Problem 59 - Calculus and its applications

Question

Find the demand function \(q=D(x),\) given each set of elasticity conditions. \(E(x)=\frac{4}{x} ; q=2\) when \(x=4\)

Step-by-Step Solution

Answer

The demand function is \(q = 2e^{1 - \frac{4}{x}}\).

Step 1: Understand the Elasticity Function

The elasticity of demand \(E(x)\) is given as \(\frac{4}{x}\). Recall that elasticity is given by \(E(x) = \frac{x}{q} \cdot \frac{dq}{dx}\). We will use this form to find the demand function \(q\).

Step 2: Set Up the Differential Equation

From the elasticity equation \(E(x) = \frac{x}{q} \cdot \frac{dq}{dx} = \frac{4}{x}\), set it equal to the given elasticity: \(\frac{x}{q} \cdot \frac{dq}{dx} = \frac{4}{x}\). Rearrange to get \(\frac{dq}{dx} = \frac{4q}{x^2}\).

Step 3: Separate the Variables

To solve \(\frac{dq}{dx} = \frac{4q}{x^2}\), separate the variables: \(\frac{dq}{q} = \frac{4}{x^2} dx\). This allows us to integrate both sides with respect to \(x\).

Step 4: Integrate Both Sides

Integrate both sides: \(\int \frac{1}{q} dq = \int \frac{4}{x^2} dx\). The integrations yield \(\ln|q| = -\frac{4}{x} + C\), where \(C\) is the integration constant.

Step 5: Solve for \(C\) Using Initial Condition

Use the given information \(q=2\) when \(x=4\) to find \(C\). Substitute \(x=4\) and \(q=2\) into the equation: \(\ln|2| = -\frac{4}{4} + C\). Calculating gives \(\ln 2 = -1 + C\), so \(C = \ln 2 + 1\).

Step 6: Solve for \(q\) as a Function of \(x\)

Substitute back \(C\) into the equation: \(\ln|q| = -\frac{4}{x} + \ln 2 + 1\) becomes \(q = e^{(-\frac{4}{x} + \ln 2 + 1)}\). Simplify to \(q = 2e^{1 - \frac{4}{x}}\).

Step 7: Final Demand Function Expression

The demand function \(q = D(x)\) is expressed as \(q = 2e^{1 - \frac{4}{x}}\). This is the equation that represents how quantity demanded \(q\) varies with price \(x\).

Key Concepts

Elasticity of Demand

The elasticity of demand measures how sensitive the demand for a good is to changes in its price. A common formula used to calculate elasticity is \(E(x) = \frac{x}{q} \cdot \frac{dq}{dx}\). In simpler terms, elasticity combines two effects:

• The effect of price change on quantity (\(\frac{dq}{dx}\)), which indicates how much the quantity demanded changes for a small price change.
• The ratio of current price to current quantity (\(\frac{x}{q}\)), honing in on how big these changes are relative to each other.

In our problem, the given elasticity \(E(x) = \frac{4}{x}\) tells us that the sensitivity decreases as the price \(x\) increases. Elasticity is crucial because it guides pricing strategies and economic predictions.

Differential Equations

A differential equation involves a function and its derivatives. It shows how the rate of change of a quantity (here, demand \(q\)) relates to other quantities (like price \(x\)). In this exercise, we constructed a differential equation from the elasticity formula given:
\[\frac{dq}{dx} = \frac{4q}{x^2}\]Solving this equation involves finding the function \(q(x)\) that satisfies this relationship. This is an essential part of mathematical modeling since it allows predictions about how changes in prices will affect demand. The process typically involves:

• Identifying the differential equation from the problem.
• Rearranging it if necessary for separation of variables.

Integration

Integration is a key mathematical tool needed to solve differential equations. It's about finding a function whose derivative matches a given function. In our problem, after separating variables, we used integration to find \(q\).
Starting with:
\[\int \frac{1}{q} \, dq = \int \frac{4}{x^2} \, dx\]This required integrating each side independently:

• The left side: Produces a natural logarithm \(\ln|q|\).
• The right side: Yields \(-\frac{4}{x}\) upon integration.

After integrating, introducing the constant \(C\) handles the unknowns introduced during integration.This step is crucial because it allows us to connect separated variables back into a complete function \(q(x)\).

Initial Condition

Initial conditions provide specific values that help determine unknown constants in solutions to differential equations. For our demand function, the initial condition was given as \(q=2\) when \(x=4\). This means at a price of 4, the demand is exactly 2.
We use this point to find the integration constant \(C\):

• Substitute \(q=2\) and \(x=4\) into the integrated form \(\ln|q| = -\frac{4}{x} + C\).
• This particular value allows solving for \(C\), stabilizing the equation.

When \(x = 4\), the equation becomes \(\ln 2 = -1 + C\), leading to \(C = \ln 2 + 1\).By determining \(C\), we ensure that the solution shows the specific behavior dictated by the initial condition— an important feature for accurately modeling real-world scenarios.