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.

Question

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

Answer 1

To find all the viable solutions for Meha's birthday party planning, we can define variables and set up corresponding equations based on the cost structure provided.

Let's define:

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

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

According to the problem:

• The total cost must be less than or equal to 300 (the maximum budget): \[ 150 + 10.50c + 5.50a \leq 300 \]

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

We can also express this in a more usable form: \[ 10.50c + 5.50a - 150 \leq 0 \]

Now we can rearrange this to express it in terms of \( a \): \[ 5.50a \leq 150 - 10.50c \] \[ a \leq \frac{150 - 10.50c}{5.50} \]

Next, to consider the conditions specified:

1. At least one adult and one child must be present: \[ c \geq 1 \] \[ a \geq 1 \]

Now let's graph the boundary of the inequality, which is defined by the line: \[ 5.50a + 10.50c = 150 \]

Finding intercepts

To graph this, we can find the intercepts:

1. When \( c = 0 \): \[ 5.50a = 150 \implies a = \frac{150}{5.50} \approx 27.27 \] (only practical values considered, so limit to \( a = 27 \))

2. When \( a = 0 \): \[ 10.50c = 150 \implies c = \frac{150}{10.50} \approx 14.29 \] (only practical values considered, so limit to \( c = 14 \))

Boundary Line

The intercepts are (0, 27) and (14, 0), and we can connect these dots to form a boundary line.

Finding the area of feasible solutions

Since both \( a \) and \( c \) must be at least 1, we are interested in the area bounded by the axes (a ≥ 1 and c ≥ 1) and below the line \( 5.50a + 10.50c = 150 \).

The equations of the lines are:

• A line reaching from approximately (0, 27) to (14, 0)
• Horizontal line \( a = 1 \)
• Vertical line \( c = 1 \)

1. Set \( c = 1 \): \[ 5.50a + 10.50(1) \leq 150 \implies 5.50a \leq 139.50 \implies a \leq \frac{139.50}{5.50} \approx 25.36 \text{ (or \( a \leq 25 \))} \]

2. Set \( a = 1 \): \[ 10.50c + 5.50(1) \leq 150 \implies 10.50c \leq 144.50 \implies c \leq \frac{144.50}{10.50} \approx 13.76 \text{ (or \( c \leq 13 \))} \]

Plotting the Graph

Finally, we can plot the defined area: Vertices of the feasible area:

• (1, 25) when \( c = 1 \)
• (13, 1) when \( a = 1 \)
• The intersection point derived from solving \( 5.50a + 10.50c = 150 \) together with \( a \geq 1 \) and \( c \geq 1 \).

Summary of Viable Solutions:

From the graph of the boundary line \( 5.50a + 10.50c = 150 \) considering \( a \geq 1 \) and \( c \geq 1 \), viable integer pairs \( (c, a) \) are the solutions within this defined area.

In conclusion, all pairs \( (c, a) \) that satisfy:

• \( 1 \leq c \leq 13 \)
• \( 1 \leq a \leq 25 \)
• \( 10.50c + 5.50a \leq 150 \), where both \( c \) and \( a \) are integers, form viable solutions for the party planning.