Optimization calculus question (Difficult)

[Wild ownz al]

A truck crossing the prairies at constant speed of 110km per hour gets 8km per litre of gas. Gas costs 0.68 dollars per litre.
The truck loses 0.10 km per litre in fuel efficiency for each km per hour increase in speed.
Drivers are paid 35 dollars per hour in wages benefits.
Fixed costs for running the truck are 15.50 dollars per hour.
If a trip of 450km is planned, what speed will minimize operating expenses?

This is the formula I made:

C = Cost of operation
v = increases of speed (1km/h)

C(v) = 0.68(450/(8 - v/10)) + (450/(110+v))(35+15.5)

 

[Wild ownz al]

If anyone also know's how to fix that font please do let me know.

 

[skeeter]

A truck crossing the prairies at constant speed of 110km per hour gets 8km per litre of gas. Gas costs \$0.68 per Litre.
The truck loses 0.10 km per litre in fuel efficiency for each km per hour increase in speed.
Drivers are paid \$35 per hour in wages 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?

This is the formula I made:

C = Cost of operation
v = increases of speed (1km/h)

C(v) = 0.68(450/(8 - v/10)) + (450/(110+v))(35+15.5)

speed = $(110+v) \text{ km/hr}$

distance = $450 \text{ km}$

trip time in hrs = $\dfrac{450 \text{ km}}{(110+v) \text{ km/hr}}$

fuel burn rate in L/hr = $\dfrac{(110+v) \text{ km/hr}}{(8-0.1v) \text{ km/L}}$cost = (driver cost rate + fixed cost rate)(hrs) + (fuel burn rate in L/hr)(hrs)(fuel cost in dollars/L)$C(v) = (35+15.5) \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} + \dfrac{(110+v) \text{ km/hr}}{(8-0.1v) \text{ km/L}} \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} \cdot 0.68$
...

 

[Wild ownz al]

speed = $(110+v) \text{ km/hr}$

distance = $450 \text{ km}$

trip time in hrs = $\dfrac{450 \text{ km}}{(110+v) \text{ km/hr}}$

fuel burn rate in L/hr = $\dfrac{(110+v) \text{ km/hr}}{(8-0.1v) \text{ km/L}}$cost = (driver cost rate + fixed cost rate)(hrs) + (fuel burn rate in L/hr)(hrs)(fuel cost in dollars/L)$C(v) = (35+15.5) \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} + \dfrac{(110+v) \text{ km/hr}}{(8-0.1v) \text{ km/L}} \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} \cdot 0.68$
...

Our functions give different outputs of the same x value. Your's looks accurate but what exactly is wrong with my function? I don't see why I can't multiply the fuel cost by the total litre's (450/(8-v/10))

 

[skeeter]

Our functions give different outputs of the same x value. Your's looks accurate but what exactly is wrong with my function? I don't see why I can't multiply the fuel cost by the total litre's (450/(8-v/10))

they're the same ... I just took an extra step

$C(v) = (35+15.5) \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} + \dfrac{\cancel{(110+v) \text{ km/hr}}}{(8-0.1v) \text{ km/L}} \cdot \dfrac{450 \text{ km}}{\cancel{(110+v) \text{ km/hr}}} \cdot 0.68$