Math Problem Statement

The cost (in dollars) of producing x units of a certain commodity is C(x) = 8000 + 13x + 0.1x^2. (a) Find the average rate of change of C with respect to x when the production level is changed from x = 100 to 104 and 101. (b) Find the instantaneous rate of change of C at x = 100.

Solution

Let's solve the problem step by step:

Problem Statement:

The cost function is given as:

C(x)=8000+13x+0.1x2C(x) = 8000 + 13x + 0.1x^2

where C(x)C(x) is the cost in dollars of producing xx units of a commodity.

Part (a): Average Rate of Change

We are asked to find the average rate of change of CC with respect to xx when xx changes from 100100 to two given values:

Formula:

The average rate of change of a function between two points x1x_1 and x2x_2 is given by:

Average Rate of Change=C(x2)C(x1)x2x1\text{Average Rate of Change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1}

(i) x2=104x_2 = 104 and x1=100x_1 = 100

  • First, we calculate C(104)C(104) and C(100)C(100).

C(104)=8000+13(104)+0.1(104)2C(104) = 8000 + 13(104) + 0.1(104)^2 C(100)=8000+13(100)+0.1(100)2C(100) = 8000 + 13(100) + 0.1(100)^2

Then, we apply the formula for the average rate of change.

(ii) x2=101x_2 = 101 and x1=100x_1 = 100

  • Similarly, calculate C(101)C(101) and C(100)C(100).

C(101)=8000+13(101)+0.1(101)2C(101) = 8000 + 13(101) + 0.1(101)^2 C(100)=8000+13(100)+0.1(100)2C(100) = 8000 + 13(100) + 0.1(100)^2

Again, apply the formula for the average rate of change.

Part (b): Instantaneous Rate of Change

The instantaneous rate of change (or marginal cost) is the derivative of the cost function C(x)C(x) with respect to xx, evaluated at x=100x = 100.

Formula for Marginal Cost:

C(x)=ddx(8000+13x+0.1x2)=13+0.2xC'(x) = \frac{d}{dx}(8000 + 13x + 0.1x^2) = 13 + 0.2x

Evaluate this at x=100x = 100:

C(100)=13+0.2(100)C'(100) = 13 + 0.2(100)

Let me calculate all of these values for you.Here are the results:

(a) Average Rate of Change:

  1. For x=104x = 104: Average rate of change=33.40dollars per unit\text{Average rate of change} = 33.40 \, \text{dollars per unit}

  2. For x=101x = 101: Average rate of change=33.10dollars per unit\text{Average rate of change} = 33.10 \, \text{dollars per unit}

(b) Instantaneous Rate of Change (Marginal Cost) at x=100x = 100:

Marginal cost=33.00dollars per unit\text{Marginal cost} = 33.00 \, \text{dollars per unit}

Would you like any further explanation or details?

Here are 5 related questions to deepen your understanding:

  1. How does the average rate of change differ from the instantaneous rate of change?
  2. What does the marginal cost represent in practical terms?
  3. How would the rates change if the cost function had a cubic term?
  4. What if we wanted to estimate costs for x=102x = 102? How would you use the marginal cost?
  5. Can the marginal cost ever be negative? Under what conditions?

Tip: The marginal cost is often used to make production decisions, as it indicates the cost of producing one additional unit.

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Math Problem Analysis

Mathematical Concepts

Rates of Change
Derivatives
Quadratic Functions

Formulas

Average Rate of Change = (C(x2) - C(x1)) / (x2 - x1)
Marginal Cost (Instantaneous Rate of Change) = d(C(x)) / dx

Theorems

Definition of Derivative

Suitable Grade Level

Grade 12 or College-Level Calculus