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New-Car Revenue The function \(R(x)\) billion dollars models revenue from new- car sales for franchised new-car dealerships in the United States when \(x\) million dollars are spent on associated advertising expenditures, data from \(1.2 \leq x \leq 6.5\) (Source: Based on data from Statistical Abstract for data between 1980 and 2000\()\) a. Use \(R(1.5)=141\) and \(R^{\prime}(1.5)=38\) to write a linearization model with respect to \(x\) at \(c=1.5 .\) b. Estimate the revenue when \(\$ 1.6\) million is spent on advertising. c. Estimate the revenue when \(\$ 2\) million is spent on advertising.

Short Answer

Expert verified
Estimated revenue is $144.8 billion for $1.6 million and $160 billion for $2 million spent on advertising.

Step by step solution

01

Understanding Linearization

Linearization is a method for approximating a function near a given point using the tangent line at that point. The linearization of a function \( R(x) \) at \( x = c \) is given by: \( L(x) = R(c) + R'(c)(x - c) \).
02

Write the Linearization Model

We're given \( R(1.5) = 141 \) and \( R'(1.5) = 38 \). Using these, the linearization model around \( x = 1.5 \) is: \[L(x) = 141 + 38(x - 1.5) \]
03

Estimate Revenue for $1.6 Million in Advertising

Substitute \( x = 1.6 \) into the linearization model \( L(x) = 141 + 38(x - 1.5) \): \[L(1.6) = 141 + 38(1.6 - 1.5) = 141 + 38(0.1) = 141 + 3.8 = 144.8\] So, the estimated revenue is \( 144.8 \) billion dollars.
04

Estimate Revenue for $2 Million in Advertising

Substitute \( x = 2 \) into the linearization model \( L(x) = 141 + 38(x - 1.5) \): \[L(2) = 141 + 38(2 - 1.5) = 141 + 38(0.5) = 141 + 19 = 160\] Thus, the estimated revenue is \( 160 \) billion dollars.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear approximation
When you're trying to understand a complex function, sometimes you need it to be a bit simpler. That's where linear approximation, or linearization, comes into play. It's a technique in calculus that helps us approximate a function near a specific point.
  • Imagine linear approximation as fitting a straight line to a function where the curve is too complex to work with easily.
  • This straight line is called the tangent line, and it passes through the function at the specific point we're interested in.
  • We use the equation of the tangent line to predict the function's behavior near this point.
The linearization of a function, represented as \( L(x) \), around a point \( c \) is given by the formula:
\[L(x) = R(c) + R'(c)(x - c)\]This formula relies on:
  • \( R(c) \), the function value at point \( c \).
  • \( R'(c) \), the derivative of the function at point \( c \), representing the slope of our tangent line.
By using linear approximation, complex changes in functions can be simplified to linear changes, making our calculations much easier. In essence, this method turns curves into straight lines only when we're looking closely enough!
Tangent line
The tangent line is a crucial concept in calculus, particularly in the context of linear approximation where it serves as the main tool. It's the straight line that just "touches" a curve at a specific point, mirroring the slope of the curve exactly at that spot.
  • Think of the tangent line as the line that best summarizes the trend of the curve at a particular point.
  • It tells us how steep the curve is right at the point where they touch, essentially giving us a "snapshot" of its behavior.
The formula of the tangent line is derived from its slope, which is determined by the derivative of the function at that point:
  • The derivative \( R'(c) \) tells us how quickly things are changing — whether the curve is climbing upwards, dipping downwards, or remaining steady.
Using both the function value and its slope at a specific point, the tangent line becomes a powerful way to approximate the function's value near that spot. Because the tangent line ideally just kisses the curve at the point of interest, it provides a close approximation of the curve itself in the immediate vicinity. This makes the tangent line indispensable, especially when solving problems like estimating car sales revenues with varying advertising budgets, where keeping things simple can lead to quick, insightful predictions!
Revenue estimation
In real-world scenarios like estimating revenue from advertisements, understanding how small changes affect outcomes can save time and resources. That's where revenue estimation becomes useful, thanks to techniques like linear approximation.
  • Revenue models, such as those predicting car dealership earnings based on advertising spend, can be complex with many variables.
  • By focusing on a small range of advertising expenses, linear approximation helps to simplify revenue predictions into manageable estimates.
For example, with our linear model:
- When advertising spends shift slightly, using the linear model helps to quickly gauge how revenues might respond. - In the given exercise, simply plugging in different values of \( x \) allows for speedy calculations without having to compute intricate, non-linear models.
By estimating revenue using a straightforward linear model, car dealerships (or any company) can better forecast outcomes and make informed financial decisions on advertising strategies without needing to dive deep into complex equations. This ability to predict with just a small amount of input — such as how much more you'll make by spending slightly more on ads — showcases the practical benefit of calculus in daily business operations.

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