Let's break this problem into two parts as per your request:

Given Profit function:

$P(p) = 5.8p - 64p^2$ Where $P(p)$ is the profit in thousands of US dollars and $p$ is the unit price in dollars per unit.

Part a: Average Rate of Change of Profit between $p = 0.30$ and $p = 0.40$

The average rate of change (AROC) of a function $P(p)$ over an interval $[p_1, p_2]$ is given by the formula:

$\text{AROC} = \frac{P(p_2) - P(p_1)}{p_2 - p_1}$

In this case, $p_1 = 0.30$ and $p_2 = 0.40$.

First, we'll compute $P(p_1)$ and $P(p_2)$:

• For $p = 0.30$:

$P(0.30) = 5.8(0.30) - 64(0.30)^2$

• For $p = 0.40$:

$P(0.40) = 5.8(0.40) - 64(0.40)^2$

Now, compute the difference $P(p_2) - P(p_1)$, and divide by the change in price, $p_2 - p_1 = 0.40 - 0.30 = 0.10$.

Let's calculate this step-by-step:

Step 1: Calculate $P(0.30)$ and $P(0.40)$

I'll compute these values now.#### Step 1 Results:

• $P(0.30) = -4.02$ thousand USD
• $P(0.40) = -7.92$ thousand USD

Step 2: Average Rate of Change (AROC)

$\text{AROC} = \frac{-7.92 - (-4.02)}{0.40 - 0.30} = \frac{-7.92 + 4.02}{0.10} = \frac{-3.90}{0.10} = -39$

Thus, the average rate of change of the profit between the unit prices of 0.30 and 0.40 is -39 thousand USD per dollar. This means that, on average, the profit decreases by 39 thousand USD for every $1 increase in unit price within this interval.

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Part b: Instantaneous Rate of Change at $p = 0.40$

The instantaneous rate of change (also the derivative of $P(p)$) at $p = 0.40$ can help us determine if the profit is increasing or decreasing at that price. Using the limit definition of derivatives:

$P'(p) = \lim_{\Delta p \to 0} \frac{P(p + \Delta p) - P(p)}{\Delta p}$

Instead of manually applying the limit definition here, we can differentiate $P(p)$ directly. Given:

$P(p) = 5.8p - 64p^2$

Differentiate this with respect to $p$:

$P'(p) = \frac{d}{dp} (5.8p - 64p^2) = 5.8 - 128p$

Now, evaluate $P'(p)$ at $p = 0.40$:

$P'(0.40) = 5.8 - 128(0.40)$

Let me compute this value next.The instantaneous rate of change of profit at $p = 0.40$ is:

$P'(0.40) = -45.4 \text{ thousand USD per dollar}$

Since the derivative is negative at $p = 0.40$, it indicates that the profit is decreasing at that price. Therefore, the profit is not instantaneously increasing when the unit price is $0.40$ USD.

Summary:

• Part a: The average rate of change of the profit from $p = 0.30$ to $p = 0.40$ is $-39$ thousand USD per dollar, meaning the profit is decreasing on average.
• Part b: The profit is instantaneously decreasing at $p = 0.40$ because the derivative at this point is negative, $P'(0.40) = -45.4$.

Would you like further details on any part of this solution?

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Here are 5 follow-up questions you might explore:

1. How does the profit change if the price increases beyond $0.40?
2. What is the critical point where the profit function reaches a maximum?
3. How would you interpret the negative profit values in this context?
4. What happens to the profit if the price increases indefinitely?
5. How would you calculate the elasticity of profit with respect to price?

Tip: Instantaneous rate of change gives a precise measure at a single point, while average rate of change gives a general trend over an interval.