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The total cost (in dollars) of ordering graduation announcements is represented by the piecewise function:

[tex]\[
c(x)=\left\{
\begin{array}{ll}
1.5x + 15, & \text{if } 0 \leq x < 25 \\
1.25x + 15, & \text{if } 25 \leq x < 40 \\
x + 15, & \text{if } x \geq 40
\end{array}
\right.
\][/tex]

a. Determine the cost of ordering 25 announcements. Then determine the cost of ordering 24 announcements.

- 25 announcements cost [tex]$\square$[/tex]

- 24 announcements cost [tex]$\square$[/tex]

b. For what number of announcements less than 25 is it financially better to purchase 25 announcements?

(0 to 25, 11 to 25, 14 to 25, or 21 to 25) [tex]$\square$[/tex]

c. For what number of announcements less than 40 is it financially better to purchase 40 announcements?

(0 to 40, 11 to 40, 25 to 40, or 33 to 40) [tex]$\square$[/tex]

Answer :

Sure! Let’s tackle each part of the problem step-by-step.

### Part a: Determining the cost

#### 1. Cost of ordering 25 announcements

First, we need to use the piecewise function to determine the cost:
[tex]\[ c(x) = \left\{ \begin{array}{ll}
1.5x + 15, & \text{if } 0 \leq x < 25 \\
1.25x + 15, & \text{if } 25 \leq x < 40 \\
x + 15, & \text{if } x \geq 40
\end{array} \right. \][/tex]

For 25 announcements:
Since [tex]\( 25 \leq x < 40 \)[/tex]:
[tex]\[ c(25) = 1.25 \times 25 + 15 \][/tex]
[tex]\[ c(25) = 31.25 + 15 \][/tex]
[tex]\[ c(25) = 46.25 \][/tex]

Therefore, the cost of ordering 25 announcements is \[tex]$46.25.

#### 2. Cost of ordering 24 announcements

For 24 announcements:
Since \( 0 \leq x < 25 \):
\[ c(24) = 1.5 \times 24 + 15 \]
\[ c(24) = 36 + 15 \]
\[ c(24) = 51 \]

Therefore, the cost of ordering 24 announcements is \$[/tex]51.

### Summary for Part a:
- 25 announcements cost \[tex]$46.25.
- 24 announcements cost \$[/tex]51.

### Part b: Announcements quantity for better cost at 25

We need to determine for what number of announcements less than 25 is it financially better to purchase 25 announcements rather than [tex]\( x \)[/tex] announcements:

[tex]\[ 1.5x + 15 > 46.25 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 1.5x + 15 > 46.25 \][/tex]
[tex]\[ 1.5x > 31.25 \][/tex]
[tex]\[ x > 20.8333 \][/tex]

Since [tex]\( x \)[/tex] must be less than 25 and an integer, financially it’s better to purchase 25 announcements for [tex]\( x = 21 \)[/tex] to 25.

### Summary for Part b:
- For announcements from 21 to 25, it’s financially better to purchase 25 announcements.

### Part c: Announcements quantity for better cost at 40

We need to determine for what number of announcements less than 40 is it financially better to purchase 40 announcements rather than [tex]\( x \)[/tex] announcements:

[tex]\[ c(x) > c(40) \][/tex]
[tex]\[ c(x) > 40 + 15 \][/tex]
[tex]\[ c(x) > 55 \][/tex]

Considering the three cases:
1. For [tex]\( 0 \leq x < 25 \)[/tex]:

[tex]\[ 1.5x + 15 > 55 \][/tex]
[tex]\[ 1.5x > 40 \][/tex]
[tex]\[ x > 26.6667 \][/tex]

2. For [tex]\( 25 \leq x < 40 \)[/tex]:

[tex]\[ 1.25x + 15 > 55 \][/tex]
[tex]\[ 1.25x > 40 \][/tex]
[tex]\[ x > 32 \][/tex]

Since [tex]\( x \)[/tex] must be less than 40 and an integer, these two inequalities show it's better to purchase 40 announcements for [tex]\( x \)[/tex] from 33 to 40.

### Summary for Part c:
- For announcements from 33 to 40, it’s financially better to purchase 40 announcements.

### Final Answers:
- Part a:
- 25 announcements cost \[tex]$46.25
- 24 announcements cost \$[/tex]51
- Part b:
- The range of announcements where it is financially better to purchase 25 announcements is 21 to 25.
- Part c:
- The range of announcements where it is financially better to purchase 40 announcements is 33 to 40.

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