We are given the cost function $C(x) = 1000 + 5.7x + 0.7x^2$, and we need to calculate the average rate of change of $C$ with respect to $x$ as the production level increases from $x = 100$ to $x = 120$.

The formula for the average rate of change of a function $C(x)$ over an interval $[a, b]$ is:

$\text{Average rate of change} = \frac{C(b) - C(a)}{b - a}$

In this case, $a = 100$ and $b = 120$.

Step 1: Calculate $C(100)$

$C(100) = 1000 + 5.7(100) + 0.7(100)^2$

Step 2: Calculate $C(120)$

$C(120) = 1000 + 5.7(120) + 0.7(120)^2$

Step 3: Find the average rate of change

$\frac{C(120) - C(100)}{120 - 100}$

I'll compute these values now.The cost at $x = 100$ is $C(100) = 8570$, and at $x = 120$, $C(120) = 11764$.

The average rate of change of $C$ from $x = 100$ to $x = 120$ is:

$\frac{C(120) - C(100)}{120 - 100} = \frac{11764 - 8570}{120 - 100} = 159.7$

Thus, the correct answer is D. 159.7.

Would you like a further explanation or have any additional questions?

Here are 5 related questions you might explore:

1. How is the derivative related to the average rate of change?
2. What is the instantaneous rate of change of the cost function at $x = 100$?
3. How would the average rate of change differ if the interval was $x = 120$ to $x = 140$?
4. Can you explain the significance of the quadratic term $0.7x^2$ in the cost function?
5. How does the constant 1000 in the cost function affect the average rate of change?

Tip: The average rate of change is essentially the slope of the secant line between two points on the graph of the function.