Meha is helping her aunt to plan a birthday party for her cousin at the science center. Her aunt tells her that the party will cost $150, plus $10.50 per child and $5.50 per adult. There needs to be at least one adult and at least one child at the party. She has budgeted $300 for the party and that is the maximum amount she wants to spend.

Question

A. Write a two-variable equation to represent this situation. Be sure to define the variables you use in your equation.
B. Is it possible for the party to include 10 children and five adults? Use your inequality from part A to check. Explain your answer.
C. Graph the solution set for the equation you wrote in part A on a piece of graph paper. Note: At this point, do not yet consider any constraints given by the situation.
D. What are the constraints of the situation? Describe them.
E. Show all the viable solutions to the problem as points on the graph.
F. Give a recommendation to Meha as to how many adults and children you think she should invite to the party. Justify your recommendation.

Answer 1

Let's break down the problem step by step:

A. Write a Two-Variable Equation

Let:

• \( c \) = number of children
• \( a \) = number of adults

The total cost of the party can be expressed as: \[ \text{Total Cost} = 150 + 10.50c + 5.50a \]

Given that the maximum budget for the party is $300, we can write the inequality: \[ 150 + 10.50c + 5.50a \leq 300 \]

Rearranging gives us: \[ 10.50c + 5.50a \leq 150 \]

B. Check if it is Possible for the Party to Include 10 Children and 5 Adults

We can substitute \( c = 10 \) and \( a = 5 \) into the inequality to check if it's satisfied.

Substituting: \[ 10.50(10) + 5.50(5) \leq 150 \] \[ 105 + 27.50 \leq 150 \] \[ 132.50 \leq 150 \]

Since \( 132.50 \) is less than \( 150 \), this means it is possible for the party to include 10 children and 5 adults.

C. Graph the Solution Set

To graph the inequality \( 10.50c + 5.50a \leq 150 \), we first determine the intercepts.

Setting \( c = 0 \): \[ 5.50a = 150 \implies a = \frac{150}{5.50} \approx 27.27 \]

Setting \( a = 0 \): \[ 10.50c = 150 \implies c = \frac{150}{10.50} \approx 14.29 \]

We plot the points (0, 27.27) and (14.29, 0) on a graph. The line connecting these points is where the total cost equals $150. We shade below this line, as we are looking for values where the cost is less than or equal to $150.

D. Constraints of the Situation

1. Existence of Participants: There must be at least 1 child (\( c \geq 1 \)) and 1 adult (\( a \geq 1 \)).
2. Budget Constraint: The total cost must not exceed $300, as represented by the inequality \( 10.50c + 5.50a \leq 150 \).

E. Show All Viable Solutions

The viable solutions are points \( (c, a) \) within the triangular region defined by:

• The axes where \( c \geq 1 \) and \( a \geq 1 \).
• The line \( 10.50c + 5.50a = 150 \).

These regions can be plotted on graph paper, noting that feasible integer combinations of \( c \) and \( a \) also need to satisfy \( c \geq 1 \) and \( a \geq 1 \).

F. Recommendation

To recommend a suitable number of adults and children, we can consider balanced ratios as well as financial limits. One logical suggestion could be 5 children and 4 adults:

1. Cost Calculation: \[ 10.50(5) + 5.50(4) + 150 = 52.50 + 22 + 150 = 224.50 \]
2. This remains well within the budget of $300, allowing for more flexibility in pricing or additional expenses.

Therefore, inviting around 5 children and 4 adults provides a fun experience without exceeding budget constraints.

Summary

• The equation representing cost constraints is \( 10.50c + 5.50a \leq 150 \).
• It is indeed possible to include 10 children and 5 adults within the budget.
• Graphing inequalities provides feasible solutions.
• As you finalize your recommendations, focus on balanced group sizes that enable a manageable and enjoyable party, staying well within budget limits.

Answer 2

what are the viable solutions to the problem?