To find the daily sales at the end of the advertising campaign, we need to evaluate the given equation when \( x = 0 \).
The daily sales function is:
\[ S = 3300 e^{-0.14x} \]
Part a:
When the advertising campaign ended, \( x = 0 \):
\[ S = 3300 e^{-0.14(0)} = 3300 e^{0} = 3300 \times 1 = 3300 \]
Thus, the daily sales when the campaign ended were $3300.00.
Part 2:
We’ll find out how many days passed after the campaign ended before daily sales fell below half of what they were at the end of the campaign. Half of the sales at the end is:
\[ \frac{3300}{2} = 1650 \]
Now, we set the sales function equal to 1650 and solve for \( x \):
\[ 3300 e^{-0.14x} = 1650 \]
Dividing both sides by 3300:
\[ e^{-0.14x} = \frac{1650}{3300} \]
This simplifies to:
\[ e^{-0.14x} = \frac{1}{2} \]
Next, we take the natural logarithm of both sides:
\[ -0.14x = \ln\left(\frac{1}{2}\right) \]
To isolate \( x \), we rearrange the equation:
\[ x = \frac{\ln\left(\frac{1}{2}\right)}{-0.14} \]
Calculating \( \ln\left(\frac{1}{2}\right) \):
\[ \ln\left(\frac{1}{2}\right) \approx -0.693147 \]
Now plug this value into the equation for \( x \):
\[ x = \frac{-0.693147}{-0.14} \approx 4.95 \]
Thus, rounding to the nearest whole number, 5 days passed after the campaign ended before daily sales were below half of what they were at the end of the campaign.
Summary of Answers:
- a. Daily sales when the campaign ended: $3300.00
- b. Days until daily sales fell below half of that: 5 days