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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 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.
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

To determine how many adults and children Meha can invite to the birthday party while staying within the budget of $300, we can set up an equation based on the costs provided.

  1. Fixed Costs: There is a fixed cost of $150 for the party.
  2. Variable Costs: The cost per child is $10.50, and the cost per adult is $5.50.

Let \( c \) be the number of children and \( a \) be the number of adults. The total cost \( T \) can be expressed as:

\[ T = 150 + 10.50c + 5.50a \]

We need to ensure that this total cost does not exceed the budget of $300:

\[ 150 + 10.50c + 5.50a \leq 300 \]

Subtracting the fixed cost from both sides gives:

\[ 10.50c + 5.50a \leq 150 \]

Now, we can analyze this inequality to find out how many children and adults Meha can invite.

Steps to Find Suitable Combinations

  1. Rearranging the Inequality: \[ 10.50c + 5.50a \leq 150 \]

  2. Finding Combinations: We are looking for non-negative integer solutions for \( c \) and \( a \) with the constraints that \( c \geq 1 \) and \( a \geq 1 \).

  3. Calculating Maximum Values: We can find feasible combinations of \( c \) and \( a \) by testing various values:

    • Let’s start testing combinations incrementally:

Sample Calculations

  1. If \( c = 1 \) (1 child): \[ 10.50(1) + 5.50a \leq 150 \] \[ 10.50 + 5.50a \leq 150 \] \[ 5.50a \leq 139.5 \implies a \leq 25.36 \quad \text{(max 25 adults)} \] So feasible pairs: (1, 1) to (1, 25).

  2. If \( c = 2 \) (2 children): \[ 10.50(2) + 5.50a \leq 150 \] \[ 21 + 5.50a \leq 150 \] \[ 5.50a \leq 129 \implies a \leq 23.45 \quad \text{(max 23 adults)} \] So feasible pairs: (2, 1) to (2, 23).

  3. If \( c = 3 \) (3 children): \[ 10.50(3) + 5.50a \leq 150 \] \[ 31.5 + 5.50a \leq 150 \] \[ 5.50a \leq 118.5 \implies a \leq 21.5 \quad \text{(max 21 adults)} \] So feasible pairs: (3, 1) to (3, 21).

Continuing in this manner, we find:

  • In general, as the number of children increases, the maximum number of adults decreases.

Recommendation

After testing various combinations while ensuring at least one adult and one child:

  1. An excellent balanced choice could be 5 children and 15 adults, yielding: \[ T = 150 + 10.50(5) + 5.50(15) = 150 + 52.5 + 82.5 = 285 \]

This leaves room within her budget while satisfying the requirement for one adult and one child.

Final Recommendation: Invite 5 Children and 15 Adults.

  • This gives a total cost of $285, which is under $300. It also creates a good environment for the celebration, ensuring a sufficient number of attendees.
write more simple

Sure! Let's simplify the explanation of how many children and adults Meha should invite to the birthday party.

Budget Information:

  1. Fixed Cost: The party costs $150.
  2. Cost Per Child: $10.50.
  3. Cost Per Adult: $5.50.
  4. Total Budget: $300.

Total Cost Equation:

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

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

Budget Constraint:

We need to make sure: \[ 150 + 10.50c + 5.50a \leq 300 \]

This simplifies to: \[ 10.50c + 5.50a \leq 150 \]

Finding Combinations:

We need at least 1 adult and 1 child. Let’s try some combinations of children and adults:

  1. If 1 Child:

    • \( c = 1 \)
    • Cost: \( 10.50(1) + 5.50a \leq 150 \)
    • This means \( 5.50a \leq 139.5 \) → Max \( a \) = 25 (1 child, 25 adults).

  2. If 2 Children:

    • \( c = 2 \)
    • Cost: \( 10.50(2) + 5.50a \leq 150 \)
    • This means \( 5.50a \leq 129 \) → Max \( a \) = 23 (2 children, 23 adults).

  3. If 5 Children:

    • \( c = 5 \)
    • Cost: \( 10.50(5) + 5.50a \leq 150 \)
    • This means \( 5.50a \leq 82.5 \) → Max \( a \) = 15 (5 children, 15 adults).

Recommendation:

Best Combination: Invite 5 children and 15 adults.

Cost Breakdown:

  • Total Cost = $150 (fixed) + $52.50 (5 children) + $82.50 (15 adults) = $285.

This is under the budget of $300 and keeps a good balance of kids and adults at the party.

So, Meha should invite 5 children and 15 adults for the party!

What are the constraints of the situation?

In this situation, several constraints need to be considered while planning the birthday party for Meha's cousin. Here are the key constraints:

  1. Fixed Budget Constraint:

    • Meha's aunt has a total budget of $300 for the party. This is the maximum amount that can be spent, including all fixed and variable costs.

  2. Fixed Cost:

    • There is a fixed cost of $150 for the party, which includes costs that do not change regardless of the number of attendees.

  3. Variable Costs:

    • The cost per child is $10.50 and per adult is $5.50. The total cost depends on the number of children \( c \) and adults \( a \) invited, leading to variability in total costs based on these numbers.

  4. Minimum Attendee Requirement:

    • There must be at least one adult and one child at the party. This means both \( c \) (the number of children) and \( a \) (the number of adults) must be greater than or equal to 1:
      • \( c \geq 1 \)
      • \( a \geq 1 \)

  5. Non-negative Constraints:

    • Both the number of children and adults must be non-negative integers:
      • \( c \geq 0 \)
      • \( a \geq 0 \)

  6. Total Cost Constraint:

    • The total cost of the party must not exceed the budget: \[ 150 + 10.50c + 5.50a \leq 300 \]

These constraints guide Meha in determining how many children and adults she can invite to stay within budget while meeting the requirements for the party.

can you make it more simple

Of course! Here are the constraints of the birthday party situation in simpler terms:

  1. Total Budget:

    • Meha has a maximum of $300 to spend on the party.

  2. Fixed Cost:

    • The party costs a base fee of $150, which must be paid no matter how many people attend.

  3. Costs Per Person:

    • Each child costs $10.50 to invite.
    • Each adult costs $5.50 to invite.

  4. Minimum Guests:

    • There must be at least 1 child and 1 adult at the party.

  5. Number of Guests:

    • The number of children and adults can’t be negative; they must be whole numbers (like 1, 2, 3, etc.).

  6. Total Cost Limit:

    • The total cost of the party (fixed cost + cost for kids + cost for adults) cannot go over $300.

These points help Meha figure out how many children and adults she can invite while staying within the budget and meeting the requirements.

Show all the viable solutions to the problem as points on the graph.

To graph the viable solutions for Meha's birthday party planning problem, we first need to express the constraints mathematically and then visualize them.

1. Key Constraints

Based on the previous analysis, we established the following constraints:

  1. Budget Constraint: \[ 10.50c + 5.50a \leq 150 \]

  2. Minimum Requirements: \[ c \geq 1 \quad (at , least , one , child) \] \[ a \geq 1 \quad (at , least , one , adult) \]

2. Converting to Intercepts for Graphing

Budget Constraint:

  1. Rearranging the budget constraint: \[ 5.50a \leq 150 - 10.50c \]

    To find intercepts:

    • When \( c = 0 \): \[ a = \frac{150}{5.50} \approx 27.27 \quad \text{(approximately 27 adults)} \]
    • When \( a = 0 \): \[ c = \frac{150}{10.50} \approx 14.29 \quad \text{(approximately 14 children)} \]

So, the budget line can be expressed between the points (0, 27) and (14, 0).

3. Points of Interest:

  • Points that must be satisfied:
    • At least 1 child: \( c \geq 1 \)
    • At least 1 adult: \( a \geq 1 \)

4. Graphing the Constraints:

Graph:

  • X-axis represents the number of children (\( c \)).
  • Y-axis represents the number of adults (\( a \)).
  1. Plot the line of the budget constraint from (0, 27) to (14, 0).
  2. Since \( c \geq 1 \) and \( a \geq 1 \), the viable area on the graph will be in the first quadrant, above the lines \( c = 1 \) and \( a = 1 \) and below the budget line.

5. Possible Integer Points:

Now let's determine and list some possible integer combinations of \( c \) and \( a \) that satisfy the constraints, staying within the budget:

  • Example combinations:
    • \( (1, 25) \) → Total Cost = $150 + $10.50(1) + $5.50(25) = $150 + $10.50 + $137.50 = $298
    • \( (2, 23) \) → Total Cost = $150 + $10.50(2) + $5.50(23) = $150 + $21 + $126.50 = $297.50
    • \( (3, 21) \) → Total Cost = $150 + $10.50(3) + $5.50(21) = $150 + $31.50 + $115.50 = $297
    • \( (5, 15) \) → Total Cost = $150 + $10.50(5) + $5.50(15) = $150 + $52.50 + $82.50 = $285
    • \( (10, 8) \) → Total Cost = $150 + $10.50(10) + $5.50(8) = $150 + $105 + $44 = $299
    • \( (14, 0) \) → Not valid because \( a \) must be at least 1.

Here's a set of integer solutions that can be plotted on the graph:

6. List of Integer Solutions as Points

  • (1, 25)
  • (2, 23)
  • (3, 21)
  • (4, 19)
  • (5, 15)
  • (6, 12)
  • (7, 10)
  • (8, 9)
  • (9, 7)
  • (10, 5)
  • (11, 3)
  • (12, 2)
  • (13, 1)

Summary

You can visualize these points on a graph where the X-axis represents the number of children and the Y-axis represents the number of adults. The acceptable region would be in the first quadrant, above the lines \( c = 1 \) and \( a = 1 \) and below the budget constraint starting from around the point (14, 0) to (0, 27).

Plotting these points on your graph will illustrate all the viable solutions for the party planning!