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Chapter 2: Problem 8

For the demand function given, find the following. a) The elasticity b) The elasticity at the given price, stating whether the demand is elastic or inelastic c) The value(s) of \(x\) for which total revenue is a maximum (assume that \(x\) is in dollars) $$ q=D(x)=\sqrt{300-x} ; \quad x=250 $$

Short Answer

Expert verified
Elasticity at \(x=250\) is -2.5 (elastic). Revenue maximizes at \(x=200\).

Step by step solution

01

Understand Elasticity

Elasticity of demand measures how the quantity demanded responds to price changes. For a demand function given by \(q = D(x)\), the elasticity can be calculated using \(E(x) = \frac{x}{q} \cdot \frac{dq}{dx}\). This helps determine if a demand is elastic or inelastic.
02

Differentiate the Demand Function

Differentiate the given demand function \(q = \sqrt{300-x}\) with respect to \(x\). Use the chain rule to find \(\frac{dq}{dx} = \frac{-1}{2\sqrt{300-x}}\).
03

Calculate Elasticity Function

Substitute \(q = \sqrt{300-x}\) and \(\frac{dq}{dx}\) into the elasticity formula: \(E(x) = \frac{x}{\sqrt{300-x}} \cdot \frac{-1}{2\sqrt{300-x}} = \frac{-x}{2(300-x)}\).
04

Evaluate Elasticity at Given Price

Substitute \(x = 250\) into the elasticity function: \(E(250) = \frac{-250}{2(300-250)} = \frac{-250}{100} = -2.5\). Since \(|E(250)| > 1\), the demand is elastic.
05

Determine Maximum Revenue Condition

Revenue \(R\) is the product of price \(x\) and quantity \(q\), i.e., \(R = x\cdot q = x \sqrt{300-x}\). Differentiate \(R\) with respect to \(x\) to find \(\frac{dR}{dx} = \sqrt{300-x} - \frac{x}{2\sqrt{300-x}}\). Set the derivative \(\frac{dR}{dx}=0\) to find maximum revenue points.
06

Solve for Maximum Revenue

Set \(\sqrt{300-x} = \frac{x}{2\sqrt{300-x}}\). This simplifies to \(2(300-x) = x\), leading to \(600 - 2x = x\). Solving gives \(3x = 600\), so \(x = 200\). Hence, total revenue is maximized when \(x = 200\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, used to find the rate at which one quantity changes with respect to another. In this exercise, we start with the demand function \( q = \sqrt{300-x} \). The aim is to determine how a small change in price \( x \) affects the quantity demanded \( q \). This involves calculating the derivative of the demand function with respect to \( x \).

Aided by the chain rule, differentiation facilitates the derivation of the derivative \( \frac{dq}{dx} \). For the function given, the derivative is \( \frac{-1}{2\sqrt{300-x}} \). This tells us how \( q \) decreases as \( x \) increases.

Differentiation plays a crucial role in elasticity and revenue analysis. By understanding how to differentiate functions, you can predict the behavior of various economic variables, especially in determining responsiveness of demand to price changes.
Revenue Maximization
Revenue maximization is a critical concept in economics, focused on finding the price level that results in the highest possible revenue. In the context of this exercise, revenue \( R \) is calculated as the product of price and quantity: \( R = x \cdot q \). For the given demand function, this becomes \( R = x \sqrt{300-x} \).

Maximizing revenue involves differentiating the revenue function with respect to \( x \). The derived formula for the derivative, \( \frac{dR}{dx} = \sqrt{300-x} - \frac{x}{2\sqrt{300-x}} \), tells us how revenue changes with changes in price. Setting \( \frac{dR}{dx} = 0 \) helps find the turning point where revenue is at its maximum.
  • At \( x = 200 \), the derivative equals zero, indicating a point of maximum revenue.
This point is particularly valuable as it hints at the ideal pricing strategy to optimize income, a key goal for businesses.
Chain Rule in Calculus
The chain rule is an essential tool in calculus, used to differentiate complex functions. It is especially useful when dealing with composite functions, as seen in this exercise with the demand function \( q = \sqrt{300-x} \).

The chain rule applies when a function is composed of two or more simple functions. Here, the expression \( \sqrt{300-x} \) is made up of the outer function \( f(u) = \sqrt{u} \) and the inner function \( u(x) = 300-x \). By the chain rule, the derivative is found by differentiating both the outer and inner functions and multiplying the results:
  • The derivative of the outer function is \( \frac{1}{2\sqrt{u}} \).
  • The derivative of the inner function is \( -1 \).
Combining these, the derivative of \( q \) becomes \( \frac{-1}{2\sqrt{300-x}} \). The chain rule is thus pivotal in solving real-world problems involving rates of change.

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Most popular questions from this chapter

One model of worldwide oil production is $$P(t)=2.69 t^{4}-63.941 t^{3}+459.895 t^{2}-688.692 t+24,150.217$$ where \(P(t)\) is the number of barrels, in thousands, produced \(t\) years after 2000. (Source: Based on data from the U.S. Energy Information Administration.) According to this model, in what year did worldwide oil production achieve an absolute minimum? What was that minimum?

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The percentage of women aged \(21-54\) in the U.S. civilian labor force can be modeled by $$f(x)=-0.0135 x^{2}+0.265 x+74.6$$ where \(x\) is the number of years since \(1992 .\) (Source: based on data from www.bls.gov.) According to this model, in what year during the period \(1992-2012\) was this percentage a maximum?

Find the absolute extrema of each function, if they exist, over the indicated interval. Also indicate the \(x\) -value at which each extremum occurs. When no interval is specified, use the real numbers, \((-\infty, \infty)\). $$f(x)=\frac{1}{3} x^{3}-3 x ; \quad[-2,2]$$

One model of oil production in the United States is given by $$\begin{aligned}P(t)=& 0.0000000219 t^{4}-0.0000167 t^{3}+0.00155 t^{2} \\\&+0.002 t+0.22, \quad 0 \leq t \leq 110\end{aligned}$$where \(P(t)\) is the number of barrels of oil, in billions, produced in a year, t years after 1910. (Source: Beyond Oil, by Kenneth S. Deffeyes, p. \(41,\) Hill and Wang, New York, \(2005 .)\) a) According to this model, what is the absolute maximum amount of oil produced in the United States and in what year did that production occur? b) According to this model, at what rate was United States oil production declining in 2010 and in \(2015 ?\)
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