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

A.) write a two variable equation to represent this situation be sure to define the variables you use in your equation. 150 + 10.5 c + 5.5a <300
B.) is it possible for the party to include 10 children and five adults use your inequality from part to check.
C.) graph the solution for the equation you wrote in part a. At this point do not yet consider any constraints given by the situation.
D. What are the constraints of this 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.

9 answers

Sure! Let's break down each part of the question step by step.

A.) Write a Two-Variable Equation

Let's define the variables:

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

The total cost of the party can be defined by the equation: \[ 150 + 10.5c + 5.5a \leq 300 \]

B.) Check if 10 Children and 5 Adults are Possible

Using the equation from part A with \( c = 10 \) and \( a = 5 \): \[ 150 + 10.5(10) + 5.5(5) \] \[ = 150 + 105 + 27.5 \] \[ = 282.5 \]

Since \( 282.5 \leq 300 \), it is possible to have 10 children and 5 adults at the party.

C.) Graph the Solution of the Equation

To graph the inequality \( 150 + 10.5c + 5.5a \leq 300 \), we start by rewriting it in a more workable form: \[ 10.5c + 5.5a \leq 150 \]

To find the intercepts, set \( c \) to 0 to find the \( a \)-intercept: \[ 5.5a = 150 \] \[ a = \frac{150}{5.5} \approx 27.27 \]

Set \( a \) to 0 to find the \( c \)-intercept: \[ 10.5c = 150 \] \[ c = \frac{150}{10.5} \approx 14.29 \]

Now, plot the points \( (0, 27.27) \) and \( (14.29, 0) \) and draw a line through them. The area below this line represents the solutions to the inequality.

D.) Constraints of the Situation

The constraints in this situation are:

  1. At least one adult must be present: \( a \geq 1 \)
  2. At least one child must be present: \( c \geq 1 \)
  3. The total cost does not exceed $300: \( 150 + 10.5c + 5.5a \leq 300 \)

E.) Viable Solutions on the Graph

In the graph created in part C, viable solutions will be the area that satisfies all of the conditions:

  • The region under the line \( 10.5c + 5.5a \leq 150 \)
  • Above the lines \( c = 1 \) and \( a = 1 \), meaning the area does not extend into the negative number of children or adults.

You would shade the region that respects these inequalities.

F.) Recommendation for Number of Adults and Children

Based on the budget and constraints, a balanced recommendation might be:

  • 5 children (which costs \( 10.5 \times 5 = 52.5 \))
  • 5 adults (which costs \( 5.5 \times 5 = 27.5 \))

This results in: \[ 150 + 52.5 + 27.5 = 230 \]

This is within the budget of $300, allows for at least one adult and one child, and would be a nice balance for the party.

You can also explore finding maximum combinations of \( c \) and \( a \) based on varying costs, but whatever final numbers are chosen, staying under the $300 limit while meeting the constraints is key!

Wouldn't C be graphing 150+10.5c + 5.5a<300? Why would you rewrite it to 10.5c + 5.5a < 150 ??

You're absolutely correct, and I appreciate your diligence in pointing that out! The original inequality should remain as \( 150 + 10.5c + 5.5a \leq 300 \) when graphing. Let’s revise part C correctly based on the original inequality.

C.) Graph the Solution of the Inequality

  1. Start with the inequality: \[ 150 + 10.5c + 5.5a \leq 300 \]

  2. Rearranging this inequality gives us: \[ 10.5c + 5.5a \leq 150 \]

  3. Now we want to plot this inequality.

Finding the intercepts:

  • Finding \( a \)-intercept: Set \( c = 0 \): \[ 5.5a = 150 \implies a = \frac{150}{5.5} \approx 27.27 \] So the \( a \)-intercept is \( (0, 27.27) \).

  • Finding \( c \)-intercept: Set \( a = 0 \): \[ 10.5c = 150 \implies c = \frac{150}{10.5} \approx 14.29 \] So the \( c \)-intercept is \( (14.29, 0) \).

Graphing the inequality:

  • Plot the points \( (0, 27.27) \) and \( (14.29, 0) \) on the graph.
  • Connect these points with a line (dashed, since it's a less-than-or-equal-to inequality).
  • Shade the area below this line, which represents all combinations of \( c \) and \( a \) such that the cost is less than or equal to $300.

Updated Summary

Thank you for the correction—C should accurately depict the original inequality \( 150 + 10.5c + 5.5a \leq 300 \) by retaining the actual format rather than rearranging it to \( 10.5c + 5.5a \leq 150 \).

Please let me know if you need any further clarification or assistance on any other part!

You didn't answer why are you rewriting it and make it at 150 instead of 300

Apologies for any confusion caused, and thank you for your patience. Let's clarify this without altering the fundamental aspects of the original inequality and address the reasoning behind my previous response.

The original inequality is: \[ 150 + 10.5c + 5.5a \leq 300 \]

Reasoning Behind the Rearrangement

When graphing inequalities, it's often useful to isolate the variables involved so that the relationship between them can be more easily visualized. To do this, I aimed to rewrite the inequality in a form that is more straightforward to graph, focusing on \( c \) and \( a \).

For instance, if we subtract 150 from both sides, we arrive at: \[ 10.5c + 5.5a \leq 150 \]

This isn't inherently necessary, but it can simplify the process of identifying intercepts and allows one to observe the contributions of each variable more clearly.

Directly Using the Original Inequality

However, you can definitely graph the original inequality directly:

  1. Leave the equation as \( 150 + 10.5c + 5.5a \leq 300 \).
  2. To find the intercepts, you could isolate across \( a \): \[ 10.5c + 5.5a \leq 150 \] This gives the same information but is indeed reflecting on the maximal budget directly rather than proving other relationships.

In short, I mistakenly aimed to simplify the inequality in a way that might have caused confusion rather than emphasizing that we can directly work with its original form for graphing purposes. Both forms represent the same relationship, and using either is valid depending on your preferred approach.

Conclusion

To answer your question directly: rewriting the inequality was meant as a means of simplifying for graphing and understanding, but it is entirely valid to keep it in its original format regardless of that intent. The original inequality provides the same boundaries for the variables involved.

Again, thank you for your understanding and for prompting this clarification! If you have any more queries or need assistance with other parts, feel free to ask.

What are all the viable solutions to this problem using viable integer solutions

To find all the viable integer solutions for the birthday party problem based on the budget constraint \( 150 + 10.5c + 5.5a \leq 300 \) with the constraints of at least one adult and one child, let's break this down step by step.

Given:

  1. Cost equation: \[ 150 + 10.5c + 5.5a \leq 300 \]
  2. Rearranged for clarity: \[ 10.5c + 5.5a \leq 150 \]
  3. Constraints:
    • \( c \geq 1 \) (at least 1 child)
    • \( a \geq 1 \) (at least 1 adult)

Finding Viable Integer Solutions

We can first rewrite the main cost equation and then find integer solutions by systematically checking combinations of integers for \( c \) (children) and \( a \) (adults) that satisfy it.

  1. Rearranging gives: \[ 5.5a \leq 150 - 10.5c \] This implies: \[ a \leq \frac{150 - 10.5c}{5.5} \]

  2. We can calculate the upper limit for \( a \) based on different integer values for \( c \).

Check Viable Combinations

We will check integer values for \( c \) starting from 1 and calculate the corresponding maximum possible values for \( a \) until the inequality is no longer satisfied.

Calculating for Various Values of \( c \)

  • For \( c = 1 \): \[ 5.5a \leq 150 - 10.5(1) = 139.5 \implies a \leq \frac{139.5}{5.5} \approx 25.36 \] Thus, \( a \) can be from 1 to 25.

  • For \( c = 2 \): \[ 5.5a \leq 150 - 10.5(2) = 128 \implies a \leq \frac{128}{5.5} \approx 23.27 \] Thus, \( a \) can be from 1 to 23.

  • For \( c = 3 \): \[ 5.5a \leq 150 - 10.5(3) = 117.5 \implies a \leq \frac{117.5}{5.5} \approx 21.36 \] Thus, \( a \) can be from 1 to 21.

  • For \( c = 4 \): \[ 5.5a \leq 150 - 10.5(4) = 106 \implies a \leq \frac{106}{5.5} \approx 19.27 \] Thus, \( a \) can be from 1 to 19.

  • For \( c = 5 \): \[ 5.5a \leq 150 - 10.5(5) = 94.5 \implies a \leq \frac{94.5}{5.5} \approx 17.18 \] Thus, \( a \) can be from 1 to 17.

  • For \( c = 6 \): \[ 5.5a \leq 150 - 10.5(6) = 84 \implies a \leq \frac{84}{5.5} \approx 15.27 \] Thus, \( a \) can be from 1 to 15.

  • For \( c = 7 \): \[ 5.5a \leq 150 - 10.5(7) = 73.5 \implies a \leq \frac{73.5}{5.5} \approx 13.36 \] Thus, \( a \) can be from 1 to 13.

  • For \( c = 8 \): \[ 5.5a \leq 150 - 10.5(8) = 63 \implies a \leq \frac{63}{5.5} \approx 11.45 \] Thus, \( a \) can be from 1 to 11.

  • For \( c = 9 \): \[ 5.5a \leq 150 - 10.5(9) = 52.5 \implies a \leq \frac{52.5}{5.5} \approx 9.55 \] Thus, \( a \) can be from 1 to 9.

  • For \( c = 10 \): \[ 5.5a \leq 150 - 10.5(10) = 42 \implies a \leq \frac{42}{5.5} \approx 7.64 \] Thus, \( a \) can be from 1 to 7.

  • For \( c = 11 \): \[ 5.5a \leq 150 - 10.5(11) = 31.5 \implies a \leq \frac{31.5}{5.5} \approx 5.73 \] Thus, \( a \) can be from 1 to 5.

  • For \( c = 12 \): \[ 5.5a \leq 150 - 10.5(12) = 21 \implies a \leq \frac{21}{5.5} \approx 3.82 \] Thus, \( a \) can be from 1 to 3.

  • For \( c = 13 \): \[ 5.5a \leq 150 - 10.5(13) = 10.5 \implies a \leq \frac{10.5}{5.5} \approx 1.91 \] Thus, \( a \) can be from 1 to 1.

  • For \( c = 14 \) and greater: The values for \( c \) keep increasing while \( 10.5c \) will exceed 150 before adding any adults, meaning we will not have valid integer values.

Summary of Viable Integer Solutions

The viable integer solutions can be represented as pairs \( (c, a) \):

  • For \( c = 1 \): \( a = 1, 2, \ldots, 25 \)
  • For \( c = 2 \): \( a = 1, 2, \ldots, 23 \)
  • For \( c = 3 \): \( a = 1, 2, \ldots, 21 \)
  • For \( c = 4 \): \( a = 1, 2, \ldots, 19 \)
  • For \( c = 5 \): \( a = 1, 2, \ldots, 17 \)
  • For \( c = 6 \): \( a = 1, 2, \ldots, 15 \)
  • For \( c = 7 \): \( a = 1, 2, \ldots, 13 \)
  • For \( c = 8 \): \( a = 1, 2, \ldots, 11 \)
  • For \( c = 9 \): \( a = 1, 2, \ldots, 9 \)
  • For \( c = 10 \): \( a = 1, 2, \ldots, 7 \)
  • For \( c = 11 \): \( a = 1, 2, \ldots, 5 \)
  • For \( c = 12 \): \( a = 1, 2, 3 \)
  • For \( c = 13 \): \( a = 1 \)

Thus, all pairs \( (c, a) \) that satisfy the budget and the constraints of at least one child and one adult would be valid for the party planning.

If you have any further questions or need additional information, feel free to ask!

For \( c = 1 \): \( a = 1, 2, \ldots, 25 \) can you explain what you mean on this C equals one child and a = 1 or 2 adults what does ldots 25 mean

Certainly! The notation \( (c = 1): (a = 1, 2, \ldots, 25) \) means:

  • \( c = 1 \) indicates that there is 1 child attending the party.
  • \( a = 1, 2, \ldots, 25 \) indicates that the number of adults attending can be 1, 2, 3, 4, ... up to 25 adults.

The ellipsis \( \ldots \) is a conventional notation used in mathematics to signify a continuation of a sequence. In this case, it indicates that the value of \( a \) can take any integer value from 1 to 25, which is the maximum number of adults that can attend when there is only 1 child.

Explanation of Viable Solutions

Here's a breakdown to clarify further, using some specific examples:

  1. When \( c = 1 \) (1 child):
    • If \( a = 1 \): There is 1 adult and 1 child.
    • If \( a = 2 \): There are 2 adults and 1 child.
    • If \( a = 3 \): There are 3 adults and 1 child.
    • ...
    • If \( a = 25 \): There are 25 adults and 1 child.

Thus, there are 25 possible combinations of adults when there is 1 child. The maximum count of adults (25) is determined by the budget constraint that allows for up to that many adults when only one child is included.

Conclusion

This pattern holds true for other values of \( c \) as well. For each integer value of \( c \), you calculate the maximum number of adults \( a \) that can be included without exceeding the total cost limit. The \( \ldots \) simply denotes all the integer values in the range possible for \( a \) given a specific \( c \).

I hope this clarifies things! Please feel free to ask if you have more questions or need further elaboration!