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Let's solve the problem step by step:

### Problem Statement:
The cost function is given as:

\[
C(x) = 8000 + 13x + 0.1x^2
\]

where \( C(x) \) is the cost in dollars of producing \( x \) units of a commodity.

### Part (a): Average Rate of Change
We are asked to find the average rate of change of \( C \) with respect to \( x \) when \( x \) changes from \( 100 \) to two given values:

#### Formula:
The average rate of change of a function between two points \( x_1 \) and \( x_2 \) is given by:

\[
\text{Average Rate of Change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1}
\]

#### (i) \( x_2 = 104 \) and \( x_1 = 100 \)
- First, we calculate \( C(104) \) and \( C(100) \).

\[
C(104) = 8000 + 13(104) + 0.1(104)^2
\]
\[
C(100) = 8000 + 13(100) + 0.1(100)^2
\]

Then, we apply the formula for the average rate of change.

#### (ii) \( x_2 = 101 \) and \( x_1 = 100 \)
- Similarly, calculate \( C(101) \) and \( C(100) \).

\[
C(101) = 8000 + 13(101) + 0.1(101)^2
\]
\[
C(100) = 8000 + 13(100) + 0.1(100)^2
\]

Again, apply the formula for the average rate of change.

### Part (b): Instantaneous Rate of Change
The instantaneous rate of change (or marginal cost) is the derivative of the cost function \( C(x) \) with respect to \( x \), evaluated at \( x = 100 \).

#### Formula for Marginal Cost:
\[
C'(x) = \frac{d}{dx}(8000 + 13x + 0.1x^2) = 13 + 0.2x
\]

Evaluate this at \( x = 100 \):

\[
C'(100) = 13 + 0.2(100)
\]

Let me calculate all of these values for you.Here are the results:

### (a) Average Rate of Change:
1. **For \( x = 104 \)**:
   \[
   \text{Average rate of change} = 33.40 \, \text{dollars per unit}
   \]

2. **For \( x = 101 \)**:
   \[
   \text{Average rate of change} = 33.10 \, \text{dollars per unit}
   \]

### (b) Instantaneous Rate of Change (Marginal Cost) at \( x = 100 \):
\[
\text{Marginal cost} = 33.00 \, \text{dollars per unit}
\]

Would you like any further explanation or details?

Here are 5 related questions to deepen your understanding:
1. How does the average rate of change differ from the instantaneous rate of change?
2. What does the marginal cost represent in practical terms?
3. How would the rates change if the cost function had a cubic term?
4. What if we wanted to estimate costs for \( x = 102 \)? How would you use the marginal cost?
5. Can the marginal cost ever be negative? Under what conditions?

**Tip**: The marginal cost is often used to make production decisions, as it indicates the cost of producing one additional unit.

The cost (in dollars) of producing x units of a certain commodity is C(x) = 8000 + 13x + 0.1x^2. (a) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to 104 and 101. (b) Find the instantaneous rate of change of C at x = 100.

Learn how to calculate the average rate of change and marginal cost for a given cost function C(x) = 8000 + 13x + 0.1x^2. This problem involves concepts from calculus, including rates of change and derivatives. Suitable for students in Grade 12 or college-level calculus.

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