfrac{\partial C}{\partial v}=0 \implies v = \dfrac{180-0.5n}{1+A} \text{ where } A = \sqrt{\dfrac{149}{445}}$

Thank you a lot for this help Skeeter, It really helped me with my work. You are a wonderful person.