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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.

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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!

A manufacturer produces and sells bolts of fabric. The quantity q of this fabric sold is a function of the selling price p, represented as q = f(p). Given f(10) = 50 and f'(10) = -2, explain the meanings of these values and find R'(10) using R(p) = p * f(p).

Learn how to interpret the derivative of a revenue function in a real-world context involving price and quantity. This problem involves applying the product rule to differentiate R(p) = p * f(p) and analyze the meaning of f(10) = 50 and f'(10) = -2. Suitable for students in Grades 11-12.

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