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The problem shown involves an exponential decay function, which models the effect of advertising on sales. The function provided is:

\[
S(t) = B + Ae^{-kt} \quad \text{for} \quad 0 \leq t \leq 4
\]

where:
- $ B = 53,000 $ (the average weekly volume of sales before the promotion),
- $ S(1) = 81,740 $ (sales at the end of the first week),
- $ S(3) = 69,010 $ (sales at the end of the third week),
- $ A $ and $ k $ are unknown constants, and $ t $ is the number of weeks since the promotion started.

### Part (a): Finding the Decay Constant $ k $

We can solve this by using the two points $ (t = 1, S = 81,740) $ and $ (t = 3, S = 69,010) $. First, we need to find $ A $ and $ k $ by setting up two equations using the given sales values:

1. **At $ t = 1 $:**
   \[
   S(1) = 53,000 + A e^{-k(1)} = 81,740
   \]
   Simplifying:
   \[
   53,000 + A e^{-k} = 81,740
   \]
   \[
   A e^{-k} = 81,740 - 53,000 = 28,740
   \]
   So:
   \[
   A e^{-k} = 28,740 \tag{1}
   \]

2. **At $ t = 3 $:**
   \[
   S(3) = 53,000 + A e^{-k(3)} = 69,010
   \]
   Simplifying:
   \[
   53,000 + A e^{-3k} = 69,010
   \]
   \[
   A e^{-3k} = 69,010 - 53,000 = 16,010
   \]
   So:
   \[
   A e^{-3k} = 16,010 \tag{2}
   \]

### Solving for $ k $:

To eliminate $ A $, divide equation (1) by equation (2):

\[
\frac{A e^{-k}}{A e^{-3k}} = \frac{28,740}{16,010}
\]
This simplifies to:
\[
e^{2k} = \frac{28,740}{16,010} \approx 1.795
\]
Take the natural logarithm of both sides:
\[
2k = \ln(1.795) \approx 0.586
\]
\[
k = \frac{0.586}{2} \approx 0.293
\]

Thus, the decay constant is $ k \approx 0.293 $ (rounded to three decimal places).

### Part (b): Finding the Sales Volume at $ t = 4 $

Using $ k = 0.293 $, we can now solve for $ A $ using equation (1):

\[
A e^{-0.293} = 28,740
\]
\[
A \times 0.746 = 28,740
\]
\[
A \approx \frac{28,740}{0.746} \approx 38,525
\]

Now, find the sales volume at the end of the fourth week $ S(4) $:

\[
S(4) = 53,000 + 38,525 e^{-0.293 \times 4}
\]
\[
S(4) = 53,000 + 38,525 e^{-1.172} \approx 53,000 + 38,525 \times 0.310
\]
\[
S(4) \approx 53,000 + 11,932 \approx 64,932
\]

So, the sales volume at the end of the fourth week is approximately **$64,932**.

---

Let me know if you need more details or clarifications! Here are some related questions:

1. How does the decay constant $ k $ affect the rate of decrease in sales?
2. What would happen if the sales promotion lasted more than four weeks?
3. Can you graph the sales volume function and describe its behavior over time?
4. How would different values of $ B $ and $ A $ affect the overall sales volume?
5. What assumptions does this exponential model make about the nature of the sales decline?

**Tip:** In exponential decay problems, a larger decay constant $ k $ means the quantity decreases more rapidly over time!

Effect of advertising on sales, find the decay constant and sales volume after a promotion.

Learn how to solve an exponential decay function to find the decay constant and future sales volume after a sales promotion. The problem involves calculating the decay rate using algebraic manipulation and applying it to predict sales after a given time.

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