Problem 55 - College Algebra and Calculus: An Applied Approach

Question

Use the demand function to find the rate of change in the demand \(x\) for the given price \(p\). $$ x=275\left(1-\frac{3 p}{5 p+1}\right), p=\$ 4 $$

Step-by-Step Solution

Answer

The rate of change of the demand x with respect to the price p when p = $4 is approximately -9.35.

Step 1: Find the Derivative of x with respect to p

First, we need to find the derivative of the function x with respect to p. This process is called differentiation. Using the chain rule and the quotient rule, the derivative of \( x = 275\left(1-\frac{3 p}{5 p+1}\right) \) with respect to p is \( \frac{dx}{dp} = 275\left(0 - \frac{15(5p+1) - 3p*5}{(5p+1)^2}\right) \). Simplify this function to get \( \frac{dx}{dp} = -\frac{4125}{(5p+1)^2} \).

Step 2: Substitute p = $4 into the derivative

Now we need to substitute p = $4 into the derivative function we just found. So for \( p = 4 \), we have \( \frac{dx}{dp} = -\frac{4125}{(5*4+1)^2} \).

Step 3: Calculate the Result

Simplify the result in step 2, you get \( \frac{dx}{dp} = -\frac{4125}{441} \), which is approximately -9.35. That indicates the rate of change of the demand x with respect to the price p when p = $4.

Key Concepts

Demand Function

The demand function is an expression that defines the relationship between the quantity demanded of a product, denoted as \( x \), and its price, \( p \). It acts as a tool that businesses use to predict how changes in price will impact demand. In this specific problem, the demand function is given as:

• \( x = 275 \left(1 - \frac{3p}{5p + 1} \right) \)

This equation tells us that the demand for a product is a complex expression involving both linear and rational components relative to price \( p \). Here, 275 is the maximum possible demand, capturing the market potential when price has negligible influence. The function inside the brackets alters this depending on price \( p \). At a price of \( p = 4 \), the function will specify the exact demand given that price input.

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function is changing at any given point. In the context of the demand function, differentiation helps us find how rapidly the demand \( x \) changes with respect to changes in price \( p \). This rate of change is known as the derivative of the demand function.

• Finding \( \frac{dx}{dp} \) means determining the slope of the demand curve at any point \( p \).


• It involves applying rules of differentiation to the given function to derive a new function that describes this rate of change.

For our demand function, differentiating with respect to \( p \) involves applying techniques such as the chain rule and quotient rule to account for the complexity of the expression. The goal is to obtain an equation that expresses how a slight increase or decrease in \( p \) will influence \( x \).

Chain Rule

The chain rule is a method used in calculus to differentiate composite functions. A composite function is one where a function is nested inside another. In our demand function, we encounter such a scenario with expressions nested within others.

• For example, \( 1 - \frac{3p}{5p + 1} \) is a composite within the larger context of the multiplier 275.


• To differentiate such expressions, the chain rule involves finding the derivative of the outer function first, then multiplying it by the derivative of the inner function.

This method enables us to systematically break down the demand function's components and handle each part in a logical sequence. The chain rule is crucial in dissecting complex functions to understand how innermost changes influence the entire function's output when we alter \( p \).

Quotient Rule

The quotient rule in calculus is specifically designed for finding the derivative of functions formed by compositions of two separate functions being divided by each other. It is necessary for tackling expressions like \( \frac{3p}{5p + 1} \) in our demand function.

• The rule is expressed as: If \( u(p) \) and \( v(p) \) are differentiable functions, then the derivative \( \frac{d}{dp}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dp} - u \frac{dv}{dp}}{v^2} \).


• Here, \( u(p) = 3p \) and \( v(p) = 5p + 1 \) in our demand function.

This systematic rule ensures that the derivative of a fraction-like demand function is accurately calculated by addressing both the numerator and the denominator's roles. It essentially captures how the rate of change of one function impacts the other, thus predicting the resulting influence on demand \( x \) as \( p \) changes.