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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?

Short Answer

Expert verified
a. The price elasticity of demand (E) when the price is set at $30 is approximately 1.0714, indicating that the demand for tissues is elastic and consumers are responsive to changes in price. b. Tissues should be sold at a price of $50 per case in order to maximize the revenue. c. Approximately 2,500 cases of tissues would be demanded at the price of $50 per case.

Step by step solution

01

Determine the price elasticity of demand (E)

The price elasticity of demand (E) can be determined using the equation: \[E = - \frac{p}{q} \cdot \frac{dq}{dp}\] First, we need to find the derivative of the demand function, which is \((dq/dp)\): \[q = (100 - p)^{2}\] Therefore, \[\frac{dq}{dp} = 2(100 - p)(-1)\] \[= -2(100 - p)\] In this case, \(p = 30\). Next, we need to find the demand (q) at this price: \[q = (100 - 30)^{2}\] \[= 70^{2}\] \[= 4900\] Now, we have all the values needed to determine the price elasticity of demand (E): \[E = - \frac{30}{4900} \cdot (-2) \cdot (100 - 30)\] \[= - \frac{30}{4900} \cdot (-2) \cdot 70\] \[E = 1.0714\]
02

Interpretation of the Price Elasticity of Demand

The price elasticity of demand (E) is 1.0714, which is greater than 1. This indicates that the demand for tissues is elastic, meaning that a small change in price will result in a more significant change in quantity demanded. Therefore, consumers are responsive to changes in price.
03

Find the Critical Price to Maximize Revenue

Revenue (R) is determined by multiplying the price (p) and the quantity demanded (q): \[R = p \cdot q\] \[= p \cdot (100 - p)^2\] To find the critical price that maximizes revenue, we can differentiate the revenue function (R) concerning the price (p), and set the derivative equal to 0. \[R = p(100 - p)^2\] \[\frac{dR}{dp} = (100 - p)^2 + 2p(100 - p)(-1)\] Setting the derivative equal to 0: \[((100 - p)^2 + 2p(100 - p)(-1) = 0\] Now, we can solve for the critical price (p): \[p = 50\]
04

Calculate the Cases Demanded Based on the Critical Price

By using the critical price (p = $50) in the demand equation, we can find the number of cases demanded (q): \[q = (100 - 50)^2\] \[= 50^{2}\] \[= 2500\] In order to maximize the revenue, the tissues should be sold at $50 per case, and approximately 2,500 cases will be demanded at that price.

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Key Concepts

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

Consumer Demand Equation
The consumer demand equation is pivotal in understanding how quantity demanded varies with price. In our tissue example, the equation is represented as \( q=(100-p)^{2} \). It’s a simple yet powerful quadratic equation, whereby \( q \) represents the quantity in demand, and \( p \) stands for the price per unit.

When graphed, this equation yields a downward-sloping curve, typical for most goods—demonstrating that as the price increases, the quantity demanded decreases, and vice versa. But what’s most interesting is how the shape of the curve can provide insights into consumer behavior and price sensitivity, which is the heart of what companies must understand to set the right prices for their products.

To see this equation in action, we look at the price set at \( \$30 \) for a case of tissues. Plugging in the price into the equation gives us the weekly demand, which we’ll use to examine the price elasticity—a crucial indicator of demand sensitivity.
Revenue Maximization
Revenue maximization is a strategy businesses employ to find the sweet spot for pricing their products. It clearly doesn’t mean simply setting the highest price, but rather pinpointing a price that, while accounting for consumer demand, maximizes the product's total revenue.

Mathematically speaking, revenue \( R \) is the product of the price \( p \) and the quantity demanded \( q \), expressed as \( R = p \cdot q \). To find the revenue-maximizing price, you need to differentiate the revenue function with respect to price and set this derivative equal to zero—a classic optimization problem.

In the case of tissues, after deriving and setting the derivative equal to zero, we discovered that the revenue is maximized when the tissues are sold at \( \$50 \) per case. At this price point, demand and price reach a balance that yields the most revenue. It's a key concept for businesses focusing on profit optimization rather than merely increasing sales volume or selling price.
Derivatives in Economics
Derivatives in economics are not about sophisticated financial instruments; it’s about the fundamental concept of rate of change. When economists talk about derivatives, they refer to the way one variable changes in response to a change in another variable.

Understanding how to apply derivatives to economic problems is a critical skill for analyzing and predicting economic behavior. In our example, the derivative is used to determine the rate at which demand for tissues changes with respect to the price. By finding \( \frac{dq}{dp} \)—the derivative of the demand function with respect to price—we can calculate the price elasticity of demand and identify the revenue-maximizing price.

To put it in practical terms, derivatives help economists and businesses to anticipate how a change in price might affect the quantity of a good that consumers will buy, and thus, how it will impact revenue. It’s an essential tool for pricing strategies and financial planning.

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