Problem 42 - Applied Calculus

Question

The quantity demanded of a certain product, \(q\), is given in terms of \(p\), the price, by $$ q=1000 e^{-0.02 p} $$ (a) Write revenue, \(R\), as a function of price. (b) Find the rate of change of revenue with respect to price. (c) Find the revenue and rate of change of revenue with respect to price when the price is \(\$ 10\). Interpret your answers in economic terms.

Step-by-Step Solution

Answer

At \$10, revenue is \$8187; increasing price by \$1 increases revenue by \$654.96.

Step 1: Define Revenue as a Function of Price

The revenue, \( R \), is defined as the product of the price, \( p \), and the quantity demanded, \( q \). Since \( q = 1000 e^{-0.02p} \), we can write the revenue function as \( R(p) = p \times q = p \times 1000 e^{-0.02p} = 1000p e^{-0.02p} \).

Step 2: Differentiate Revenue Function with Respect to Price

We need to find \( \frac{dR}{dp} \). Let \( R(p) = 1000p e^{-0.02p} \). To differentiate this, use the product rule: if \( R(p) = u \times v \) where \( u = 1000p \) and \( v = e^{-0.02p} \), then \( \frac{dR}{dp} = u'v + uv' \). Here \( u' = 1000 \) and \( v' = e^{-0.02p} \times (-0.02) \). Hence, \( \frac{dR}{dp} = 1000 e^{-0.02p} + 1000p (-0.02) e^{-0.02p} = 1000 e^{-0.02p} - 20pe^{-0.02p} = (1000 - 20p)e^{-0.02p} \).

Step 3: Calculate Revenue and Rate of Change at Specific Price

First, substitute \( p = 10 \) into the revenue function: \( R(10) = 1000 \times 10 \times e^{-0.02 \times 10} = 10000 e^{-0.2} \approx 10000 \times 0.8187 = 8187 \). Next, substitute \( p = 10 \) into the rate of change formula: \( \frac{dR}{dp}\Big|_{p=10} = (1000 - 20 \times 10)e^{-0.02 \times 10} = (1000 - 200)e^{-0.2} = 800 \times e^{-0.2} \approx 800 \times 0.8187 = 654.96 \). This means that at a price of \( \\(10 \), the revenue is approximately \( \\)8187 \), and increasing the price slightly will increase the revenue by approximately \( \$654.96 \) per unit price increase.

Key Concepts

Quantity Demanded

In economics, **quantity demanded** refers to the number of units of a product consumers are willing to purchase at a given price. The relationship between price and quantity demanded is usually inverse, meaning as the price decreases, the quantity demanded generally increases, and vice versa.
In this exercise, we have a function that expresses quantity demanded, \( q \), in terms of price, \( p \). Specifically, the function is \( q = 1000 e^{-0.02 p} \). Here, \( e \) represents the mathematical constant approximately equal to 2.71828, which is the base of the natural logarithm.
This equation shows an exponential relationship where:\

• When the price \( p \) increases, \( e^{-0.02 p} \) decreases, leading to a drop in \( q \).
• As the price \( p \) decreases, \( e^{-0.02 p} \) increases, resulting in a rise in \( q \).

Understanding this concept helps determine how sensitive the market is to changes in the price of a product.

Rate of Change

The **rate of change** is a central concept in calculus, especially when analyzing functions related to economics, such as revenue functions. It tells us how a function (in this case, revenue) changes concerning another variable (here, the price).
In mathematical terms, it is the derivative of a function. For this exercise, we need to find the derivative of the revenue function \( R(p) = 1000p e^{-0.02p} \) with respect to the price, \( p \).
After performing differentiation, we get the rate of change of revenue with respect to price as \( \frac{dR}{dp} = (1000 - 20p)e^{-0.02p} \). This equation tells us how the revenue changes when the price changes. For example:\

• If \( \frac{dR}{dp} > 0 \), the revenue increases as the price increases.
• If \( \frac{dR}{dp} < 0 \), the revenue decreases as the price increases.

Using this concept, businesses can strategize setting optimal prices to maximize their revenues.

Product Rule

The **product rule** is a technique used in calculus to find the derivative of a product of two functions. When dealing with the derivative of revenue functions, it often comes into play since revenue is typically the product of price and quantity demanded.
For example, given \( R(p) = 1000p e^{-0.02p} \), we treat this as a product of two components: \( u = 1000p \) and \( v = e^{-0.02p} \). To find \( \frac{dR}{dp} \), we apply the product rule: \( \frac{dR}{dp} = u'v + uv' \). Thus:

• First, differentiate \( u \) with respect to \( p \), which is \( u' = 1000 \).
• Then, differentiate \( v \) with respect to \( p \), giving \( v' = -0.02e^{-0.02p} \).
• Combine these as \( 1000 e^{-0.02p} - 20p e^{-0.02p} \).

By mastering the product rule, students can easily manage complex derivative operations involving multiplied functions and enhance their problem-solving toolkits.