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The image presents the following mathematical problem for HW#5:

- A product's total cost function \( C(x) \) is given as:
  \[
  C(x) = \frac{x^2}{100} + 100x + 40
  \]
  where \( x \) is the number of units produced.

- The price at which each unit can be sold, \( p(x) \), is given by:
  \[
  p(x) = 200 - \frac{x}{400}
  \]

The tasks are as follows:

1. **Find the marginal cost and interpret the result.**
2. **Find the average cost.**
3. **Find the exact cost of the 21st item and compare that with the marginal cost at \( x = 20 \).**
4. **Graph the two equations \( C(x) \) and \( p(x) \), find the intersection points and interpret.**
5. **Determine \( x \) for maximum profit and find the maximum profit.**

Let’s solve these step by step.

### 1. Marginal Cost
The marginal cost is the derivative of the cost function \( C(x) \):
\[
C'(x) = \frac{d}{dx} \left( \frac{x^2}{100} + 100x + 40 \right)
\]
\[
C'(x) = \frac{2x}{100} + 100
\]
\[
C'(x) = \frac{x}{50} + 100
\]

The marginal cost represents the cost of producing one additional unit of the product.

### 2. Average Cost
The average cost is given by:
\[
\text{Average Cost} = \frac{C(x)}{x}
\]
Substituting \( C(x) \):
\[
\text{Average Cost} = \frac{\frac{x^2}{100} + 100x + 40}{x}
\]
Simplifying:
\[
\text{Average Cost} = \frac{x}{100} + 100 + \frac{40}{x}
\]

### 3. Exact Cost of the 21st Item vs. Marginal Cost at \( x = 20 \)

- **Exact cost of the 21st item** is:
  \[
  C(21) - C(20)
  \]
  First, calculate \( C(21) \) and \( C(20) \):

  \[
  C(21) = \frac{21^2}{100} + 100(21) + 40 = \frac{441}{100} + 2100 + 40 = 4.41 + 2100 + 40 = 2144.41
  \]
  \[
  C(20) = \frac{20^2}{100} + 100(20) + 40 = \frac{400}{100} + 2000 + 40 = 4 + 2000 + 40 = 2044
  \]
  Therefore, the exact cost of the 21st item:
  \[
  2144.41 - 2044 = 100.41
  \]

- **Marginal cost at \( x = 20 \)** is:
  \[
  C'(20) = \frac{20}{50} + 100 = 0.4 + 100 = 100.4
  \]

Thus, the exact cost of the 21st item is \( 100.41 \), and the marginal cost at \( x = 20 \) is \( 100.4 \).

### 4. Graphing \( C(x) \) and \( p(x) \)
The two functions are:

- \( C(x) = \frac{x^2}{100} + 100x + 40 \)
- \( p(x) = 200 - \frac{x}{400} \)

To find the intersection points, set \( C(x) = p(x) \) and solve for \( x \):
\[
\frac{x^2}{100} + 100x + 40 = 200 - \frac{x}{400}
\]
Multiply by 400 to eliminate fractions:
\[
4x^2 + 40000x + 16000 = 80000 - x
\]
Simplify the equation and solve for \( x \) (this step involves solving a quadratic equation).

### 5. Maximizing Profit
Profit \( P(x) \) is given by:
\[
P(x) = x \cdot p(x) - C(x)
\]
Substitute \( p(x) \) and \( C(x) \), then differentiate \( P(x) \) with respect to \( x \), set the derivative to zero, and solve for \( x \) to find the maximum profit.

---

Would you like a detailed graphing process, quadratic solving, or more specific interpretations of the results?

### 5 Related Questions:
1. How does the marginal cost function behave as production increases?
2. What is the economic significance of average cost and how does it compare to marginal cost?
3. How can the intersection points of \( C(x) \) and \( p(x) \) inform pricing strategies?
4. How do you derive the profit function in terms of \( x \), and why is maximizing it important?
5. Can the maximum profit occur at a production level where marginal cost exceeds price?

### Tip:
Marginal cost should always be compared to marginal revenue when determining the optimal production level for maximizing profit.

A given product can be manufactured at a total cost C(x) = (x^2 / 100) + 100x + 40, where x is the number of units produced. If p(x) = 200 - (x / 400) is the price at which each unit can be sold, then: 

1. Find the marginal cost and interpret the result. 
2. Find the average cost. 
3. Find the exact cost of 21st item and compare that with marginal cost at x = 20. 
4. Graph the two equations C(x) and p(x), find the intersection points and interpret. 
5. Determine x for maximum profit and find the maximum profit.

Learn how to calculate marginal cost, average cost, and maximize profit for a product where the total cost function is C(x) = (x^2 / 100) + 100x + 40 and the price function is p(x) = 200 - (x / 400). This problem involves calculus concepts including differentiation and quadratic equations. Suitable for IB and advanced high school students.

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