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.

Let's define the variables for the number of children and adults attending the party:

• Let \( c \) represent the number of children.
• Let \( a \) represent the number of adults.

According to the problem, we have the following information regarding the costs:

• The fixed cost for the party is $150.
• The cost per child is $10.50, so for \( c \) children, the cost is \( 10.50c \).
• The cost per adult is $5.50, so for \( a \) adults, the cost is \( 5.50a \).

The total cost of the party can be expressed as:

\[ \text{Total Cost} = 150 + 10.50c + 5.50a \]

Since Meha's aunt has a budget of $300, the total cost must not exceed this budget:

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

Next, we also need to account for the requirement that there must be at least one adult and at least one child at the party. This can be expressed with the following inequalities:

\[ c \geq 1 \quad (at , least , one , child) \] \[ a \geq 1 \quad (at , least , one , adult) \]

Combining everything together, the set of inequalities representing this situation is:

1. \( 150 + 10.50c + 5.50a \leq 300 \)
2. \( c \geq 1 \)
3. \( a \geq 1 \)

Thus, the two-variable inequality that represents the cost constraint is:

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

with the conditions \( c \geq 1 \) and \( a \geq 1 \).

shorter

Sure! Here’s a more concise representation:

Variables:

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

Inequalities:

1. Cost constraint: \[ 10.50c + 5.50a \leq 150 \]
2. Minimum requirements: \[ c \geq 1 \quad \text{and} \quad a \geq 1 \]

Is it possible for the party to include 10 children and five adults? Use your inequality from part A to check. Explain your answer.

Let's check if the party can include 10 children and 5 adults by substituting \( c = 10 \) and \( a = 5 \) into the cost constraint inequality:

Cost Constraint: \[ 10.50c + 5.50a \leq 150 \]

Substituting the values: \[ 10.50(10) + 5.50(5) \]

Calculating the individual costs: \[ 10.50 \times 10 = 105 \] \[ 5.50 \times 5 = 27.50 \]

Now, add these amounts together: \[ 105 + 27.50 = 132.50 \]

Now we check the inequality: \[ 132.50 \leq 150 \]

Since \( 132.50 \) is indeed less than \( 150 \), the cost constraint is satisfied.

Conclusion:

Yes, it is possible to include 10 children and 5 adults at the party, as the total cost of $132.50 is within the budget limit of $150.

What are the constraints of the situation? Describe them.

In this situation, the following constraints define the requirements and limitations for planning the birthday party:

1. Cost Constraint:

   • The total cost of the party must not exceed the budget. The cost includes a fixed fee of $150, plus $10.50 per child and $5.50 per adult. This can be expressed as: \[ 10.50c + 5.50a \leq 150 \]

2. Minimum Attendee Constraints:

   • There must be at least one child attending the party. This condition ensures that the party has the intended demographic, which is captured by the inequality: \[ c \geq 1 \]

   • There must also be at least one adult attending the party for supervision and support. This condition is expressed as: \[ a \geq 1 \]

Summary of Constraints:

• Cost Limit: The total spending on the party (including fixed costs and variable costs based on the number of children and adults) must be within the budget of $150.
• Minimum Child Attendance: At least one child is required for the party to take place.
• Minimum Adult Attendance: At least one adult is also required for the party.

These constraints together ensure that the party remains within budget while fulfilling essential requirements for attendance.

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

To illustrate the viable solutions to the problem as points on a graph, we'll focus on the two-variable inequality derived from the cost constraint and the conditions on the number of children and adults.

Inequalities:

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

   Rearranging this inequality: \[ 5.50a \leq 150 - 10.50c \quad \Rightarrow \quad a \leq \frac{150 - 10.50c}{5.50} \]

2. Minimum Attendance Constraints:

   • \( c \geq 1 \)
   • \( a \geq 1 \)

Steps to Graph the Inequalities:

1. Graph the Cost Constraint: To graph the line represented by \( 10.50c + 5.50a = 150 \):

   • Find the intercepts.
   • Setting \( c = 0 \): \[ 5.50a = 150 \quad \Rightarrow \quad a = \frac{150}{5.50} \approx 27.27 , \text{(approximately 27)} \]
   • Setting \( a = 0 \): \[ 10.50c = 150 \quad \Rightarrow \quad c = \frac{150}{10.50} \approx 14.29 , \text{(approximately 14)} \]

   The line will intersect the axes at \( (14, 0) \) and \( (0, 27) \).

2. Shading the Feasible Region:

   • Since \( a \) and \( c \) must be non-negative, we will consider the area in the first quadrant.
   • Additionally, since \( 10.50c + 5.50a \leq 150 \), shade below the line \( 10.50c + 5.50a = 150 \).

3. Adding Minimum Attendance Constraints:

   • Draw the vertical line \( c = 1 \) and the horizontal line \( a = 1 \).
   • The feasible solutions must be in the shaded region in the first quadrant that is bounded by these vertical and horizontal lines, as well as below the line \( 10.50c + 5.50a = 150 \).

Viable Solutions as Points:

To find specific viable integer solutions, we can check for integer combinations of \( c \) and \( a \):

1. \( c = 1 \): \[ 10.50(1) + 5.50a \leq 150 \quad \Rightarrow \quad 5.50a \leq 139.50 \quad \Rightarrow \quad a \leq 25.36 \Rightarrow \text{(integer values: 1 to 25)} \]

2. \( c = 2 \): \[ 10.50(2) + 5.50a \leq 150 \quad \Rightarrow \quad 5.50a \leq 129 \quad \Rightarrow \quad a \leq 23.45 \Rightarrow \text{(integer values: 1 to 23)} \]

3. Continuing this process for values of \( c \) from 1 up to 14 gives us the valid \( a \) range.

Example Points:

Some example viable solutions (integer points \( (c, a) \)) might include:

• \( (1, 25) \)
• \( (2, 23) \)
• \( (3, 21) \)
• \( (4, 19) \)
• \( (5, 17) \)
• \( (6, 15) \)
• \( (7, 14) \)
• \( (8, 12) \)
• \( (9, 10) \)
• \( (10, 8) \)
• \( (11, 7) \)
• \( (12, 5) \)
• \( (13, 3) \)
• \( (14, 1) \)

These points can be plotted on the graph in the viable shaded region that meets all constraints.

To visualize this graphically, you may use graphing software or graph paper, where you can plot the inequalities and shade the feasible region accordingly. The \( (c, a) \) points indicated above will all fall within that shaded area.

I need all the viable solutions

To find all the viable solutions for the number of children \(c\) and adults \(a\) that meet the given constraints, we'll systematically check integer values for \(c\) and calculate corresponding \(a\) values based on the inequality derived from the cost constraint.

Cost Constraint Recap

The cost constraint based on the variables is: \[ 10.50c + 5.50a \leq 150 \]

We need to find suitable pairs of \( (c, a) \) such that:

1. \( c \geq 1 \)
2. \( a \geq 1 \)
3. The total cost does not exceed $150.

Solving for All Viable Solutions

1. Rearranging the cost constraint to find \(a\): \[ 5.50a \leq 150 - 10.50c \] \[ a \leq \frac{150 - 10.50c}{5.50} \]

2. We will calculate \(a\) for integer values of \(c\) from 1 to the maximum feasible value, which can be found from when \(a\) is at least 1:

Calculate Viable Solutions:

We will compute feasible \(a\) values for each integer \(c\) starting from 1.

Calculating Maximum Value of \(c\): To find the maximum \(c\) where \(a \geq 1\): \[ 5.50a \geq 0 \ a \geq 1 \rightarrow 10.50c + 5.50(1) \leq 150 \ 10.50c + 5.50 \leq 150 \ 10.50c \leq 144.50 \ c \leq \frac{144.50}{10.50} \approx 13.76 \Rightarrow c \leq 13. \]

Now, we find \(a\) for every integer \(c\) from 1 to 13:

Summary of Viable Solutions

• For \(c = 1\): \[ a \leq \frac{150 - 10.50(1)}{5.50} = \frac{139.5}{5.50} \approx 25.36 \rightarrow a \text{ can be } 1 \text{ to } 25. \]

• For \(c = 2\): \[ a \leq \frac{150 - 10.50(2)}{5.50} = \frac{129}{5.50} \approx 23.36 \rightarrow a \text{ can be } 1 \text{ to } 23. \]

• For \(c = 3\): \[ a \leq \frac{150 - 10.50(3)}{5.50} = \frac{118.5}{5.50} \approx 21.55 \rightarrow a \text{ can be } 1 \text{ to } 21. \]

• For \(c = 4\): \[ a \leq \frac{150 - 10.50(4)}{5.50} = \frac{108}{5.50} \approx 19.64 \rightarrow a \text{ can be } 1 \text{ to } 19. \]

• For \(c = 5\): \[ a \leq \frac{150 - 10.50(5)}{5.50} = \frac{97.5}{5.50} \approx 17.73 \rightarrow a \text{ can be } 1 \text{ to } 17. \]

• For \(c = 6\): \[ a \leq \frac{150 - 10.50(6)}{5.50} = \frac{87}{5.50} \approx 15.82 \rightarrow a \text{ can be } 1 \text{ to } 15. \]

• For \(c = 7\): \[ a \leq \frac{150 - 10.50(7)}{5.50} = \frac{76.5}{5.50} \approx 13.91 \rightarrow a \text{ can be } 1 \text{ to } 13. \]

• For \(c = 8\): \[ a \leq \frac{150 - 10.50(8)}{5.50} = \frac{66}{5.50} \approx 12 \rightarrow a \text{ can be } 1 \text{ to } 12. \]

• For \(c = 9\): \[ a \leq \frac{150 - 10.50(9)}{5.50} = \frac{55.5}{5.50} \approx 10.09 \rightarrow a \text{ can be } 1 \text{ to } 10. \]

• For \(c = 10\): \[ a \leq \frac{150 - 10.50(10)}{5.50} = \frac{45}{5.50} \approx 8.18 \rightarrow a \text{ can be } 1 \text{ to } 8. \]

• For \(c = 11\): \[ a \leq \frac{150 - 10.50(11)}{5.50} = \frac{34.5}{5.50} \approx 6.27 \rightarrow a \text{ can be } 1 \text{ to } 6. \]

• For \(c = 12\): \[ a \leq \frac{150 - 10.50(12)}{5.50} = \frac{24}{5.50} \approx 4.36 \rightarrow a \text{ can be } 1 \text{ to } 4. \]

• For \(c = 13\): \[ a \leq \frac{150 - 10.50(13)}{5.50} = \frac{13.5}{5.50} \approx 2.45 \rightarrow a \text{ can be } 1 \text{ to } 2. \]

Complete List of Viable Solutions:

Here are all the viable solutions as \( (c, a) \) pairs:

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

This list encompasses all the integer pairs \( (c, a) \) that satisfy the constraints of the problem.