Suppose it has been determined that the sales at a gallery decline after the end of an advertising campaign, with daily sales given by
S = 8000(2−0.1x)
where S is in dollars and x is the number of days after the campaign ends. What are the daily sales 50 days after the end of the campaign?

To find the daily sales 50 days after the end of the advertising campaign, we substitute \( x = 50 \) into the given sales function:

\[
S = 8000(2 - 0.1x)
\]

Substitute \( x = 50 \):

\[
S = 8000 \left(2 - 0.1 \times 50\right)
\]

Calculate the expression inside the parentheses:

\[
2 - 0.1 \times 50 = 2 - 5 = -3
\]

Now, multiply by 8000:

\[
S = 8000 \times (-3) = -24,000 \text{ dollars}
\]

Thus, the daily sales 50 days after the end of the campaign are \(-24,000\) dollars. However, since negative sales don't make practical sense, this result suggests that the model might not be valid for \( x = 50 \) or later days, as it predicts a decline to zero or negative values.

### Would you like to explore why the model might predict negative sales, or do you have any questions about this problem?

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**Related Questions:**

1. How would the sales be calculated 10 days after the campaign ends?
2. What does the expression \( 8000(2 - 0.1x) \) suggest about the initial sales when \( x = 0 \)?
3. What happens to the sales if the number of days, \( x \), continues to increase indefinitely?
4. Could the sales model be adjusted to prevent negative sales? How?
5. How does the decline rate of sales change with time in this model?

**Tip:** Always check the practical implications of a mathematical model, especially when dealing with real-world situations.

Explore how to calculate daily sales 50 days after the end of an advertising campaign using a linear function model. Understand the implications of negative sales predictions and the practical validity of the model. Suitable for students in Grades 7-9.

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