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Let's go through the problem step by step.

### The given function:
The total sales $ S(x) $ in relation to the advertising expenditure $ x $ (both in thousands of dollars) is given by:
\[
S(x) = -0.002x^3 + 0.8x^2 + 4x + 500
\]
where $ x $ is in the range $ 0 \leq x \leq 200 $.

### Part 1: Finding $ S'(x) $
We need to find the derivative $ S'(x) $, which represents the rate of change of the sales with respect to the amount of money spent on advertising.

Using the power rule:
\[
S'(x) = \frac{d}{dx}(-0.002x^3 + 0.8x^2 + 4x + 500)
\]
Differentiating each term:
- $ \frac{d}{dx}(-0.002x^3) = -0.006x^2 $
- $ \frac{d}{dx}(0.8x^2) = 1.6x $
- $ \frac{d}{dx}(4x) = 4 $
- The constant 500 vanishes since its derivative is zero.

So, the derivative is:
\[
S'(x) = -0.006x^2 + 1.6x + 4
\]

### Part 2: Calculating $ S'(x) $ at specific points
- **For $ x = 110 $ (representing $110,000 of advertising):**
  \[
  S'(110) = -0.006(110)^2 + 1.6(110) + 4
  \]
  Let's calculate this.

- **For $ x = 160 $ (representing $160,000 of advertising):**
  \[
  S'(160) = -0.006(160)^2 + 1.6(160) + 4
  \]
  Let's calculate this as well.

I'll now compute these values.The rate of change of sales with respect to advertising is:

- For $ x = 110 $ (when \$110,000 is spent on advertising), $ S'(110) = 107.4 $ dollars per thousand dollars of advertising.
- For $ x = 160 $ (when \$160,000 is spent on advertising), $ S'(160) = 106.4 $ dollars per thousand dollars of advertising.

### Part 3: Comparison
Since $ S'(110) = 107.4 $ is greater than $ S'(160) = 106.4 $, we conclude that Cannon's total sales are increasing at a faster rate when \$110,000 is spent on advertising compared to when \$160,000 is spent.

Let me know if you'd like further details or if you have any questions!

Here are five related questions for further exploration:
1. How would the rate of change behave at $ x = 0 $ (no money spent on advertising)?
2. At what advertising expenditure does the rate of change of sales peak?
3. How would increasing the budget beyond $ x = 200 $ impact sales?
4. Can you find the critical points of $ S'(x) $ to determine any maximum or minimum rates of change?
5. How does the cubic term in $ S(x) $ affect the long-term behavior of the sales function?

**Tip:** Derivatives help you analyze how a function behaves at different points—use them to find optimal values!

The relationship between the amount of money x that Cannon Precision Instruments spends on advertising and the company's total sales S(x) is given by the following function where S(x) and x are both measured in thousands of dollars: S(x) = -0.002x^3 + 0.8x^2 + 4x + 500 (0 <= x <= 200). Find S'(x), the rate of change of the sales with respect to the amount of money spent on advertising.

Learn how to find the rate of change of sales, S'(x), given a cubic polynomial function for Cannon Precision Instruments' advertising expenditures. The solution covers polynomial differentiation and includes step-by-step calculations for specific advertising values, $110,000 and $160,000. Suitable for high school students in Grades 10-12.

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