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We are given a cost function $ C(x) = 4000 + 12x + 0.05x^2 $, where $ C(x) $ represents the cost in dollars of producing $ x $ units of a commodity.

### Part (a): Finding the Average Rate of Change
The **average rate of change** of $ C $ with respect to $ x $ between two points $ x = 100 $ and $ x = a $ is calculated using the formula:
\[
\text{Average rate of change} = \frac{C(a) - C(100)}{a - 100}
\]
We need to compute this for two values of $ a $: 105 and 101.

#### (i) For $ x = 105 $:
1. Compute $ C(100) $ and $ C(105) $:
\[
C(100) = 4000 + 12(100) + 0.05(100)^2
\]
\[
C(105) = 4000 + 12(105) + 0.05(105)^2
\]
2. Calculate the average rate of change between $ x = 100 $ and $ x = 105 $.

#### (ii) For $ x = 101 $:
1. Similarly, compute $ C(101) $:
\[
C(101) = 4000 + 12(101) + 0.05(101)^2
\]
2. Calculate the average rate of change between $ x = 100 $ and $ x = 101 $.

### Part (b): Finding the Instantaneous Rate of Change
The **instantaneous rate of change** of $ C $ with respect to $ x $ at $ x = 100 $ is given by the derivative $ C'(x) $.

To find the derivative of the cost function $ C(x) $:
\[
C'(x) = \frac{d}{dx} (4000 + 12x + 0.05x^2) = 12 + 0.1x
\]
At $ x = 100 $:
\[
C'(100) = 12 + 0.1(100) = 12 + 10 = 22
\]
So, the marginal cost (instantaneous rate of change) at $ x = 100 $ is $22 per unit.

Let me calculate the missing values for the average rate of change.Here are the computed values:

### Part (a) Average Rate of Change:
(i) For $ x = 105 $:
\[
\text{Average rate of change} = 22.25 \, \text{dollars per unit}
\]

(ii) For $ x = 101 $:
\[
\text{Average rate of change} = 22.05 \, \text{dollars per unit}
\]

### Part (b) Instantaneous Rate of Change (Marginal Cost) at $ x = 100 $:
\[
\text{Marginal cost} = 22 \, \text{dollars per unit}
\]

Do you need further details or explanations on any of the steps?

Here are 5 related questions:
1. What is the significance of the average rate of change in this context?
2. How does the marginal cost relate to production decisions?
3. How do you interpret the difference between average and instantaneous rates of change?
4. Could we calculate the marginal cost at other values of $ x $?
5. How would increasing the power of $ x $ in the cost function affect the rate of change?

**Tip**: The marginal cost represents the cost of producing one additional unit of the commodity and is key for optimizing production levels.

The cost (in dollars) of producing x units of a certain commodity is C(x) = 4000 + 12x + 0.05x^2. Find the average and instantaneous rates of change of C with respect to x for specific values of x.

This problem involves finding the average and instantaneous rates of change for the cost function C(x) = 4000 + 12x + 0.05x^2. Key mathematical concepts include derivatives, cost functions, and rates of change. Suitable for college-level students or advanced high school calculus learners.

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