Carnival Tickets: Inequalities For Alana's Spending

Let's break down how to create a system of inequalities to represent Alana's carnival ticket situation. This involves understanding the constraints given: Alana's budget, the cost of each type of ticket, and the minimum number of tickets she intends to purchase. By translating these conditions into mathematical inequalities, we can accurately model the possible combinations of ride and food tickets Alana can buy.

Understanding the Problem

Alana is heading to the carnival, and she needs to be smart about her spending. She has a maximum of $40 to spend on tickets. There are two types of tickets: ride tickets, which cost $4 each, and food tickets, which cost $2 each. Alana also knows she wants to buy at least 16 tickets in total. Our goal is to create a system of inequalities that represents these conditions, using ${ r }$ for the number of ride tickets and ${ f }$ for the number of food tickets.

Setting Up the Inequalities

To solve this problem, we need to create inequalities based on the information provided.

Inequality 1: The Budget Constraint

The total amount Alana spends on ride tickets and food tickets must be less than or equal to $40. Since each ride ticket costs $4 and each food ticket costs $2, the total cost can be represented as ${ 4r + 2f }$. Therefore, the first inequality is:

${ 4r + 2f \leq 40 }$

This inequality ensures that Alana does not exceed her $40 budget. It accounts for the combined cost of ride tickets and food tickets, keeping the total spending within the limit.

Inequality 2: The Minimum Number of Tickets

Alana wants to buy at least 16 tickets in total. This means the sum of ride tickets and food tickets must be greater than or equal to 16. The second inequality is:

${ r + f \geq 16 }$

This inequality reflects Alana's intention to purchase a minimum of 16 tickets, combining both ride and food tickets to meet this requirement.

Non-Negative Constraints

Since Alana cannot buy a negative number of tickets, we also have two non-negative constraints:

${ r \geq 0 }$

${ f \geq 0 }$

These constraints ensure that the number of ride tickets and food tickets are non-negative, as it is impossible to buy a negative amount of tickets.

The System of Inequalities

Combining all the inequalities, we get the following system:

${ \begin{cases} 4r + 2f \leq 40 \\ r + f \geq 16 \\ r \geq 0 \\ f \geq 0 \end{cases} }$

This system of inequalities represents all the constraints on the number of ride tickets ${ r }$ and food tickets ${ f }$ that Alana can buy. The first inequality represents the budget constraint, the second inequality represents the minimum number of tickets, and the last two inequalities represent the non-negative constraints.

Simplifying the Inequalities

To make the inequalities easier to work with, we can simplify the first inequality by dividing all terms by 2:

${ 2r + f \leq 20 }$

So the simplified system of inequalities is:

${ \begin{cases} 2r + f \leq 20 \\ r + f \geq 16 \\ r \geq 0 \\ f \geq 0 \end{cases} }$

This simplified system is equivalent to the original system and can be used to find the possible combinations of ride and food tickets that Alana can buy.

Graphing the Inequalities

To visualize the solution set, we can graph these inequalities on a coordinate plane. The x-axis represents the number of ride tickets ${ r }$, and the y-axis represents the number of food tickets ${ f }$. First, we'll graph each inequality separately.

Graphing ${ 2r + f \leq 20 }$

To graph this inequality, we first treat it as an equation: ${ 2r + f = 20 }$. We can find two points on this line by setting ${ r = 0 }$ and ${ f = 0 }$ respectively.

• If ${ r = 0 }$, then ${ f = 20 }$. So, the point ${ (0, 20) }$ is on the line.
• If ${ f = 0 }$, then ${ 2r = 20 }$, so ${ r = 10 }$. Thus, the point ${ (10, 0) }$ is on the line.

Plot these points and draw a line through them. Since the inequality is ${ 2r + f \leq 20 }$, we shade the region below the line.

Graphing ${ r + f \geq 16 }$

Similarly, we treat this inequality as an equation: ${ r + f = 16 }$. Find two points on this line:

• If ${ r = 0 }$, then ${ f = 16 }$. So, the point ${ (0, 16) }$ is on the line.
• If ${ f = 0 }$, then ${ r = 16 }$. Thus, the point ${ (16, 0) }$ is on the line.

Plot these points and draw a line through them. Since the inequality is ${ r + f \geq 16 }$, we shade the region above the line.

Graphing ${ r \geq 0 }$ and ${ f \geq 0 }$

These inequalities simply restrict the solution to the first quadrant, where both ${ r }$ and ${ f }$ are non-negative.

Finding the Feasible Region

The feasible region is the area where all shaded regions overlap. This region represents all possible combinations of ride tickets and food tickets that satisfy all the inequalities. The vertices of the feasible region are the points where the boundary lines intersect.

Finding the Vertices of the Feasible Region

To find the vertices, we need to solve the systems of equations formed by the intersecting lines.

1. Intersection of ${ 2r + f = 20 }$ and ${ r + f = 16 }$:

   Subtract the second equation from the first:

   ${ (2r + f) - (r + f) = 20 - 16 }$

   ${ r = 4 }$

   Substitute ${ r = 4 }$ into ${ r + f = 16 }$:

   ${ 4 + f = 16 }$

   ${ f = 12 }$

   So, the intersection point is ${ (4, 12) }$.

2. Intersection of ${ 2r + f = 20 }$ and ${ f = 0 }$:

   Substitute ${ f = 0 }$ into ${ 2r + f = 20 }$:

   ${ 2r + 0 = 20 }$

   ${ r = 10 }$

   So, the intersection point is ${ (10, 0) }$.

3. Intersection of ${ r + f = 16 }$ and ${ r = 0 }$:

   Substitute ${ r = 0 }$ into ${ r + f = 16 }$:

   ${ 0 + f = 16 }$

   ${ f = 16 }$

   So, the intersection point is ${ (0, 16) }$.

4. Intersection of ${ r + f = 16 }$ and ${ f = 0 }$:

   Substitute ${ f = 0 }$ into ${ r + f = 16 }$:

   ${ r + 0 = 16 }$

   ${ r = 16 }$

   So, the intersection point is ${ (16, 0) }$.

5. Intersection of ${ 2r + f = 20 }$ and ${ r = 0 }$:

   Substitute ${ r = 0 }$ into ${ 2r + f = 20 }$:

   ${ 2(0) + f = 20 }$

   ${ f = 20 }$

   So, the intersection point is ${ (0, 20) }$.

However, we must consider only the points that lie within the feasible region defined by all inequalities. The vertices of the feasible region are ${ (4, 12) }$, ${ (10, 0) }$, ${ (0, 16) }$, and ${ (16, 0) }$. The intersection points ${ (0, 20) }$ and ${ (16, 0) }$ are not part of the feasible region, as they do not satisfy all inequalities.

Conclusion

By setting up and solving this system of inequalities, Alana can determine the possible combinations of ride and food tickets she can purchase while staying within her budget and meeting her minimum ticket requirement. Understanding how to create and interpret these inequalities is a valuable skill that can be applied to various real-world scenarios involving constraints and optimization.

For further reading on inequalities and their applications, you might find helpful resources on websites like Khan Academy's Algebra section.