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 \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\). 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 \(m=31,\) and \(p\) changes from \(p=\$ 3.80\) to \(p=\$ 3.85\) 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 \(L,\) in \(m,\) or in \(p\) (assuming the other two variables are fixed)? Explain.

Short answer

A: The total cost of the trip is equally sensitive to a 1% change in distance traveled (L), mileage of the car (m), or gas price (p) when the other two variables are fixed, since the sensitivities are approximately equal at 1.78 for all three variables.

Step-by-step solution

Derivation of the Cost Function

Let's first derive the cost function \(C (L, m, p)\). We know that the cost of a trip is directly proportional to the price of gas and the distance traveled, and inversely proportional to the mileage of the car. This can be represented as: \(C = L \cdot \frac{p}{m}\) This is the cost function, where \(C\) is the total cost, \(L\) is the distance traveled in miles, \(m\) is the mileage of the car in miles/gallon, and \(p\) is the gas cost in dollars/gallon. #b. Compute the partial derivatives \(C_{L}, C_{m^{\prime}}\) and \(C_{p^{\prime}}\).# In order to find the partial derivatives of \(C\), we will differentiate the given function with respect to \(L\), \(m\), and \(p\).

Partial Derivative with respect to \(L\)

\(\frac{\partial C}{\partial L} = \frac{p}{m}\). The sign of this partial derivative is positive, which signifies that the total cost increases as the distance traveled increases (keeping other variables constant).

Partial Derivative with respect to \(m\)

\(\frac{\partial C}{\partial m} = -L \frac{p}{m^2}\). The sign of this partial derivative is negative, which indicates that the total cost decreases as the mileage of the car increases (keeping other variables constant).

Partial Derivative with respect to \(p\)

\(\frac{\partial C}{\partial p} = \frac{L}{m}\). The sign of this partial derivative is positive, which shows that the total cost increases as the gas price increases (keeping other variables constant). #c. Estimate the change in the total cost of the trip.#

Estimating the Change in Total Cost

To estimate the change in the total cost of the trip, we can use the partial derivatives and the given changes in \(L\), \(m\), and \(p\). Let’s denote the changes as \(\Delta L = 20\), \(\Delta m = -1\), \(\Delta p = 0.05\). Using the values of the partial derivatives, we have: \(\Delta C = \left(\frac{\partial C}{\partial L}\right) \Delta L + \left(\frac{\partial C}{\partial m}\right)\Delta m + \left(\frac{\partial C}{\partial p}\right) \Delta p\) \(\Delta C = \left(\frac{p}{m}\right)(20) - \left(L\frac{p}{m^2}\right)(-1) + \left(\frac{L}{m}\right)(0.05)\) Now plug in the values for \(L=1500\), \(m=32\), \(p=3.80\): \(\Delta C = \left(\frac{3.80}{32}\right)(20) - \left(1500\frac{3.80}{32^2}\right)(-1) + \left(\frac{1500}{32}\right)(0.05)\) Simplifying the expression, we have: \(\Delta C \approx 6.88\) Therefore, the estimated change in the total cost of the trip is approximately \(6.88\) dollars. #d. Is the total cost of the trip more sensitive to changes in \(L\), \(m\), or \(p\)?#

Sensitivity Analysis

The sensitivity of the total cost to a 1% change in any variable can be calculated using their respective partial derivatives and their corresponding variable value. For \(L\): Sensitivity \(= \left|\left(\frac{\partial C}{\partial L}\right) \cdot \frac{1}{100} L\right| = \left|\left(\frac{p}{m}\right) \cdot \frac{1}{100} L\right|\) For \(m\): Sensitivity \(= \left|\left(\frac{\partial C}{\partial m}\right) \cdot \frac{1}{100} m\right| = \left|\left(- L \frac{p}{m^2}\right) \cdot \frac{1}{100} m\right|\) For \(p\): Sensitivity \(= \left|\left(\frac{\partial C}{\partial p}\right) \cdot \frac{1}{100} p\right| = \left|\left(\frac{L}{m}\right) \cdot \frac{1}{100} p\right|\) Plugging in the values \(L=1500\), \(m=32\), and \(p=3.80\), we get: Sensitivity for \(L \approx 1.78\), for \(m \approx 1.78\), and for \(p \approx 1.78\) The total cost of the trip is equally sensitive to a 1% change in \(L\), \(m\), or \(p\) when the other two variables are fixed.

Key concepts

Partial Derivatives

Partial derivatives are like the magnifying glass of calculus, allowing us to see how one part of a multivariable function changes while keeping the others constant. Imagine you have a big, multi-flavored cake, and you want to know how adding a little more of one ingredient affects the taste without changing anything else – that's the essence of partial derivatives! In the cost function for a trip, we take partial derivatives to find out how the cost changes with respect to each factor: distance (L), mileage (m), and gas price (p).
- For the distance, the partial derivative \( \frac{\partial C}{\partial L} = \frac{p}{m} \) tells us the cost increases as the trip gets longer, sort of like pouring more syrup on a stack of pancakes.
- When it comes to mileage (\frac{\partial C}{\partial m}), the negative sign \( -L\frac{p}{m^2} \) suggests that better mileage makes the trip cheaper. Think of it as using a lighter touch with your ingredients, saving money at each step.
- For the gas price (\frac{\partial C}{\partial p}), \( \frac{L}{m} \) is positive, implying that a rise in gas price would increase your costs, like everyone else charging more at the same booth. This tool helps us see the creeping changes in cost when individual components shift.

Sensitivity Analysis

Sensitivity analysis is like testing the waters before diving in. It's used to understand which factors have the biggest impact on the outcome – in our case, the trip cost. Imagine you're cooking a dish and want to know if it's the salt, pepper, or garlic that's crucial to the flavor. Sensitivity tests give this insight for cost functions.

Here, we calculate sensitivity to minor percentage changes in distance, mileage, and gas price. It involves combining the partial derivatives with the actual values to see which change shakes things up the most.
- For each variable, like distance, the formula becomes: \[ \left|\frac{\partial C}{\partial L} \cdot \frac{1}{100} \times L \right|\]. Do this for mileage and gas price too. It's like fiddling a 1% change knob and noting the shift.
- In actual steps, all variables (L, m, p) turned out equally sensitive. This means tweaking any of them by 1% affects the trip cost similarly. A rare balance!

Trip Cost Calculation

Trip cost calculation forms the backbone of this exercise. It's like creating a budget map for your whole journey from start to end. The cost function \( C = \frac{Lp}{m} \) is more than just math; think about a plan for exactly how much you'll shell out for gas on this trip.
- You put in your journey details: distance (1500 miles), mileage (32 miles per gallon), and cost per gallon (\$3.80). All these together give the cost, guiding the feasibility of your trip financially.

- Estimating change in costs with tweaks is like adjusting your GPS when wandering. If your distance adjustment is 20 miles, mileage drops by 1 mpg, and gas price ticks up by \(0.05\), these all require re-evaluating your budget.
- Finally, a neat calculation based on partial derivatives predicts you'll need an extra \( \approx 6.88 \) dollars. Whether planning an actual trip or not, this equips you with the power of foresight over budget.