Math Problem Statement

A manufacturer produces bolts of a fabric with a fixed width. The quantity π‘ž 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 π‘ž = 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 β€² (20) = βˆ’300? (b) Assuming the values in part (a), find 𝑅 β€² (20) and interpret your answer

Solution

Let’s break down the problem part by part.

(a) Interpretation of f(20)=15,000f(20) = 15,000 and fβ€²(20)=βˆ’300f'(20) = -300:

  • f(p)f(p) represents the quantity of fabric sold as a function of the price per yard, pp.

  • f(20)=15,000f(20) = 15,000 means that when the price of the fabric is $20 per yard, the manufacturer sells 15,000 yards of fabric.

  • fβ€²(20)=βˆ’300f'(20) = -300 represents the rate of change of the quantity sold with respect to the price at p=20p = 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 Rβ€²(20)R'(20) and its interpretation:

The total revenue function R(p)R(p) is given by: R(p)=pβ‹…f(p)R(p) = p \cdot f(p) To find the rate of change of revenue with respect to price, Rβ€²(p)R'(p), we use the product rule for differentiation: Rβ€²(p)=f(p)+pβ‹…fβ€²(p)R'(p) = f(p) + p \cdot f'(p) Substitute the given values f(20)=15,000f(20) = 15,000, fβ€²(20)=βˆ’300f'(20) = -300, and p=20p = 20: Rβ€²(20)=f(20)+20β‹…fβ€²(20)R'(20) = f(20) + 20 \cdot f'(20) Rβ€²(20)=15,000+20β‹…(βˆ’300)R'(20) = 15,000 + 20 \cdot (-300) Rβ€²(20)=15,000βˆ’6,000=9,000R'(20) = 15,000 - 6,000 = 9,000

Interpretation of Rβ€²(20)=9,000R'(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)=15,000f(20) = 15,000 means 15,000 yards are sold when the price is $20.
  • fβ€²(20)=βˆ’300f'(20) = -300 means that for every $1 increase in price, 300 fewer yards are sold.
  • Rβ€²(20)=9,000R'(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 fβ€²(p)f'(p) were positive instead of negative?
  2. How is elasticity of demand related to the values of fβ€²(p)f'(p)?
  3. What would happen to Rβ€²(p)R'(p) if fβ€²(p)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 fβ€²(p)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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Revenue Function
Product Rule

Formulas

Revenue Function: R(p) = p * f(p)
Derivative using Product Rule: R'(p) = f(p) + p * f'(p)

Theorems

Product Rule for Differentiation

Suitable Grade Level

Grades 11-12, College