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Chapter 5: Problem 60

Find the demand function \(q=D(x),\) given each set of elasticity conditions. \(E(x)=\frac{x}{200-x} ; q=190\) when \(x=10\)

Short Answer

Expert verified
The demand function is \(D(x) = \frac{36100}{200-x}\).

Step by step solution

01

Understand Elasticity

Elasticity of demand, denoted by \(E(x)\), is given by \(E(x) = -x \cdot \frac{D'(x)}{D(x)}\). We need to utilize this definition to find the relationship between \(x\) and \(q = D(x)\) given \(E(x)\).
02

Set Up the Given Elasticity Equation

We are given \(E(x) = \frac{x}{200-x}\). According to the elasticity formula, this must equal \(-x \cdot \frac{D'(x)}{D(x)}\). Thus, the equation becomes \(\frac{x}{200-x} = -x \cdot \frac{D'(x)}{D(x)}\).
03

Simplify and Integrate

Rearranging the equation from Step 2 gives \(\frac{D'(x)}{D(x)} = -\frac{1}{200-x}\). This is separable, so we can integrate both sides: \(\int \frac{1}{D(x)} \, dD(x) = \int -\frac{1}{200-x} \, dx\).
04

Solve the Integrals

The left side integrates to \(\ln|D(x)|\) and the right side integrates to \(\ln|200-x| + C\) (where \(C\) is the constant of integration). Thus, \(\ln|D(x)| = -\ln|200-x| + C\).
05

Exponentiate to Solve for D(x)

Exponentiating both sides to solve for \(D(x)\) gives \(D(x) = A \cdot (200-x)^{-1}\), where \(A = e^C\).
06

Use the Initial Condition

We are given that \(q = 190\) when \(x = 10\). So, substitute \(D(10) = 190\) and \(x = 10\) into the equation from Step 5: \(190 = A \cdot (200-10)^{-1}\), which simplifies to \(190 = A \cdot \frac{1}{190}\).
07

Solve for Constant A

From Step 6, we solve \(190 = \frac{A}{190}\) for \(A\), which gives \(A = 190^2\).
08

Final Demand Function

Substitute \(A = 36100\) back into the demand function: \(D(x) = \frac{36100}{200-x}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Demand Function
The demand function is a critical concept in economics. It describes the relationship between the quantity demanded of a good and its price. In our exercise, we're trying to express the demand function in terms of price, that is, find a function \(q = D(x)\), where \(x\) is the price.
This involves using the elasticity of demand, which measures how much the quantity demanded changes in response to a change in price, as a guide. This demand function can help businesses understand consumer behavior, plan production, and set prices that maximize revenue or market share.
Integration in Calculus
Integration in calculus is the process of finding the integral of a function. It's the opposite of differentiation, which involves finding the derivative. In this problem, integration helps transform the elasticity equation into the demand function.
We rearranged the elasticity formula to
  • \( \int \frac{1}{D(x)} \, dD(x) = \int -\frac{1}{200-x} \, dx \).
This step is essential because it simplifies our problem and leads us directly to an expression involving the logarithms, which can then be exponentiated to solve for \(D(x)\). By integrating, we convert the differential equation involving \(D'(x)\) into a solvable function of \(D(x)\). Thus, integration allows us to move from rate of change to actual amounts or levels.
Initial Conditions
Initial conditions provide specific values to determine unknown constants in mathematical equations, ensuring a unique solution. In our problem, the initial condition is "\(q = 190\) when \(x = 10\)." We use it after finding a general solution to the differential equation.
This origin of specific values allows us to solve for the constant \(A\) in our demand function. Consequently, we substitute the known values into the expression:
  • \(190 = A \cdot (200-10)^{-1}\)
Solving this gives us the necessary constant to finalize our demand function \(D(x)\). Hence, initial conditions enable us to pin down a specific demand function that fits our scenario.
Exponential Functions
Exponential functions are those where the variable appears in the exponent. These functions are important in describing a variety of physical phenomena and financial models.
In this exercise, exponential equations come into play when we exponentiate both sides of the integrated logarithmic equation:
  • \( \ln|D(x)| = -\ln|200-x| + C \).
This transformation leads to the equation \(D(x) = A \cdot (200-x)^{-1}\), representing an exponential-like relationship between demand and price. In business scenarios, such functions help model growth rates and understand how rapidly factors like demand change with price. Recognizing these exponential relationships is crucial for accurate predictions and strategic planning.

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Most popular questions from this chapter

Domar's capital expansion model is $$ \frac{d I}{d t}=h k I $$ where \(I\) is the investment, \(h\) is the investment productivity (constant), \(k\) is the marginal productivity to the consumer (constant), and \(t\) is the time. a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the condition \(I_{0}=I(0)\)

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The capitalized cost, \(c,\) of an asset over its lifetime is the total of the initial cost and the present value of all maintenance expenses that will occur in the future. It is computed with the formula $$ c=c_{0}+\int_{0}^{L} m(t) e^{-k t} d t $$ where \(c_{0}\) is the initial cost of the asset, \(L\) is the lifetime (in years), \(k\) is the interest rate (compounded continuously), and \(m(t)\) is the annual cost of maintenance. Find the capitalized cost under each set of assumptions. $$ c_{0}=\$ 400,000, k=5.5 \%, m(t)=\$ 10,000, L=25 $$
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