Complete The Equation That Models This Scenario, Where C ( X C(x C ( X ] Is The Revenue Generated And X X X Is The Number Of $ 2 \$2 $2 Fee Increases.John Hosts An Art Workshop On The Weekends. He Has An Average Of 14 Students In Each Session

Introduction

John hosts an art workshop on the weekends, and he has an average of 14 students in each session. To increase revenue, John is considering implementing a fee increase of $2 for each student. In this scenario, we need to model the revenue generated by the art workshop as a function of the number of fee increases. The revenue generated is denoted by $c(x)$, where $x$ represents the number of $2 fee increases.

Understanding the Problem

The problem requires us to find the equation that models the revenue generated by the art workshop as a function of the number of fee increases. We are given that the revenue generated is denoted by $c(x)$, where $x$ represents the number of $2 fee increases. To solve this problem, we need to understand the relationship between the revenue generated and the number of fee increases.

Revenue as a Function of Fee Increases

Let's assume that the initial fee for each student is $x$. When John implements a fee increase of $2, the new fee becomes $x + 2$. The revenue generated by each student is then $(x + 2) \times 14$, where 14 is the average number of students in each session. Since the fee increase is $2, the number of fee increases is denoted by $x$.

Modeling Revenue as a Function of Fee Increases

To model the revenue generated as a function of the number of fee increases, we can use the following equation:

$$c(x) = (x + 2) \times 14 \times (x + 1)$$

where $c(x)$ is the revenue generated, and $x$ is the number of $2 fee increases.

Simplifying the Equation

We can simplify the equation by expanding the product:

$$c(x) = 14x^2 + 42x + 28$$

This equation represents the revenue generated by the art workshop as a function of the number of fee increases.

Interpreting the Equation

The equation $c(x) = 14x^2 + 42x + 28$ represents a quadratic function, which is a polynomial of degree 2. The graph of this function is a parabola that opens upwards. This means that as the number of fee increases ($x$) increases, the revenue generated ($c(x)$) also increases.

Conclusion

In this scenario, we have modeled the revenue generated by the art workshop as a function of the number of fee increases. The equation $c(x) = 14x^2 + 42x + 28$ represents a quadratic function that models the revenue generated. This equation can be used to determine the revenue generated by the art workshop for a given number of fee increases.

Example Use Cases

1. Determining Revenue for a Given Number of Fee Increases: Suppose John wants to determine the revenue generated by the art workshop for 5 fee increases. We can plug in $x = 5$ into the equation $c(x) = 14x^2 + 42x + 28$ to get:

$$c(5) = 14(5)^2 + 42(5) + 28$$

$$c(5) = 14(25) + 210 + 28$$

$$c(5) = 350 + 210 + 28$$

$$c(5) = 588$$

Therefore, the revenue generated by the art workshop for 5 fee increases is $588.

2. Determining the Number of Fee Increases Required to Reach a Certain Revenue: Suppose John wants to determine the number of fee increases required to reach a revenue of $1000. We can set up the equation $c(x) = 1000$ and solve for $x$:

$$14x^2 + 42x + 28 = 1000$$

$$14x^2 + 42x - 972 = 0$$

We can solve this quadratic equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 14$, $b = 42$, and $c = -972$. Plugging in these values, we get:

$$x = \frac{-42 \pm \sqrt{42^2 - 4(14)(-972)}}{2(14)}$$

$$x = \frac{-42 \pm \sqrt{1764 + 54432}}{28}$$

$$x = \frac{-42 \pm \sqrt{56796}}{28}$$

$$x = \frac{-42 \pm 238}{28}$$

We have two possible solutions for $x$:

$$x = \frac{-42 + 238}{28} = \frac{196}{28} = 7$$

$$x = \frac{-42 - 238}{28} = \frac{-280}{28} = -10$$

Since the number of fee increases cannot be negative, we discard the solution $x = -10$. Therefore, the number of fee increases required to reach a revenue of $1000 is $x = 7$.

Conclusion

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Introduction

In our previous article, we modeled the revenue generated by the art workshop as a function of the number of fee increases. The equation $c(x) = 14x^2 + 42x + 28$ represents a quadratic function that models the revenue generated. In this article, we will answer some frequently asked questions about modeling revenue from art workshop fee increases.

Q: What is the initial fee for each student?

A: The initial fee for each student is not specified in the problem. However, we can assume that the initial fee is $x$, where $x$ is the initial fee.

Q: How do I determine the revenue generated for a given number of fee increases?

A: To determine the revenue generated for a given number of fee increases, you can plug in the value of $x$ into the equation $c(x) = 14x^2 + 42x + 28$. For example, if you want to determine the revenue generated for 5 fee increases, you can plug in $x = 5$ into the equation:

$$c(5) = 14(5)^2 + 42(5) + 28$$

$$c(5) = 14(25) + 210 + 28$$

$$c(5) = 350 + 210 + 28$$

$$c(5) = 588$$

Therefore, the revenue generated by the art workshop for 5 fee increases is $588.

Q: How do I determine the number of fee increases required to reach a certain revenue?

A: To determine the number of fee increases required to reach a certain revenue, you can set up the equation $c(x) = revenue$ and solve for $x$. For example, if you want to determine the number of fee increases required to reach a revenue of $1000, you can set up the equation:

$$14x^2 + 42x + 28 = 1000$$

$$14x^2 + 42x - 972 = 0$$

We can solve this quadratic equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 14$, $b = 42$, and $c = -972$. Plugging in these values, we get:

$$x = \frac{-42 \pm \sqrt{42^2 - 4(14)(-972)}}{2(14)}$$

$$x = \frac{-42 \pm \sqrt{1764 + 54432}}{28}$$

$$x = \frac{-42 \pm \sqrt{56796}}{28}$$

$$x = \frac{-42 \pm 238}{28}$$

We have two possible solutions for $x$:

$$x = \frac{-42 + 238}{28} = \frac{196}{28} = 7$$

$$x = \frac{-42 - 238}{28} = \frac{-280}{28} = -10$$

Since the number of fee increases cannot be negative, we discard the solution $x = -10$. Therefore, the number of fee increases required to reach a revenue of $1000 is $x = 7$.

Q: What is the significance of the quadratic function in modeling revenue from art workshop fee increases?

A: The quadratic function in modeling revenue from art workshop fee increases represents the relationship between the revenue generated and the number of fee increases. The graph of this function is a parabola that opens upwards, indicating that as the number of fee increases increases, the revenue generated also increases.

Q: Can I use this model to determine the revenue generated by other types of workshops or events?

A: While this model is specific to art workshops, you can modify it to determine the revenue generated by other types of workshops or events. However, you will need to adjust the equation to reflect the specific characteristics of the workshop or event.

Q: How accurate is this model in predicting revenue from art workshop fee increases?

A: The accuracy of this model depends on the assumptions made and the data used to develop the equation. While this model provides a good approximation of the revenue generated by art workshops, it may not accurately predict revenue in all cases.

Conclusion

In this article, we have answered some frequently asked questions about modeling revenue from art workshop fee increases. We have provided examples of how to use the equation $c(x) = 14x^2 + 42x + 28$ to determine the revenue generated for a given number of fee increases and to determine the number of fee increases required to reach a certain revenue. We have also discussed the significance of the quadratic function in modeling revenue from art workshop fee increases and the limitations of this model.