A Price \[$ P \$\] (in Dollars) And Demand \[$ X \$\] For A Product Are Related By The Equation:$\[ 2x^2 + 8xp + 50p^2 = 6000 \\]If The Price Is Increasing At A Rate Of 2 Dollars Per Month When The Price Is 10 Dollars, Find The

Introduction

In economics, the relationship between the price of a product and its demand is a crucial concept. The price-demand equation is a mathematical representation of this relationship, which can be used to analyze and predict the behavior of consumers. In this article, we will explore a specific price-demand equation and use it to find the rate of change of demand when the price is increasing at a certain rate.

The Price-Demand Equation

The price-demand equation is given by:

${ 2x^2 + 8xp + 50p^2 = 6000 }$

where $p$ is the price of the product in dollars and $x$ is the demand for the product.

Understanding the Equation

To understand the equation, let's break it down into its components. The equation is a quadratic equation in terms of $x$, with the price $p$ being a parameter. The coefficients of the quadratic terms are $2$ and $50$, which represent the sensitivity of demand to price changes. The linear term $8xp$ represents the interaction between price and demand.

Finding the Rate of Change of Demand

We are given that the price is increasing at a rate of 2 dollars per month when the price is 10 dollars. We need to find the rate of change of demand at this point. To do this, we will use the concept of implicit differentiation.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate an equation that is not in the form of a function. In this case, we have an equation with two variables, $x$ and $p$. We will differentiate both sides of the equation with respect to $p$.

${ \frac{d}{dp} (2x^2 + 8xp + 50p^2) = \frac{d}{dp} (6000) }$

Using the chain rule and the product rule, we get:

${ 4x \frac{dx}{dp} + 8x + 8p \frac{dx}{dp} + 100p = 0 }$

Solving for dx/dp

We are interested in finding the rate of change of demand, which is given by $\frac{dx}{dp}$. To solve for $\frac{dx}{dp}$, we will isolate it on one side of the equation.

${ 4x \frac{dx}{dp} + 8p \frac{dx}{dp} = -8x - 100p }$

Factoring out $\frac{dx}{dp}$, we get:

${ (4x + 8p) \frac{dx}{dp} = -8x - 100p }$

Dividing both sides by $(4x + 8p)$, we get:

${ \frac{dx}{dp} = \frac{-8x - 100p}{4x + 8p} }$

Substituting the Given Values

We are given that the price is increasing at a rate of 2 dollars per month when the price is 10 dollars. We need to find the rate of change of demand at this point. To do this, we will substitute $p = 10$ into the equation.

${ \frac{dx}{dp} = \frac{-8x - 100(10)}{4x + 8(10)} }$

Simplifying the equation, we get:

${ \frac{dx}{dp} = \frac{-8x - 1000}{4x + 80} }$

Finding the Value of x

We need to find the value of $x$ when the price is 10 dollars. To do this, we will substitute $p = 10$ into the original equation.

${ 2x^2 + 8(10)x + 50(10)^2 = 6000 }$

Simplifying the equation, we get:

${ 2x^2 + 80x + 5000 = 6000 }$

Subtracting 6000 from both sides, we get:

${ 2x^2 + 80x - 1000 = 0 }$

Dividing both sides by 2, we get:

${ x^2 + 40x - 500 = 0 }$

Factoring the quadratic equation, we get:

${ (x + 50)(x - 10) = 0 }$

Solving for $x$, we get:

${ x = -50 \text{ or } x = 10 }$

Since the demand cannot be negative, we take $x = 10$.

Substituting the Value of x

We have found the value of $x$ when the price is 10 dollars. We will substitute this value into the equation for $\frac{dx}{dp}$.

${ \frac{dx}{dp} = \frac{-8(10) - 1000}{4(10) + 80} }$

Simplifying the equation, we get:

${ \frac{dx}{dp} = \frac{-80 - 1000}{40 + 80} }$

${ \frac{dx}{dp} = \frac{-1080}{120} }$

${ \frac{dx}{dp} = -9 }$

Conclusion

In this article, we have used the price-demand equation to find the rate of change of demand when the price is increasing at a certain rate. We have used implicit differentiation to differentiate the equation with respect to price and have solved for the rate of change of demand. We have found that the rate of change of demand is -9 when the price is 10 dollars.

References

• [1] Economics: Principles, Problems, and Policies. By Campbell R. McConnell, Stanley L. Brue, and Sean M. Maloney.
• [2] Microeconomics. By Paul Krugman and Robin Wells.

Appendix

The following is a list of formulas and theorems used in this article:

• Implicit Differentiation: If $F(x,y) = 0$ is an implicit function, then $\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
• Quadratic Formula: If $ax^2 + bx + c = 0$ is a quadratic equation, then $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  A Price-Demand Equation: Q&A =============================

Introduction

In our previous article, we explored a price-demand equation and used it to find the rate of change of demand when the price is increasing at a certain rate. In this article, we will answer some common questions related to the price-demand equation and provide additional insights into the topic.

Q: What is the price-demand equation?

A: The price-demand equation is a mathematical representation of the relationship between the price of a product and its demand. It is a quadratic equation in terms of demand, with the price being a parameter.

Q: How is the price-demand equation used in economics?

A: The price-demand equation is used in economics to analyze and predict the behavior of consumers. It can be used to determine the optimal price of a product, the demand for a product at a given price, and the rate of change of demand with respect to price.

Q: What is implicit differentiation?

A: Implicit differentiation is a technique used to differentiate an equation that is not in the form of a function. It is used to find the derivative of an implicit function, which is a function that is defined implicitly by an equation.

Q: How is implicit differentiation used in the price-demand equation?

A: Implicit differentiation is used to differentiate the price-demand equation with respect to price. This allows us to find the rate of change of demand with respect to price, which is a crucial concept in economics.

Q: What is the rate of change of demand?

A: The rate of change of demand is the rate at which demand changes with respect to price. It is a measure of how sensitive demand is to changes in price.

Q: How is the rate of change of demand used in economics?

A: The rate of change of demand is used in economics to determine the optimal price of a product, the demand for a product at a given price, and the impact of price changes on demand.

Q: What are some common applications of the price-demand equation?

A: Some common applications of the price-demand equation include:

• Determining the optimal price of a product
• Analyzing the demand for a product at a given price
• Predicting the impact of price changes on demand
• Determining the rate of change of demand with respect to price

Q: What are some common challenges associated with the price-demand equation?

A: Some common challenges associated with the price-demand equation include:

• Estimating the parameters of the equation
• Handling non-linear relationships between price and demand
• Accounting for external factors that affect demand

Q: How can the price-demand equation be used in real-world applications?

A: The price-demand equation can be used in a variety of real-world applications, including:

• Pricing strategy development
• Demand forecasting
• Market analysis
• Policy evaluation

Conclusion

In this article, we have answered some common questions related to the price-demand equation and provided additional insights into the topic. We have discussed the importance of the price-demand equation in economics, the use of implicit differentiation, and the rate of change of demand. We have also highlighted some common applications and challenges associated with the price-demand equation.

References

• [1] Economics: Principles, Problems, and Policies. By Campbell R. McConnell, Stanley L. Brue, and Sean M. Maloney.
• [2] Microeconomics. By Paul Krugman and Robin Wells.

Appendix

The following is a list of formulas and theorems used in this article:

• Implicit Differentiation: If $F(x,y) = 0$ is an implicit function, then $\frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
• Quadratic Formula: If $ax^2 + bx + c = 0$ is a quadratic equation, then $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$