Problem 44 - Calculus Early Transcendentals

Question

Travel cost The cost of a trip that is \(L\) miles long, driving a car that gets \(m\) miles per gallon, with gas costs of \(\$ p /\) gal is \(C=L p / m\) dollars. Suppose you plan a trip of \(L=1500 \mathrm{mi}\) in a car that gets \(m=32 \mathrm{mi} / \mathrm{gal},\) with gas costs of \(p=\$ 3.80 / \mathrm{gal}\) a. Explain how the cost function is derived. b. Compute the partial derivatives \(C_{L}, C_{m},\) and \(C_{p} .\) Explain the meaning of the signs of the derivatives in the context of this problem. c. Estimate the change in the total cost of the trip if \(L\) changes from \(L=1500\) to \(L=1520, m\) changes from \(m=32\) to 31 and \(p\) changes from \(\$ 3.80\) to \(\$ 3.85\) d. Is the total cost of the trip (with \(L=1500 \mathrm{mi}\) \(m=32 \mathrm{mi} / \mathrm{gal},\) and \(p=\$ 3.80\) ) more sensitive to a \(1 \%\) change in \(L, m,\) or \(p\) (assuming the other two variables are fixed)? Explain.

Step-by-Step Solution

Answer

#Answer# 0.13

Step 1: Derivation of the cost function

To derive the cost function, we need to incorporate the distance \(L\) in miles, the car's fuel efficiency \(m\) in miles/gallon, and the gas cost \(p\) in dollars/gallon. The total number of gallons needed for the trip is given by the distance/efficiency; i.e., \(L / m\). Multiplying this by the cost of each gallon, \(p\), gives the total cost of the trip: $$ C = \frac{Lp}{m}. $$

Step 2: Partial derivatives of the cost function

Now, we need to compute the partial derivatives \(C_L, C_m\), and \(C_p\). \(C_L = \frac{\partial C}{\partial L} = \frac{p}{m}\) \(C_m = \frac{\partial C}{\partial m} = -\frac{Lp}{m^2}\) \(C_p = \frac{\partial C}{\partial p} = \frac{L}{m}\)

Step 3: Meaning of the partial derivatives

The signs of the partial derivatives provide insight into how the total cost would be affected by changes in each variable: \(C_L\): This is a positive value, meaning that an increase in the distance of the trip (\(L\)) will result in a higher cost. \(C_m\): This is a negative value, meaning that an increase in the car's fuel efficiency (\(m\)) will result in a lower cost. \(C_p\): This is a positive value, meaning that an increase in the gas cost (\(p\)) will result in a higher cost.

Step 4: Estimating the change in total cost

We are asked to estimate the change in the total cost of the trip if \(L\) changes from 1500 to 1520, \(m\) changes from 32 to 31, and \(p\) changes from \(3.80 to \)3.85. We can use the partial derivatives to estimate this change by summing their products with the respective changes: \(\Delta C \approx C_L \Delta L + C_m \Delta m + C_p \Delta p\) Plugging in the given values we have: \(\Delta C \approx (\frac{3.80}{32})(20) - (\frac{1500(3.80)}{32^2})(1) + (\frac{1500}{32})(0.05)\) \(\Delta C \approx 0.13\)

Step 5: Sensitivity of the total cost

Finally, we need to determine which of the variables is more sensitive to a 1% change, assuming the other two variables are fixed. We can do this by comparing the magnitude of the partial derivatives: \(C_L = \frac{3.80}{32} \approx 0.119\) \(C_m = -\frac{1500(3.80)}{32^2} \approx -0.446\) \(C_p = \frac{1500}{32} \approx 46.875\) From the above comparisons, we can see that the total cost of the trip is most sensitive to a 1% change in the gas cost, as \(C_p\) has the highest value.

Key Concepts

Cost Function

The cost function is a mathematical formula that helps calculate the total cost of a car trip based on specific variables. The variables involved include:

• Distance of the trip (\(L\)) in miles.
• Fuel efficiency of the car (\(m\)) in miles per gallon.
• Cost of gas (\(p\)) in dollars per gallon.

By combining these, we arrive at the cost function, which is given by the formula:\[\C = \frac{Lp}{m}.\]
This equation calculates how much you spend on gas by determining how many gallons are needed and multiplying this by the price per gallon. It provides a clear view of how changes in distance, fuel efficiency, and gas price can affect overall travel expenses.

Fuel Efficiency

Fuel efficiency measures how effectively a car uses fuel, expressed as miles per gallon (\(m\)). In the context of the cost function, higher fuel efficiency lowers the total cost of the trip because:

• The cost function shows that when \(m\) increases, the denominator of \(\frac{Lp}{m}\) becomes larger.
• This decrease in cost per gallon needed results in less fuel used, meaning less money spent.

Understanding fuel efficiency's role helps identify economic benefits of driving more fuel-efficient vehicles, such as saving money and reducing environmental impact.

Sensitivity Analysis

Sensitivity analysis is about understanding how changes in different variables affect the outcome, here being the cost of the trip.

• Partial derivatives \(C_L\), \(C_m\), and \(C_p\) help measure sensitivity. They indicate how the cost responds to small changes in distance, efficiency, and gas price.
• In this problem:
  • \(C_L\), positive, shows cost increase with longer trips.
  • \(C_m\), negative, reveals cost decreases with better efficiency.
  • \(C_p\), positive, illustrates cost rises with gas price.

Conducting a sensitivity analysis helps prioritize decisions, such as whether improving fuel efficiency is more beneficial than seeking cheaper gas prices.

Distance and Cost Relationship

The relationship between distance and cost is proportional when other factors are constant. In the cost function \(C = \frac{Lp}{m}\), distance (\(L\)) directly affects the cost:

• Longer distances increase the total cost, as more fuel is required.
• The partial derivative \(C_L = \frac{p}{m}\) is positive, confirming this direct relationship.

Whenever planning a trip, this relationship is crucial for budget estimation. Understanding it helps in making informed decisions about the route, potential stops, or even considering alternative transportation methods if costs are high. Implementing such analysis ensures efficiency and helps avoid unexpected expenses during travel.