Given: Profit function P(p) = - 2p ^ 2 + 100p - 500
a) Find optimal price p using gradient ascent.
b) Perform 3 iterations with p = 20 a = 0.5
c) Analyze impact of different learning rates (0.2, 0.8) on convergence

To solve this problem, we'll proceed with each part step by step.

### Part (a): Find the Optimal Price \( p \) Using Gradient Ascent

The given profit function is:

\[
P(p) = -2p^2 + 100p - 500
\]

To perform gradient ascent, we need to calculate the derivative of \( P(p) \) with respect to \( p \), which will give us the gradient.

The derivative \( P'(p) \) is:

\[
P'(p) = \frac{d}{dp}\left(-2p^2 + 100p - 500\right) = -4p + 100
\]

To find the optimal price \( p \), we set the derivative \( P'(p) \) to zero and solve for \( p \):

\[
-4p + 100 = 0
\]

\[
p = \frac{100}{4} = 25
\]

So, the optimal price \( p \) using gradient ascent is \( p = 25 \).

### Part (b): Perform 3 Iterations with \( p = 20 \) and Learning Rate \( \alpha = 0.5 \)

Starting with an initial price \( p_0 = 20 \) and learning rate \( \alpha = 0.5 \), we perform the iterations as follows:

The update rule for gradient ascent is:

\[
p_{n+1} = p_n + \alpha \cdot P'(p_n)
\]

1. **Iteration 1:**
   \[
   p_1 = p_0 + 0.5 \cdot P'(p_0)
   \]
   \[
   p_1 = 20 + 0.5 \cdot (-4 \times 20 + 100)
   \]
   \[
   p_1 = 20 + 0.5 \cdot (20) = 20 + 10 = 30
   \]

2. **Iteration 2:**
   \[
   p_2 = p_1 + 0.5 \cdot P'(p_1)
   \]
   \[
   p_2 = 30 + 0.5 \cdot (-4 \times 30 + 100)
   \]
   \[
   p_2 = 30 + 0.5 \cdot (-20) = 30 - 10 = 20
   \]

3. **Iteration 3:**
   \[
   p_3 = p_2 + 0.5 \cdot P'(p_2)
   \]
   \[
   p_3 = 20 + 0.5 \cdot (-4 \times 20 + 100)
   \]
   \[
   p_3 = 20 + 0.5 \cdot 20 = 20 + 10 = 30
   \]

So, after 3 iterations, the price oscillates between 20 and 30.

### Part (c): Analyze Impact of Different Learning Rates (\( \alpha = 0.2 \) and \( \alpha = 0.8 \)) on Convergence

To analyze the impact, we need to perform similar iterations with different learning rates.

1. **Learning Rate \( \alpha = 0.2 \):**
   \[
   p_1 = 20 + 0.2 \cdot P'(20) = 20 + 0.2 \cdot 20 = 20 + 4 = 24
   \]
   \[
   p_2 = 24 + 0.2 \cdot P'(24) = 24 + 0.2 \cdot 4 = 24 + 0.8 = 24.8
   \]
   \[
   p_3 = 24.8 + 0.2 \cdot P'(24.8) = 24.8 + 0.2 \cdot 0.8 = 24.8 + 0.16 = 24.96
   \]
   With \( \alpha = 0.2 \), the price converges slowly towards the optimal price 25.

2. **Learning Rate \( \alpha = 0.8 \):**
   \[
   p_1 = 20 + 0.8 \cdot P'(20) = 20 + 0.8 \cdot 20 = 20 + 16 = 36
   \]
   \[
   p_2 = 36 + 0.8 \cdot P'(36) = 36 + 0.8 \cdot (-44) = 36 - 35.2 = 0.8
   \]
   \[
   p_3 = 0.8 + 0.8 \cdot P'(0.8) = 0.8 + 0.8 \cdot 96.8 = 0.8 + 77.44 = 78.24
   \]
   With \( \alpha = 0.8 \), the price oscillates wildly, showing that a large learning rate can lead to instability.

### Summary of Analysis:
- **\( \alpha = 0.2 \):** Slow convergence towards the optimal price.
- **\( \alpha = 0.5 \):** Oscillation between prices but still somewhat controlled.
- **\( \alpha = 0.8 \):** Unstable and wild oscillations, which might never converge.

Do you want further details or have any questions about this process?

**Related Questions:**
1. How does gradient ascent differ from gradient descent?
2. What is the significance of the learning rate in optimization algorithms?
3. What other methods can be used to find the maximum or minimum of a function?
4. Can gradient ascent be used for non-convex functions? Why or why not?
5. How does the choice of the initial point \( p_0 \) affect the convergence?

**Tip:** Always carefully choose the learning rate in gradient ascent, as too large or too small values can lead to poor convergence behavior.

Learn how to optimize a profit function using gradient ascent with a detailed step-by-step guide. Analyze the impact of different learning rates (0.2, 0.5, 0.8) on convergence. Suitable for advanced undergraduate students and practitioners.

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