Problem 79
Question
The demand function for a product is given by \(p=5000\left(1-\frac{4}{4+e^{-0.002 x}}\right)\) where \(p\) is the price per unit and \(x\) is the number of units sold. Find the numbers of units sold for prices of (a) \(p=\$ 200\) and (b) \(p=\$ 800\).
Step-by-Step Solution
Verified Answer
Given a proper calculator or computing software, the student should get the number of units sold at prices \$200 and \$800 from this solution.
1Step 1: Substitute Price into Demand Function for Case (a)
Replace \(p\) in the demand function with \(200\): \[200=5000\left(1-\frac{4}{4+e^{-0.002 x}}\right)\] Solving this, you will get a equation on the variable \(x\).
2Step 2: Solve for x in Case (a)
Solve the equation for \(x\). It may be necessary to use a calculator due to the presence of e, the base of natural logarithms. This will give you the number of units sold at a price of \$200.
3Step 3: Substitute Price into Demand Function for Case (b)
Now, replace \(p\) in the demand function with \(800\): \[800=5000\left(1-\frac{4}{4+e^{-0.002 x}}\right)\] Similar to Step 1, this will give you an equation in terms of \(x\).
4Step 4: Solve for x in Case (b)
Again, solve the equation for \(x\). The solution will give you the number of units sold at a price of \$800.
Key Concepts
Price ElasticityExponential FunctionsSolving Equations
Price Elasticity
Price elasticity is a core concept in understanding how the quantity demanded of a product responds to price changes. Specifically, it measures the percentage change in quantity demanded when there's a one percent change in price.
It's crucial since businesses need to know if a price increase will result in a small or large reduction in sales. If the demand for a product is elastic, a small change in price will cause a large change in the volume of sales. Conversely, if it is inelastic, the quantity demanded doesn't change much with the price.
In our example, understanding the demand function can help predict how the number of units sold changes with different product prices like $200 or $800. Calculating price elasticity involves using the derivatives of the demand function at a particular price and quantity, which gives insight into sales strategies.
It's crucial since businesses need to know if a price increase will result in a small or large reduction in sales. If the demand for a product is elastic, a small change in price will cause a large change in the volume of sales. Conversely, if it is inelastic, the quantity demanded doesn't change much with the price.
In our example, understanding the demand function can help predict how the number of units sold changes with different product prices like $200 or $800. Calculating price elasticity involves using the derivatives of the demand function at a particular price and quantity, which gives insight into sales strategies.
Exponential Functions
Exponential functions are a type of mathematical expression where a constant base is raised to a variable exponent. They’re prevalent in modeling scenarios such as population growth, radioactive decay, and financial interests.
In the demand function described, we have an exponential component, \(e^{-0.002 x}\), where \(e\) is Euler's number, approximately equal to 2.718. The exponential part of our function helps model how decreases in price can accelerate the demand for a product.
Understanding the behavior of exponential functions is essential since even small changes in an exponent can have significant impacts on the outcome, often leading to rapid increases or decreases as seen in our demand model. In this scenario, \(x\) represents the number of units, and changes in \(x\) will exponentially influence the demand due to this component.
In the demand function described, we have an exponential component, \(e^{-0.002 x}\), where \(e\) is Euler's number, approximately equal to 2.718. The exponential part of our function helps model how decreases in price can accelerate the demand for a product.
Understanding the behavior of exponential functions is essential since even small changes in an exponent can have significant impacts on the outcome, often leading to rapid increases or decreases as seen in our demand model. In this scenario, \(x\) represents the number of units, and changes in \(x\) will exponentially influence the demand due to this component.
Solving Equations
Solving equations is a fundamental skill in mathematics. It's about finding the value(s) of the variable(s) that make the equation true. In the context of the demand equation, the task is to isolate \(x\) to determine how many units are sold at specific prices.
Let's take the solution process for \(p = \\(200\) or \(\\)800\).
Let's take the solution process for \(p = \\(200\) or \(\\)800\).
- First, substitute the given price into the demand equation.
- Next, simplify the equation by isolating \(x\) using algebraic techniques. Sometimes, because of the exponential part, using logarithms is necessary.
- Finally, for a solution that involves \(e\), it may be helpful to use a calculator to compute the exact value of \(x\).
Other exercises in this chapter
Problem 78
Use a spreadsheet to complete the table, which shows the time \(t\) necessary for \(P\) dollars to triple if the interest is compounded continuously at the rate
View solution Problem 79
In Exercises, find \(d x / d p\) for the demand function. Interpret this rate of change when the price is \(\$ 10\). $$ x=\ln \frac{1000}{p} $$
View solution Problem 80
In Exercises, find \(d x / d p\) for the demand function. Interpret this rate of change when the price is \(\$ 10\). $$ x=\frac{500}{\ln \left(p^{2}+1\right)} $
View solution Problem 80
The demand function for a product is given by \(p=10,000\left(1-\frac{3}{3+e^{-0.001 x}}\right)\) where \(p\) is the price per unit and \(x\) is the number of u
View solution