Optimization calculus question (Difficult)

In summary, we can use the formula C(v) = 0.68(450/(8 - v/10)) + (450/(110+v))(35+15.5) to calculate the cost of operation for a truck crossing the prairies at different speeds. The truck gets 8km per litre of gas at a constant speed of 110km per hour, but loses 0.10 km per litre in fuel efficiency for each km per hour increase in speed. In addition, drivers are paid \$35 per hour in wages benefits and fixed costs for running the truck are \$15.50 per hour. To minimize operating expenses for a trip of 450km, we can use the function C(v) = (35
  • #1
Wild ownz al
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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)
 
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  • #2
If anyone also know's how to fix that font please do let me know.
 
  • #3
Wild ownz al said:
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)$C(v) = (35+15.5) \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} + \dfrac{(110+v) \text{ km/hr}}{(8-0.1v) \text{ km/L}} \cdot \dfrac{450 \text{ km}}{(110+v) \text{ km/hr}} \cdot 0.68$
...
 
  • #4
skeeter said:
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))
 
  • #5
Wild ownz al said:
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

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

FAQ: Optimization calculus question (Difficult)

1. What is optimization calculus?

Optimization calculus is a branch of mathematics that deals with finding the maximum or minimum value of a function. It involves using calculus techniques such as differentiation and integration to solve problems related to optimization.

2. How do you solve optimization calculus problems?

To solve optimization calculus problems, you need to follow a few steps. First, you need to identify the objective function and the constraints. Then, you need to use calculus techniques to find the critical points of the function. Finally, you need to evaluate the critical points and determine which one gives the maximum or minimum value of the function.

3. What are the common applications of optimization calculus?

Optimization calculus has various applications in real-life situations, such as in economics, engineering, and physics. It can be used to maximize profits, minimize costs, optimize production processes, and find the most efficient design for a structure or system.

4. What are the challenges of solving difficult optimization calculus problems?

Solving difficult optimization calculus problems can be challenging because it requires a strong understanding of calculus concepts and techniques. It also involves making assumptions and simplifications, which may not always be accurate in real-life scenarios. Additionally, finding the global maximum or minimum of a function can be time-consuming and may require trial and error.

5. How can I improve my skills in solving optimization calculus problems?

To improve your skills in solving optimization calculus problems, practice is key. Make sure to have a strong foundation in calculus concepts and techniques. It can also be helpful to work on a variety of problems and seek guidance from a teacher or tutor when needed. Additionally, understanding the real-life applications of optimization calculus can help you approach problems more effectively.

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