Optimization calculus question (Difficult)

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    Calculus Optimization
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Discussion Overview

The discussion revolves around an optimization problem related to minimizing operating expenses for a truck trip of 450 km, considering factors such as fuel efficiency, fuel costs, driver wages, and fixed costs. Participants are analyzing the mathematical formulation of the cost function and exploring different approaches to derive the optimal speed for the trip.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a cost function C(v) that incorporates fuel costs, driver wages, and fixed costs, with variables defined for speed increases and distance.
  • Another participant questions the accuracy of their own cost function compared to a peer's, expressing confusion over the multiplication of fuel costs by total liters consumed.
  • There is a reiteration of the cost function with an emphasis on the components involved, including trip time and fuel burn rate, suggesting a detailed breakdown of the calculations.
  • Participants note that their functions yield different outputs for the same input values, prompting a discussion on the correctness of each formulation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the different cost functions presented. There is ongoing debate regarding the formulation of the cost function and the implications of their respective approaches.

Contextual Notes

Some assumptions regarding fuel efficiency and cost calculations may not be fully articulated, and there are unresolved mathematical steps in the derivation of the cost functions. The discussion reflects varying interpretations of the problem parameters.

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

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

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