Let's break down the problem provided in the image.

Part (a):

We are given:

• $f(10) = 50$
• $f'(10) = -2$

Contextual explanation:

• $f(p)$ is the quantity $q$ of fabric sold in thousands of yards when the price per yard is $p$ dollars.
• $f(10) = 50$ means that when the price of the fabric is $10 per yard, the quantity sold is 50,000 yards (since $q$ is in thousands of yards).
• $f'(10) = -2$ indicates that at a price of $10 per yard, increasing the price by $1 (a small increase) would decrease the quantity sold by 2,000 yards. In other words, for every $1 increase in price at $10, the manufacturer will sell 2,000 fewer yards of fabric.

Part (b):

The revenue function is given by:

$R(p) = p \cdot f(p)$

Where:

• $R(p)$ is the total revenue, which is the product of the price per yard and the quantity of fabric sold at that price.

We are tasked with finding $R'(10)$, which represents the rate of change of revenue when the price is $10.

To do this, we use the product rule of differentiation: $R'(p) = f(p) + p \cdot f'(p)$

Substituting $p = 10$, $f(10) = 50$, and $f'(10) = -2$:

$R'(10) = f(10) + 10 \cdot f'(10)$ $R'(10) = 50 + 10 \cdot (-2)$ $R'(10) = 50 - 20 = 30$

Meaning of $R'(10) = 30$:

This means that at a price of $10 per yard, a small increase in the price will increase the revenue by $30 for each $1 increase in the price. Therefore, at this price point, raising the price slightly would lead to a net increase in revenue.

Do you want further details or have any questions?

Here are 5 related questions that expand on this problem:

1. How would the meaning change if $f'(10)$ were positive instead of negative?
2. What happens to the revenue if $f(p)$ is a linear function?
3. How does the second derivative $R''(p)$ relate to changes in the rate of revenue growth?
4. How would you interpret $f(15)$ and $f'(15)$ in this context?
5. Can you find the price that maximizes revenue using the function $R(p) = p \cdot f(p)$?

Tip: The product rule is essential when differentiating functions that involve a product of two variables. Always check if the product rule is needed before applying simple differentiation!