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!