Problem 71 - Calculus and its applications

Question

Total revenue. A total-revenue function is given by $$ R(x)=1000 \sqrt{x^{2}-0.1 x} $$ where \(R(x)\) is the total revenue, in thousands of dollars, from the sale of \(x\) airplanes. Find the rate at which total revenue is changing when 20 airplanes have been sold.

Step-by-Step Solution

Answer

The rate of change of total revenue when 20 airplanes are sold is $1,000,000 per airplane.

Step 1: Identify the Given Function and Differentiate

We are given the total-revenue function as \( R(x) = 1000 \sqrt{x^2 - 0.1x} \). To find the rate at which revenue changes, we need to compute the derivative \( R'(x) \). First, let's apply the chain rule to differentiate \( \sqrt{x^2 - 0.1x} \).

Step 2: Apply the Chain Rule

The derivative of \( \sqrt{x^2 - 0.1x} \) is \( \frac{1}{2} (x^2 - 0.1x)^{-1/2} \cdot (2x - 0.1) \). Thus, \( R'(x) = 1000 \cdot \frac{1}{2} \cdot (x^2 - 0.1x)^{-1/2} \cdot (2x - 0.1) \).

Step 3: Simplify the Derivative

Simplifying, we get \( R'(x) = 500 (x^2 - 0.1x)^{-1/2} (2x - 0.1) \). This is the derivative function that gives us the rate of change of revenue with respect to airplanes sold.

Step 4: Evaluate the Derivative at Specific Point

We need to find \( R'(20) \), so we substitute \( x = 20 \) into the derivative. Calculate:1. \( x^2 - 0.1x = 20^2 - 0.1 \times 20 = 400 - 2 = 398 \). 2. \( (x^2 - 0.1x)^{-1/2} = (398)^{-1/2} \).3. \( 2x - 0.1 = 2 \times 20 - 0.1 = 40 - 0.1 = 39.9 \).Thus, \( R'(20) = 500 \times 39.9 \times (398)^{-1/2} \).

Step 5: Calculate R'(20)

Using a calculator to compute, we find:1. \( (398)^{1/2} \approx 19.95 \).2. \( 500 \times 39.9 / 19.95 \approx 1000 \).Therefore, \( R'(20) \approx 1000 \).

Step 6: Interpret the Result

The result \( R'(20) = 1000 \) means that when 20 airplanes have been sold, the rate at which total revenue is changing is approximately \(1000,000 per airplane, or \)1,000,000 per airplane.

Key Concepts

Derivative Applications

The application of derivatives is fundamental in understanding how a function behaves, especially in terms of its rate of change. In the context of economics, the derivative of a revenue function indicates how revenue changes with respect to varying levels of production or sales.

In this exercise, we are working with the revenue function \( R(x) = 1000 \sqrt{x^2 - 0.1x} \), where \( x \) is the number of airplanes sold. To determine how the total revenue changes as more airplanes are sold, we compute the derivative, \( R'(x) \).

Derivatives tell us:
- The instantaneous rate of change of the function.
- Slope of the tangent line at any given point.

These insights are crucial for businesses because understanding the rate of change in revenue can inform strategic decisions, like pricing or production adjustments. The exercise specifically asks for \( R'(20) \), indicating the rate at which revenue changes when 20 airplanes are sold.

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. In simple terms, it helps us differentiate functions that are nested within others.

In the case of our revenue function, \( R(x) = 1000 \sqrt{x^2 - 0.1x} \), we deal with a composite function. The outer function is the square root, and the inner function is \( x^2 - 0.1x \). To differentiate \( R(x) \), we apply the chain rule:

• Differentiate the outer function, treating the inner function as a single variable. This gives \( \frac{1}{2}(x^2 - 0.1x)^{-1/2} \).
• Multiply by the derivative of the inner function, \( 2x - 0.1 \).

Using the chain rule like this allows us to calculate the rate of change of the revenue, which becomes \( R'(x) = 500 (x^2 - 0.1x)^{-1/2}(2x - 0.1) \). This method reveals how intertwined rates of change are calculated efficiently.

Revenue Function Analysis

Analyzing a revenue function involves examining how revenue varies as output changes. In business, these analyses can provide pivotal insights into optimal production levels and pricing strategies.

For \( R(x) = 1000 \sqrt{x^2 - 0.1x} \), our goal was to determine \( R'(20) \), essentially measuring how receptive revenue is to change when exactly 20 airplanes are sold. Calculations involved finding \( R'(x) \), and then evaluating it at \( x = 20 \).

After applying the chain rule and simplifying, the derivative function \( R'(x) \) was found. Plugging in \( x = 20 \) gives us specific values:

• Calculate \( 20^2 - 0.1 \times 20 = 398 \).
• Find the inverse square root of 398, \( (398)^{-1/2} \).
• Compute \( 2 \times 20 - 0.1 = 39.9 \).

Thus, \( R'(20) = 1000 \), indicating that when 20 airplanes are sold, the revenue increases by $1,000,000 per additional airplane sold. This type of analysis helps businesses make proactive decisions driven by data, ultimately improving profitability.