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Let's solve the problem step by step.

### Part (b):
#### Given:
Revenue function: \( R(x) = 100x - 0.2x^2 \)  
Cost function: \( C(x) = 5000 + 20x \)  

---

#### i) The profit function \( P(x) \):  
Profit function is defined as:  
\[
P(x) = R(x) - C(x)
\]  
Substituting the given functions:  
\[
P(x) = (100x - 0.2x^2) - (5000 + 20x)
\]
\[
P(x) = 100x - 0.2x^2 - 5000 - 20x
\]
\[
P(x) = -0.2x^2 + 80x - 5000
\]  

---

#### ii) The marginal profit function \( P'(x) \):  
The marginal profit is the derivative of \( P(x) \):  
\[
P'(x) = \frac{d}{dx}(-0.2x^2 + 80x - 5000)
\]
\[
P'(x) = -0.4x + 80
\]

---

#### iii) The level of demand \( x \) which maximizes the firm's profit:  
To find the maximum, set \( P'(x) = 0 \):  
\[
-0.4x + 80 = 0
\]
\[
0.4x = 80
\]
\[
x = 200
\]

---

#### iv) Determine the firm's profit at \( x = 200 \):  
Substitute \( x = 200 \) into \( P(x) \):  
\[
P(200) = -0.2(200)^2 + 80(200) - 5000
\]
\[
P(200) = -0.2(40000) + 16000 - 5000
\]
\[
P(200) = -8000 + 16000 - 5000
\]
\[
P(200) = 3000
\]

---

#### v) Verify that the level of demand \( x = 200 \) is indeed a maximum:  
To verify, check the second derivative of \( P(x) \):  
\[
P''(x) = \frac{d}{dx}(-0.4x + 80)
\]
\[
P''(x) = -0.4
\]
Since \( P''(x) < 0 \), the profit function is concave down, and \( x = 200 \) is indeed a maximum.

---

### Part (c):  
Find and classify the stationary points of \( f(x) = x^3 - 3x^2 + 6 \).  

#### Step 1: Find the derivative \( f'(x) \):  
\[
f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 6)
\]
\[
f'(x) = 3x^2 - 6x
\]

#### Step 2: Find the stationary points by setting \( f'(x) = 0 \):  
\[
3x^2 - 6x = 0
\]
Factorize:  
\[
3x(x - 2) = 0
\]
\[
x = 0 \quad \text{or} \quad x = 2
\]

#### Step 3: Classify the stationary points using the second derivative:  
\[
f''(x) = \frac{d}{dx}(3x^2 - 6x)
\]
\[
f''(x) = 6x - 6
\]

Evaluate \( f''(x) \) at each stationary point:  
- At \( x = 0 \):  
\[
f''(0) = 6(0) - 6 = -6
\]
Since \( f''(0) < 0 \), \( x = 0 \) is a **local maximum**.

- At \( x = 2 \):  
\[
f''(2) = 6(2) - 6 = 6
\]
Since \( f''(2) > 0 \), \( x = 2 \) is a **local minimum**.

#### Stationary points:  
- Local maximum at \( x = 0 \) with \( f(0) = 6 \)  
- Local minimum at \( x = 2 \) with \( f(2) = 2^3 - 3(2)^2 + 6 = 8 - 12 + 6 = 2 \)

---

Let me know if you have questions about these steps or need further details! Here are 5 related questions:

1. How would you interpret marginal profit in real-world terms?
2. Can you extend the process to find the break-even point where profit is zero?
3. How can we confirm global maxima or minima for \( f(x) \)?
4. What is the economic significance of \( P''(x) \)?
5. How do the stationary points of \( f(x) \) relate to its graph?

**Tip:** Always double-check your derivatives to ensure accuracy when solving optimization problems!

For a particular firm, the revenue function for a certain item x is given by R(x) = 100x − 0.2x^2, and the cost function is C(x) = 5000 + 20x. Based on this, use calculus techniques to compute the profit function, marginal profit function, level of demand x that maximizes the profit, profit at maximum demand, and verify if it is a maximum. Additionally, classify the stationary points of f(x) = x^3 − 3x^2 + 6.

Solve a calculus problem involving a firm's revenue and cost functions to determine profit maximization and stationary points. Key topics include optimization, derivatives, and the second derivative test, suitable for advanced high school or introductory college calculus students.

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