Given the cost function $C(x) = 0.04x^2 + 5.1x + 40$, where $C$ is in thousands of dollars and $x$ represents the number of units produced, let’s proceed with each part of the question.

Part (a): Change in Cost from 10 to 11 Units

To find the change in cost as the production increases from 10 to 11 units, calculate $C(11) - C(10)$.

1. Calculate $C(10)$: $C(10) = 0.04(10)^2 + 5.1(10) + 40 = 0.04(100) + 51 + 40 = 4 + 51 + 40 = 95 \text{ (thousands of dollars)}$

2. Calculate $C(11)$: $C(11) = 0.04(11)^2 + 5.1(11) + 40 = 0.04(121) + 56.1 + 40 = 4.84 + 56.1 + 40 = 100.94 \text{ (thousands of dollars)}$

3. Change in Cost: $C(11) - C(10) = 100.94 - 95 = 5.94 \text{ (thousands of dollars)}$

The change in cost as production increases from 10 to 11 units is 5.94 thousand dollars.

---

Part (b): Average Cost from 10 to 15 Units

The average cost over the interval $x = 10$ to $x = 15$ is given by: $\frac{C(15) - C(10)}{15 - 10}$

1. Calculate $C(15)$: $C(15) = 0.04(15)^2 + 5.1(15) + 40 = 0.04(225) + 76.5 + 40 = 9 + 76.5 + 40 = 125.5 \text{ (thousands of dollars)}$

2. Average Cost: $\frac{C(15) - C(10)}{5} = \frac{125.5 - 95}{5} = \frac{30.5}{5} = 6.1 \text{ (thousands of dollars per unit)}$

The average cost as production changes from 10 to 15 units is 6.1 thousand dollars per unit.

---

Part (c): Instantaneous Rate of Change of Cost at 10 Units

To find the instantaneous rate of change of the cost with respect to the production level when $x = 10$, we compute the derivative $C'(x)$ and evaluate it at $x = 10$.

1. Find $C'(x)$: $C(x) = 0.04x^2 + 5.1x + 40$ Differentiate with respect to $x$: $C'(x) = 0.08x + 5.1$

2. Evaluate $C'(10)$: $C'(10) = 0.08(10) + 5.1 = 0.8 + 5.1 = 5.9 \text{ (thousands of dollars per unit)}$

The instantaneous rate of change of the cost when 10 units are produced is 5.9 thousand dollars per unit. Since this value is positive, the cost is increasing at this production level.

---

Summary:

• (a) Change in cost from 10 to 11 units: $5.94$ thousand dollars.
• (b) Average cost from 10 to 15 units: $6.1$ thousand dollars per unit.
• (c) Instantaneous rate of change at 10 units: $5.9$ thousand dollars per unit (cost is increasing).

Would you like further clarification or have any questions?

---

Further Questions:

1. How would the change in cost differ if the production increased from 15 to 16 units?
2. What is the total cost if the production level is 20 units?
3. How does the average cost change as production increases from 15 to 20 units?
4. How could we determine the level of production that minimizes the cost?
5. What does the rate of change tell us about production efficiency?

---

Tip:

The instantaneous rate of change, or derivative, is particularly useful for determining the exact behavior of costs at specific production levels—helpful for cost control in business planning!