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
Write a two-variable inequality to represent this situation. Be sure to define the variables you use in your inequality find all viable solutions
Let's define the variables:
- Let c be the number of children attending the party.
- Let a be the number of adults attending the party.
According to Meha's aunt, the cost of the party is $150 for the base fee, $10.50 per child, and $5.50 per adult. Meha's aunt has a budget of $300 for the party, and she does not want to exceed this amount.
The cost of the party can be represented with the following equation:
Cost = Base fee + (Cost per child * Number of children) + (Cost per adult * Number of adults)
Using the variables and the cost information, we get the inequality:
150 + 10.50c + 5.50a ≤ 300
where:
- c >= 1 (There has to be at least one child)
- a >= 1 (There has to be at least one adult)
Now, to explore all viable solutions for c (children) and a (adults) that satisfy the inequality and constraints, we can do the following:
First, let's rearrange the inequality in terms of one of the variables, we will choose 'a' (the number of adults) for example:
5.50a ≤ 300 - 150 - 10.50c
5.50a ≤ 150 - 10.50c
a ≤ (150 - 10.50c) / 5.50
Now, since c and a must be integers (there are no half children or half adults in attendance), we will list some possible values for c and calculate the corresponding value of a. Remember that the result for a must be such that a >= 1.
If c = 1 (minimum one child):
a ≤ (150 - 10.50 * 1) / 5.50
a ≤ (150 - 10.50) / 5.50
a ≤ 139.50 / 5.50
a ≤ 25 (approximately, but must be an integer, so it has to be 25 or less)
If c = 2:
a ≤ (150 - 10.50 * 2) / 5.50
a ≤ (150 - 21) / 5.50
a ≤ 129 / 5.50
a ≤ 23 (approximately, but must be an integer, so it has to be 23 or less)
You can continue this process to find different combinations that satisfy the inequality 150 + 10.50c + 5.50a ≤ 300. You can do this methodically by increasing the number of children by 1 each time and calculating the maximum number of adults allowed within the budget until you reach a value of c where no adults can attend (a < 1), which would mean the number of children is too high to satisfy the budget constraint.
Remember: because the costs per child and adult are fixed, you could end up with a situation where you have non-integer solutions for the number of adults. In real-life applications, like this party, you would round down to the nearest whole number because you can't have a fraction of an adult attend the party.
What are the viable solutions
To find the viable solutions, we need to determine all the pairs of numbers of children (c) and adults (a) that satisfy the inequality:
\[ 150 + 10.50c + 5.50a \leq 300 \]
Provided that there is at least one adult and at least one child at the party, we are looking for integer solutions of the inequality where \( c \geq 1 \) and \( a \geq 1 \).
Let's simplify the inequality to find how many adults can attend for a given number of children.
\[ 10.50c + 5.50a \leq 300 - 150 \]
\[ 10.50c + 5.50a \leq 150 \]
Now let’s express a in terms of c, and then we'll find integer solutions for different values of c.
\[ a \leq \frac{150 - 10.50c}{5.50} \]
Here are the viable solutions within the budget:
For \( c = 1 \):
\[ a \leq \frac{150 - 10.50 \times 1}{5.50} \]
\[ a \leq \frac{139.50}{5.50} \]
\[ a \leq 25.36 \]
Since a must be an integer, the maximum number of adults is 25.
For \( c = 2 \):
\[ a \leq \frac{150 - 10.50 \times 2}{5.50} \]
\[ a \leq \frac{129}{5.50} \]
\[ a \leq 23.45 \]
The maximum number of adults is 23.
We can continue this process until we find a value for c where a would be less than 1 (thus going over budget). Let's find a couple more solutions:
For \( c = 3 \):
\[ a \leq \frac{150 - 10.50 \times 3}{5.50} \]
\[ a \leq 21 \]
For \( c = 4 \):
\[ a \leq \frac{150 - 10.50 \times 4}{5.50} \]
\[ a \leq 18.82 \]
The maximum number of adults is 18.
We can continue in this manner until we reach the value of c that uses up the budget for children alone. It's also important to remember that there must be at least one adult present. Therefore, we need to adjust for this in our calculation to ensure that the final dollar amount allows for at least one adult ticket ($5.50):
For \( c = \) maximum possible children without adults,
\[ 150 + 10.50c \leq 300 - 5.50 \]
\[ 10.50c \leq 144.50 \]
\[ c \leq 13.76 \]
Since c must be an integer, the maximum number of children without considering adults is 13.
Let's calculate a maximum number of adults when c = 13:
\[ a \leq \frac{150 - 10.50 \times 13}{5.50} \]
\[ a \leq \frac{150 - 136.50}{5.50} \]
\[ a \leq \frac{13.50}{5.50} \]
\[ a \leq 2.45 \]
The maximum number of adults is 2, but since we're not actually using up all the budget with 13 children (because the cost is slightly less than $144.50), there may indeed be a little more room in the budget to include more adults if we have fewer children.
Based on the range of solutions, you can pair different numbers of children with corresponding maximum numbers of adults until the budget is reached, keeping in mind that as you decrease the number of children, the number of adults can increase accordingly.
Without listing all pairs explicitly, the general idea is that for every additional child you add (costing an additional $10.50), you can expect to subtract roughly two adults (since $10.50 is almost double the cost per adult, $5.50). You keep adjusting until no more pairs fit the budget of $300 with the constraints that there's at least one child and at least one adult.
Let's define the variables:
Let's use the variable 'c' to represent the number of children attending the party.
Let's use the variable 'a' to represent the number of adults attending the party.
Now we can write the two-variable inequality to represent the situation:
The cost of the party is given by the following equation:
Cost = $150 + ($10.50 x c) + ($5.50 x a)
We know that there needs to be at least one child and at least one adult, so c ≥ 1 and a ≥ 1.
The maximum amount Meha's aunt wants to spend is $300. Therefore, the inequality can be written as:
$150 + ($10.50 x c) + ($5.50 x a) ≤ $300
Simplifying the inequality, we get:
10.50c + 5.50a ≤ 150
So, the two-variable inequality representing this situation is:
10.50c + 5.50a ≤ 150, where c represents the number of children and a represents the number of adults.