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.

To determine how many adults and children Meha can invite to the party within her budget of $300, we need to set up the total cost equation based on the information provided.

1. Fixed costs: $150 (the base cost of the party).
2. Variable costs:
   • $10.50 per child (let's denote the number of children as \( c \)).
   • $5.50 per adult (let's denote the number of adults as \( a \)).

The total cost \( T \) can be represented as: \[ T = 150 + 10.50c + 5.50a \]

Meha's aunt has set a budget of $300, so we want: \[ 150 + 10.50c + 5.50a \leq 300 \]

To simplify, we can subtract 150 from both sides: \[ 10.50c + 5.50a \leq 150 \]

Next, we can further simplify by dividing all terms by 5.50 to make the numbers easier to work with: \[ \frac{10.50}{5.50}c + a \leq \frac{150}{5.50} \] Calculating the division gives us: \[ 1.909c + a \leq 27.27 \]

Since the number of children and adults must be whole numbers, we can multiply through by 100 to eliminate the decimals: \[ 190.9c + 100a \leq 2727 \]

Now, we need to find integer values for \( a \) and \( c \) such that \( a \geq 1 \), \( c \geq 1 \), and the budget constraint holds.

Recommendations:

Let’s try a few integer combinations:

1. Case 1: Assume \( c = 1 \) (1 child) \[ 190.9(1) + 100a \leq 2727 \] \[ 190.9 + 100a \leq 2727 \] \[ 100a \leq 2536.1 \] \[ a \leq 25.36 \] This means Meha can invite a maximum of 25 adults if there is 1 child.

2. Calculating total costs with \( c = 1 \) and \( a = 25 \): \[ T = 150 + 10.50(1) + 5.50(25) = 150 + 10.50 + 137.50 = 298 \] This combination fits the budget.

3. Case 2: Let’s try with \( c = 2 \) (2 children) \[ 190.9(2) + 100a \leq 2727 \] \[ 381.8 + 100a \leq 2727 \implies 100a \leq 2345.2 \implies a \leq 23.45 \] The maximum adults here can be 23. \[ T = 150 + 10.50(2) + 5.50(23) = 150 + 21 + 126.50 = 297.50 \] This also fits the budget.

Conclusion:

To maximize the number of people at the party while staying within the budget, Meha could consider the following options:

• Option 1: 25 adults and 1 child ($298 total).
• Option 2: 23 adults and 2 children ($297.50 total).
• Option 3: 20 adults and 5 children, for an estimate ($295 total) if she prefers a balanced group.

Based on this analysis, Meha should invite at least 2 children and 20 adults as it's a balanced option within the budget and allows for a fun party atmosphere without sacrificing adult supervision or child participation. This way she could ensure everyone has a delightful time!