To help Meha plan her party within her budget of $150, we need to consider the cost of adults and children she might invite.
Let:
- \( x \) = number of adults
- \( y \) = number of children
The cost equation can be represented as: \[ 5.50x + 10.50y \leq 150 \]
To maximize both adults and children within this budget, we should look at various combinations of \( x \) and \( y \).
Cost Analysis:
-
If Meha invites 10 adults (which costs $55) and no children:
- Total cost: \( 10 \times 5.50 = 55 \)
- Remaining budget: \( 150 - 55 = 95 \)
- Meha could then invite approximately \( 95 / 10.50 ≈ 9 \) children.
- Total invited: 10 adults + 9 children = 19 guests.
-
If Meha invites 5 adults (which costs $27.50) and no children:
- Total cost: \( 5 \times 5.50 = 27.50 \)
- Remaining budget: \( 150 - 27.50 = 122.50 \)
- Meha could then invite approximately \( 122.50 / 10.50 ≈ 11 \) children.
- Total invited: 5 adults + 11 children = 16 guests.
-
If Meha invites 6 adults (which costs $33):
- Total cost: \( 6 \times 5.50 = 33 \)
- Remaining budget: \( 150 - 33 = 117 \)
- She can then invite approximately \( 117 / 10.50 ≈ 11 \) children.
- Total invited: 6 adults + 11 children = 17 guests.
Conclusion:
Based on the budget, a balanced approach could be to invite:
- 6 adults ($33) and 9 children (which costs $94.50).
- Total cost: \( 33 + 94.50 = 127.50 \)
- Total invited: 6 adults + 9 children = 15 guests.
This maximizes the number of guests while keeping within the budget and allowing for extra expenses if needed. Additionally, adjusting for more children or fewer adults could further be evaluated based on the nature of the party and who Meha wishes to invite. Therefore, I recommend inviting 6 adults and 9 children for a total of 15 guests.