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Chapter 10: Problem 31

Relationship between Price and Sales A model that describes the relationship between the price and the weekly sales of a product might have a form such as $$\frac{d y}{d p}=-\frac{1}{2}\left(\frac{y}{p+3}\right)$$ where \(y\) is the volume of sales and \(p\) is the price per unit. That is, at any time, the rate of decrease of sales with respect to price is directly proportional to the sales level and inversely proportional to the sales price plus a constant. Solve this differential equation. (Figure 6 shows several typical solutions.)

Short Answer

Expert verified
The solution is \[y = \frac{K}{\sqrt{p+3}}\].

Step by step solution

01

Rewrite the differential equation

The given differential equation is \[ \frac{d y}{d p} = -\frac{1}{2} \left( \frac{y}{p+3} \right) \] Rewrite it as \[ \frac{d y}{y} = -\frac{1}{2} \frac{d p}{p+3} \]
02

Integrate both sides

Integrate both sides of the equation. \[ \int \frac{1}{y} dy = \int -\frac{1}{2} \frac{1}{p+3} dp \] The left side becomes \[ \ln |y| = \] while the right side integrates to \[ -\frac{1}{2} \ln |p+3| \]
03

Combine constant of integration

Include the constant of integration, \(C\): \[ \ln |y| = -\frac{1}{2} \ln |p+3| + C \]
04

Exponentiate both sides

Exponentiate both sides to solve for \(y\): \[ e^{\ln |y|} = e^{-\frac{1}{2} \ln |p+3| + C} \] This simplifies to \[ |y| = e^C e^{-\frac{1}{2} \ln |p+3| } \] Let \(e^C = K\), so \[ |y| = K (p+3)^{-\frac{1}{2}} \]
05

Remove absolute value sign

Since \(K\) can be positive or negative, we get \[ y = K (p+3)^{-\frac{1}{2}} \] or equivalently \[ y = \frac{K}{\sqrt{p+3}} \]
06

Solution of the differential equation

Therefore, the solution to the differential equation is \[ y = \frac{K}{\sqrt{p+3}} \]

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Key Concepts

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

price and sales relationship
Understanding the relationship between price and sales is crucial for businesses. This model tells us how changes in price impact sales volume over time. It provides a tangible way to anticipate how a price increase might reduce sales, and vice versa. The given differential equation \/[ \frac{d y}{d p} = -\frac{1}{2} \frac{y}{p+3} \/] suggests that as the price per unit (\(p\)) increases, the volume of sales (\(y\)) decreases at a rate dependent on both the current sales volume and a modified price term. This kind of insight helps businesses make informed pricing decisions to optimize their sales and revenue.
separable differential equations
Separable differential equations are a class that can be written in the form \[ g(y) dy = h(p) dp. \] Here, the equation is already given in a separable form \[ \frac{d y}{y} = -\frac{1}{2} \frac{d p}{p+3}. \] Separating the variables allows us to integrate each side independently, thus simplifying the process of finding a solution. The key advantage of separable differential equations is their simplicity once the initial separation is done. Integrating each part separately is typically straightforward.
integration of logarithmic functions
Logarithmic functions often arise in the process of integrating separable differential equations. In this exercise, we encounter \[ \int \frac{1}{y} dy = \int -\frac{1}{2} \frac{1}{p+3} dp. \] Integrating the left side gives us \[ \ln |y|, \] while integrating the right side yields \[ -\frac{1}{2} \ln |p+3|. \] Logarithmic integration is a powerful tool, particularly when dealing with rates of change that are inversely proportional to another variable. This method captures the essence of exponential growth and decay, which is prevalent in models that describe natural phenomena.
exponential functions
Exponentiation is used to solve for the dependent variable after integrating a logarithmic equation. By exponentiating both sides, we transform the logarithmic expressions back into their original forms. Given \[ \ln |y| = -\frac{1}{2} \ln |p+3| + C, \] we exponentiate to find \[ e^{\ln |y|} = e^{-\frac{1}{2} \ln |p+3| + C}. \] Simplifying, we get \[ |y| = e^C (p+3)^{-\frac{1}{2}}. \] Letting \[ e^C = K, y = \frac{K}{\sqrt{p+3}}. \] Exponential functions like this one are foundational in mathematics because they describe a wide range of processes including growth, decay, and more complex dynamic systems.

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Most popular questions from this chapter

Solve the following differential equations with the given initial conditions. $$y^{\prime}=\frac{t^{2}}{y}, y(0)=-5$$

A person planning for her retirement arranges to make continuous deposits into a savings account at the rate of \(\$ 3600\) per year. The savings account earns \(5 \%\) interest compounded continuously. (a) Set up a differential equation that is satisfied by \(f(t),\) the amount of money in the account at time \(t.\) (b) Solve the differential equation in part (a), assuming that \(f(0)=0,\) and determine how much money will be in the account at the end of 25 years.

Solve the following differential equations with the given initial conditions. $$3 y^{2} y^{\prime}=-\sin t, y\left(\frac{\pi}{2}\right)=1$$

Solve the following differential equations with the given initial conditions. $$\frac{d N}{d t}=2 t N^{2}, N(0)=5$$

The National Automobile Dealers Association reported that the average retail selling price of a new vehicle was \(\$ 30,303\) in 2012. A person purchased a new car at the average price and financed the entire amount. Suppose that the person can only afford to pay \(\$ 500\) per month. Assume that the payments are made at a continuous annual rate and that interest is compounded continuously at the rate of \(3.5 \%\). (Source: The National Automobile Dealers Association, www.nada.com.) (a) Set up a differential equation that is satisfied by the amount \(f(t)\) of money owed on the car loan at time \(t.\) (b) How long will it take to pay off the car loan?
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