A manufacturer produces bolts of a fabric with a fixed width. The quantity 𝑞 of
this fabric (measured in yards) that is sold is a function of the selling price 𝑝 (in
dollars per yard), so we can write 𝑞 = 𝑓 (𝑝). Then the total revenue earned with
selling price 𝑝 is 𝑅(𝑝) = 𝑝 𝑓 (𝑝).
(a) What does it mean to say that 𝑓 (20) = 15, 000 and 𝑓
′
(20) = −300?
(b) Assuming the values in part (a), find 𝑅
′
(20) and interpret your answer

Let’s break down the problem part by part.

### (a) Interpretation of $ f(20) = 15,000 $ and $ f'(20) = -300 $:

- $ f(p) $ represents the quantity of fabric sold as a function of the price per yard, $ p $. 
- $ f(20) = 15,000 $ means that when the price of the fabric is $20 per yard, the manufacturer sells 15,000 yards of fabric.
  
- $ f'(20) = -300 $ represents the rate of change of the quantity sold with respect to the price at $ p = 20 $. This means that when the price is $20 per yard, an increase of $1 in price will result in a decrease in the quantity sold by 300 yards. In other words, the demand is decreasing as the price increases.

### (b) Finding $ R'(20) $ and its interpretation:

The total revenue function $ R(p) $ is given by:
\[
R(p) = p \cdot f(p)
\]
To find the rate of change of revenue with respect to price, $ R'(p) $, we use the product rule for differentiation:
\[
R'(p) = f(p) + p \cdot f'(p)
\]
Substitute the given values $ f(20) = 15,000 $, $ f'(20) = -300 $, and $ p = 20 $:
\[
R'(20) = f(20) + 20 \cdot f'(20)
\]
\[
R'(20) = 15,000 + 20 \cdot (-300)
\]
\[
R'(20) = 15,000 - 6,000 = 9,000
\]

#### Interpretation of $ R'(20) = 9,000 $:

This means that when the price is $20 per yard, the revenue is increasing at a rate of $9,000 per dollar increase in price. Thus, increasing the price by $1 from $20 will result in a $9,000 increase in total revenue.

### Summary:
- $ f(20) = 15,000 $ means 15,000 yards are sold when the price is $20.
- $ f'(20) = -300 $ means that for every $1 increase in price, 300 fewer yards are sold.
- $ R'(20) = 9,000 $ means that the revenue is increasing by $9,000 for each $1 increase in price when the price is $20.

---

Would you like more details or have any further questions? Here are five related questions for deeper understanding:

1. How would the total revenue behave if $ f'(p) $ were positive instead of negative?
2. How is elasticity of demand related to the values of $ f'(p) $?
3. What would happen to $ R'(p) $ if $ f'(p) $ were smaller in magnitude?
4. Can you determine the price that maximizes revenue based on this information?
5. How does the interpretation of $ f'(p) $ change if we measure price in different units?

**Tip**: The product rule is crucial when dealing with the derivative of functions involving both price and quantity in revenue or cost models.

A manufacturer produces bolts of a fabric with a fixed width. The quantity 𝑞 of this fabric (measured in yards) that is sold is a function of the selling price 𝑝 (in dollars per yard), so we can write 𝑞 = 𝑓(𝑝). Then the total revenue earned with selling price 𝑝 is 𝑅(𝑝) = 𝑝𝑓(𝑝). (a) What does it mean to say that 𝑓(20) = 15,000 and 𝑓′(20) = −300? (b) Assuming the values in part (a), find 𝑅′(20) and interpret your answer.

Learn how to interpret the derivative of a revenue function where sales quantity is a function of price. Given f(20) = 15,000 and f'(20) = -300, find R'(20) and understand the meaning of these values. This detailed breakdown covers key calculus concepts like differentiation and product rule, suitable for students in Grades 11-12 and college.

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