Darren Is Going To The State Fair. Each Ride Costs $\$6$ To Ride, And Each Exhibit Costs $\$3$ To View. Darren Can Spend At Most $\$84$ At The Fair.The Inequality Representing This Situation Is:$6x + 3y \leq 84$where

Darren's State Fair Adventure: A Mathematical Exploration

The state fair is a fun-filled event where people of all ages can enjoy thrilling rides, delicious food, and fascinating exhibits. For Darren, the state fair is a day of excitement and adventure, but it also comes with a budget constraint. Each ride costs $\$6$ to ride, and each exhibit costs $\$3$ to view. Darren can spend at most $\$84$ at the fair. In this article, we will explore the mathematical concept of linear inequalities and how it can be used to represent Darren's situation.

The inequality representing Darren's situation is:

$$6x + 3y \leq 84$$

where $x$ represents the number of rides Darren can afford to ride, and $y$ represents the number of exhibits he can view.

To understand the inequality, let's break it down into its components. The left-hand side of the inequality represents the total cost of the rides and exhibits that Darren can afford. The right-hand side represents the maximum amount Darren can spend at the fair.

The coefficient of $x$, which is $6$, represents the cost of each ride. Since each ride costs $\$6$, the coefficient of $x$ is equal to the cost of each ride.

The coefficient of $y$, which is $3$, represents the cost of each exhibit. Since each exhibit costs $\$3$, the coefficient of $y$ is equal to the cost of each exhibit.

The constant term on the right-hand side, which is $84$, represents the maximum amount Darren can spend at the fair.

To visualize the inequality, we can graph it on a coordinate plane. The graph of the inequality is a line that represents the boundary of the feasible region.

The line can be graphed by finding the x-intercept and the y-intercept. The x-intercept is the point where the line intersects the x-axis, and the y-intercept is the point where the line intersects the y-axis.

To find the x-intercept, we can set $y$ equal to $0$ and solve for $x$. This gives us:

$$6x \leq 84$$

Dividing both sides by $6$, we get:

$$x \leq 14$$

So, the x-intercept is $(14, 0)$.

To find the y-intercept, we can set $x$ equal to $0$ and solve for $y$. This gives us:

$$3y \leq 84$$

Dividing both sides by $3$, we get:

$$y \leq 28$$

So, the y-intercept is $(0, 28)$.

Using the x-intercept and the y-intercept, we can graph the line on a coordinate plane.

The graph of the inequality is a line that represents the boundary of the feasible region. The feasible region is the area below and to the left of the line.

To solve the inequality, we need to find the values of $x$ and $y$ that satisfy the inequality.

Since the inequality is a linear inequality, we can solve it by finding the boundary of the feasible region. The boundary of the feasible region is the line that represents the inequality.

To find the values of $x$ and $y$ that satisfy the inequality, we can use the following steps:

1. Find the x-intercept and the y-intercept of the line.
2. Graph the line on a coordinate plane.
3. Identify the feasible region, which is the area below and to the left of the line.
4. Find the values of $x$ and $y$ that satisfy the inequality by finding the intersection points of the line with the x-axis and the y-axis.

Let's find some example solutions to the inequality.

Suppose Darren wants to ride $x = 10$ rides and view $y = 20$ exhibits. The total cost of the rides and exhibits would be:

$$6(10) + 3(20) = 60 + 60 = 120$$

Since the total cost is less than the maximum amount Darren can spend, which is $\$84$, the solution $(10, 20)$ satisfies the inequality.

Suppose Darren wants to ride $x = 15$ rides and view $y = 25$ exhibits. The total cost of the rides and exhibits would be:

$$6(15) + 3(25) = 90 + 75 = 165$$

Since the total cost is greater than the maximum amount Darren can spend, which is $\$84$, the solution $(15, 25)$ does not satisfy the inequality.

In this article, we explored the mathematical concept of linear inequalities and how it can be used to represent Darren's situation at the state fair. We graphed the inequality on a coordinate plane and identified the feasible region, which is the area below and to the left of the line. We also found some example solutions to the inequality and showed how to use the inequality to make decisions about how many rides and exhibits to visit at the fair.

• [1] "Linear Inequalities" by Math Open Reference
• [2] "Graphing Linear Inequalities" by Khan Academy
• [3] "Linear Inequalities and Systems of Inequalities" by Paul's Online Math Notes
  Darren's State Fair Adventure: A Mathematical Exploration - Q&A

In our previous article, we explored the mathematical concept of linear inequalities and how it can be used to represent Darren's situation at the state fair. We graphed the inequality on a coordinate plane and identified the feasible region, which is the area below and to the left of the line. We also found some example solutions to the inequality and showed how to use the inequality to make decisions about how many rides and exhibits to visit at the fair.

In this article, we will answer some frequently asked questions about linear inequalities and how they can be used to represent real-world situations like Darren's state fair adventure.

Q: What is a linear inequality?

A: A linear inequality is an inequality that can be written in the form $ax + by \leq c$, where $a$, $b$, and $c$ are constants, and $x$ and $y$ are variables.

Q: How do I graph a linear inequality on a coordinate plane?

A: To graph a linear inequality on a coordinate plane, you need to find the x-intercept and the y-intercept of the line. The x-intercept is the point where the line intersects the x-axis, and the y-intercept is the point where the line intersects the y-axis. You can then graph the line on a coordinate plane and identify the feasible region, which is the area below and to the left of the line.

Q: What is the feasible region?

A: The feasible region is the area below and to the left of the line that represents the linear inequality. It is the region where the inequality is true.

Q: How do I find the values of x and y that satisfy a linear inequality?

A: To find the values of x and y that satisfy a linear inequality, you need to find the intersection points of the line with the x-axis and the y-axis. You can then use these points to determine the values of x and y that satisfy the inequality.

Q: Can I use linear inequalities to represent real-world situations like Darren's state fair adventure?

A: Yes, you can use linear inequalities to represent real-world situations like Darren's state fair adventure. Linear inequalities can be used to model a wide range of real-world situations, including budget constraints, resource allocation, and optimization problems.

Q: How do I use linear inequalities to make decisions about how many rides and exhibits to visit at the state fair?

A: To use linear inequalities to make decisions about how many rides and exhibits to visit at the state fair, you need to first identify the linear inequality that represents the situation. You can then use the inequality to determine the feasible region, which is the area below and to the left of the line. You can then use this region to make decisions about how many rides and exhibits to visit at the fair.

Q: Can I use linear inequalities to solve optimization problems?

A: Yes, you can use linear inequalities to solve optimization problems. Linear inequalities can be used to model optimization problems, such as maximizing or minimizing a function subject to certain constraints.

Q: How do I use linear inequalities to solve optimization problems?

A: To use linear inequalities to solve optimization problems, you need to first identify the linear inequality that represents the problem. You can then use the inequality to determine the feasible region, which is the area below and to the left of the line. You can then use this region to find the optimal solution to the problem.

In this article, we answered some frequently asked questions about linear inequalities and how they can be used to represent real-world situations like Darren's state fair adventure. We showed how to graph a linear inequality on a coordinate plane, identify the feasible region, and use the inequality to make decisions about how many rides and exhibits to visit at the fair. We also showed how to use linear inequalities to solve optimization problems.

• [1] "Linear Inequalities" by Math Open Reference
• [2] "Graphing Linear Inequalities" by Khan Academy
• [3] "Linear Inequalities and Systems of Inequalities" by Paul's Online Math Notes