Math Problem Statement

p = f(x) = −0.1x2 − 0.3x + 30 where x is the quantity demanded in thousands and p is the unit price in dollars. (a) Find f '(x). f '(x) =

(b) What is the rate of change of the unit price (in dollars per 1,000 lamps) when the quantity demanded is 2,000 units (x = 2)? $ per 1,000 lamps What is the unit price (in dollars) at that level of demand?

Solution

Part (a): Find the derivative of f(x)f(x)

Given the function:
f(x)=0.1x20.3x+30f(x) = -0.1x^2 - 0.3x + 30

To find f(x)f'(x), we will differentiate each term of f(x)f(x) with respect to xx:

  • The derivative of 0.1x2-0.1x^2 is 0.2x-0.2x (using the power rule: ddx(axn)=naxn1\frac{d}{dx}(ax^n) = n \cdot ax^{n-1}).
  • The derivative of 0.3x-0.3x is 0.3-0.3 (since the derivative of xx is 1).
  • The derivative of the constant 30 is 0 (since the derivative of a constant is always 0).

Thus, the derivative is: f(x)=0.2x0.3f'(x) = -0.2x - 0.3

Part (b): Rate of change when x=2x = 2

To find the rate of change when the quantity demanded is 2,000 units (which corresponds to x=2x = 2 because the units are in thousands), we will substitute x=2x = 2 into the derivative f(x)f'(x).

Substituting x=2x = 2 into f(x)f'(x): f(2)=0.2(2)0.3=0.40.3=0.7f'(2) = -0.2(2) - 0.3 = -0.4 - 0.3 = -0.7

Thus, the rate of change of the unit price when the quantity demanded is 2,000 units is 0.7-0.7 dollars per 1,000 lamps. This means the price decreases by $0.70 for each additional 1,000 lamps demanded.

Unit price at x=2x = 2

To find the unit price at x=2x = 2, we substitute x=2x = 2 into the original price function f(x)f(x):

f(2)=0.1(2)20.3(2)+30f(2) = -0.1(2)^2 - 0.3(2) + 30 f(2)=0.1(4)0.6+30=0.40.6+30=29f(2) = -0.1(4) - 0.6 + 30 = -0.4 - 0.6 + 30 = 29

Thus, the unit price when the quantity demanded is 2,000 units is $29.

Summary of Answers:

  • (a) f(x)=0.2x0.3f'(x) = -0.2x - 0.3
  • (b) The rate of change of the unit price at x=2x = 2 is 0.7-0.7 dollars per 1,000 lamps.
  • The unit price at x=2x = 2 (2,000 units) is $29.

Let me know if you'd like more details or have further questions!

Related Questions:

  1. What happens to the rate of change of price if xx increases further (e.g., at x=3x = 3)?
  2. How does the rate of change of price impact a firm's pricing strategy in real-life scenarios?
  3. What would be the price when the demand is at 1,000 units (x=1x = 1)?
  4. How can the second derivative f(x)f''(x) help analyze the concavity of the price function?
  5. If the price increases by $0.70 per 1,000 units, what effect would that have on total revenue?

Tip: When interpreting the rate of change in a function like this, think of it as a slope that tells you how much the price will increase or decrease as demand changes.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quadratic Functions
Rate of Change

Formulas

f'(x) = d(ax^2 + bx + c)/dx = 2ax + b
f'(2) = -0.2(2) - 0.3
f(x) = ax^2 + bx + c for calculating price when demand level is known

Theorems

Power Rule for Derivatives
Linear Approximation for Rates of Change

Suitable Grade Level

Grades 10-12