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A gencral cxponcntial demand function has the form \(q=A e^{-b p}(A\) and \(b\) nonzero constants). a. Obtain a formula for the price elasticity of demand at a unit price of \(p\) b. Obtain a formula for the price that maximizes revenue.

Short Answer

Expert verified
a. The price elasticity of demand at a unit price of p is given by: \( E = -bp \) b. The price that maximizes revenue is given by: \( p = \frac{1}{b} \)

Step by step solution

01

Derivative of demand function with respect to price

To calculate the derivative, we can use the chain rule. The derivative of an exponential function is itself multiplied by the derivative of the exponent. So, the derivative of the demand function with respect to price is: \( \frac{dq}{dp} = -Abe^{-bp} \) Step 2: Calculate the price elasticity of demand.
02

Price elasticity of demand formula

The price elasticity of demand (E) is calculated as: \(E = \frac{\% \Delta q}{\% \Delta p} = \frac{\frac{dq}{dp}}{q} \cdot \frac{p}{1}\) Substitute the demand function q and its derivative dq/dp into the formula: \(E = \frac{-Abe^{-bp}}{Ae^{-bp}}\cdot p\) Since the exponential terms cancel out, we get: \(E = -bp\) Step 3: Find the price that maximizes revenue.
03

Revenue function

The revenue function (R) is the product of price (p) and quantity (q): \(R = pq = pAe^{-bp}\)
04

Derivative of the revenue function

Next, we need to find the derivative of the revenue function with respect to the price. Using the product rule and the chain rule, we get: \(\frac{dR}{dp} = A e^{-bp} - bpAe^{-bp} \) Step 4: Solve for the price that sets the derivative of the revenue function to zero.
05

Finding the price to maximize revenue

To find the price that maximizes revenue, set the derivative of the revenue function equal to zero and solve for p: \(0 = A e^{-bp} (1-bp)\) Divide both sides by the nonzero term \(A e^{-bp}\): \(0 = 1-bp\) Now, solve for p: \(bp = 1\) Therefore, the price that maximizes revenue is: \(p = \frac{1}{b}\) To summarize: a. The price elasticity of demand at a unit price of p is given by: \( E = -bp \) b. The price that maximizes revenue is given by: \( p = \frac{1}{b} \)

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Key Concepts

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

Exponential Demand Function
The exponential demand function, expressed as \( q = A e^{-bp} \), where \( A \) and \( b \) are nonzero constants, provides insight into how the demand for a product changes with its price. This function implies that demand decreases exponentially as the price increases. A common characteristic of such functions is that they never hit zero; demand will decrease but not cease entirely with increasing prices, reflecting a realistic scenario in many economic situations.

In this context, \( A \) can be thought of as the initial demand when the price is zero, and the constant \( b \) indicates the rate at which demand decreases as the price increases. The negative exponent \( -bp \) ensures a decline in demand with an increase in price, modeling the inverse relationship typically observed between price and quantity demanded in real-life markets.
Revenue Maximization
Revenue maximization is a crucial goal for many businesses, and understanding how to achieve this through pricing is a key strategic component. The revenue of a product is found by multiplying the price per unit, \( p \), by the quantity sold, \( q \), thus \( R = pq \). When we're working with an exponential demand function, our revenue function takes the specific form \( R = pAe^{-bp} \).

To maximize revenue, we need to find the optimal price point that leads to the highest possible revenue. Mathematically, this involves taking the derivative of the revenue function with respect to price and then finding the price \( p \) that makes this derivative zero, indicating a local maximum. The steps provided in the solution guide us through this process. It's also important to note that while setting the derivative to zero helps identify potential maximum points, verification through second derivative or other means may be necessary to ensure that the critical point is indeed a maximum for certain more complex functions.
Derivative of Demand Function
Taking the derivative of the demand function with respect to price is essential for understanding how a slight change in price can affect the quantity demanded, a concept known as price elasticity of demand. This measure helps businesses determine how sensitive or responsive consumers are to changes in prices.

The derivative \( \frac{dq}{dp} = -Abe^{-bp} \) represents the rate of change of quantity demanded as the price changes. The negative sign indicates an inverse relationship; as the price increases, the quantity demanded typically decreases, which is consistent with the law of demand. When coupled with the demand function, the derivative helps us calculate the price elasticity, which is economically significant because it informs the pricing strategy: if the demand is elastic, a small decrease in price can lead to a significant increase in demand, which could potentially increase revenue.

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Most popular questions from this chapter

The Physics Club sells \(E=m c^{2}\) T-shirts at the local flea market. Unfortunately, the club's previous administration has been losing money for years, so you decide to do an analysis of the sales. A quadratic regression based on old sales data reveals the following demand equation for the T-shirts: $$q=-2 p^{2}+33 p . \quad(9 \leq p \leq 15)$$ Here, \(p\) is the price the club charges per T-shirt, and \(q\) is the number it can sell each day at the flea market. a. Obtain a formula for the price elasticity of demand for \(E=m c^{2}\) T-shirts. b. Compute the elasticity of demand if the price is set at \(\$ 10\) per shirt. Interpret the result. c. How much should the Physics Club charge for the T-shirts in order to obtain the maximum daily revenue? What will this revenue be?

The following graph shows the approximate value of the U.S. Consumer Price Index (CPI) from September 2004 through November \(2005.32\) The approximating curve shown on the figure is given by $$I(t)=-0.005 t^{3}+0.12 t^{2}-0.01 t+190 \quad(0 \leq t \leq 14)$$ where \(t\) is time in months ( \(t=0\) represents September 2004). a. Use the model to estimate the monthly inflation rate in July \(2005(t=10)\). [Recall that the inflation rate is \(\left.I^{\prime}(t) / I(t) .\right]\) b. Was inflation slowing or speeding up in July \(2005 ?\) c. When was inflation speeding up? When was inflation slowing? HINT [See Example 3.]

The consumer demand equation for tissues is given by \(q=(100-p)^{2}\), where \(p\) is the price per case of tissues and \(q\) is the demand in weekly sales. a. Determine the price elasticity of demand \(E\) when the price is set at \(\$ 30\), and interpret your answer. b. At what price should tissues be sold in order to maximize the revenue? c. Approximately how many cases of tissues would be demanded at that price?

Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(g(x)=x^{3} /\left(x^{2}-3\right)\)

Assume that the demand equation for tuna in a small coastal town is $$p q^{1.5}=50,000$$ where \(q\) is the number of pounds of tuna that can be sold in one month at the price of \(p\) dollars per pound. The town's fishery finds that the demand for tuna is currently 900 pounds per month and is increasing at a rate of 100 pounds per month each month. How fast is the price changing?

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