Problem 55 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

When the price of a certain commodity is \(p\) dollars per unit, consumers demand \(x\) hundred units of the commodity, where $$ 75 x^{2}+17 p^{2}=5,300 $$ How fast is the demand \(x\) changing with respect to time when the price is \( 7\) and is decreasing at the rate of 75 cents per months? (That is, \(\frac{d p}{d t}=-0.75\).)

Step-by-Step Solution

Answer

\( \frac{dx}{dt} \approx 0.168 \text{ units/month} \text{ when } p = 7 \text{ dollars} \)

Step 1: Identify the given equation

The given relationship between price ( p ) and demand ( x ) is: \( 75 x^2 + 17 p^2 = 5,300 \)

Step 2: Differentiate implicitly with respect to time

To find how fast the demand ( x ) is changing with respect to time ( t ), implicitly differentiate the equation with respect to t : \( \frac{d}{dt}[75 x^2 + 17 p^2] = \frac{d}{dt}[5,300] \). This yields: \( 75 \frac{d}{dt} [x^2] + 17 \frac{d}{dt} [p^2] = 0 \)

Step 3: Apply the chain rule

Using the chain rule for differentiation, we get: \( 75 \times 2x \frac{dx}{dt} + 17 \times 2p \frac{dp}{dt} = 0 \). Simplify the expression to: \( 150x \frac{dx}{dt} + 34p \frac{dp}{dt} = 0 \)

Step 4: Substitute the known values

We know that p = 7 dollars and \( \frac{dp}{dt} = -0.75 \). Substitute these values into the equation: \( 150x \frac{dx}{dt} + 34 \times 7 \times (-0.75) = 0 \), which simplifies to: \( 150x \frac{dx}{dt} - 178.5 = 0 \)

Step 5: Solve for the derivative of demand

Rearrange the equation to solve for \( \frac{dx}{dt} \): \( 150x \frac{dx}{dt} = 178.5 \), then \( \frac{dx}{dt} = \frac{178.5}{150x} \). To find x when p is 7, use the original equation.

Step 6: Find x when p = 7

Substitute p = 7 into the original equation: \( 75x^2 + 17(7^2) = 5,300 \), which simplifies to: \( 75x^2 + 17 \times 49 = 5,300 \), then \( 75x^2 + 833 = 5,300 \), and finally: \( 75x^2 = 4,467 \) so \( x^2 = \frac{4,467}{75} \) and \( x = \frac{\text{sqrt}(4,467)}{75} \)

Step 7: Calculate the final rate of demand

After calculating x , substitute back to find \( \frac{dx}{dt} \): \( \frac{dx}{dt} = \frac{178.5}{150x} \). Plug in the value of x to get the final rate.

Key Concepts

Chain Rule

The chain rule is a technique used in calculus for differentiating compositions of functions. It's essential when dealing with implicit differentiation. In this problem, we need to use the chain rule to find how fast the demand, represented by variable x, changes with respect to time when the price p is changing.
When differentiating an equation implicitly, each term that involves a variable being implicitly changed with respect to time also requires multiplying by the derivative of that variable. For example, in our problem: \[ 75 \cdot 2x \frac{dx}{dt} + 17 \cdot 2p \frac{dp}{dt} = 0 \]
Here, we differentiate each term separately and then apply the chain rule.

Rate of Change

The rate of change defines how a quantity changes over time. Here, we are focused on how the demand for a commodity (x) changes as time progresses and the price (p) changes.
Given that the price is decreasing at the rate of 0.75 dollars per month (\frac{dp}{dt} = -0.75), we need to figure out the corresponding rate of change in demand (x).
After differentiating the equation and substituting the known values, we found: \[ \frac{dx}{dt} = \frac{178.5}{150x} \]
This equation tells us the rate at which demand x is changing with respect to time t. However, we need to determine the specific value of x when p is 7 to solve the rate accurately.

Demand Function

The demand function shows the relationship between the price of a commodity and the quantity demanded by consumers. For our problem, it is represented by the equation:
\[ 75 x^2 + 17 p^2 = 5,300 \]
This equation means the quantity demanded (x) changes as the price per unit (p) changes. By solving this equation for specific values of p, we can find the corresponding quantity x. When the price p is 7, we simplify to find x: \[ 75x^2 = 4,467 \] Therefore, the precise quantity x can then be used to determine the rate of change in demand.

Calculus

Calculus provides tools like differentiation and integration to understand changes between variables. In this problem, by using concepts from calculus, we can determine the relationship between price and demand over time. Implicit differentiation and chain rule are crucial calculus techniques applied here to find the rate of change in demand.
Our goal is to determine \[ \frac{dx}{dt} \] which quantifies how the demand for a commodity varies as the price changes over time. Calculus allows us to systematically approach the solution and understand the dynamics between these quantities.