Math Problem Statement

The cost (in dollars) of producing x units of a certain commodity is C(x) = 4000 + 12x + 0.05x^2. Find the average and instantaneous rates of change of C with respect to x for specific values of x.

Solution

We are given a cost function C(x)=4000+12x+0.05x2C(x) = 4000 + 12x + 0.05x^2, where C(x)C(x) represents the cost in dollars of producing xx units of a commodity.

Part (a): Finding the Average Rate of Change

The average rate of change of CC with respect to xx between two points x=100x = 100 and x=ax = a is calculated using the formula: Average rate of change=C(a)C(100)a100\text{Average rate of change} = \frac{C(a) - C(100)}{a - 100} We need to compute this for two values of aa: 105 and 101.

(i) For x=105x = 105:

  1. Compute C(100)C(100) and C(105)C(105): C(100)=4000+12(100)+0.05(100)2C(100) = 4000 + 12(100) + 0.05(100)^2 C(105)=4000+12(105)+0.05(105)2C(105) = 4000 + 12(105) + 0.05(105)^2
  2. Calculate the average rate of change between x=100x = 100 and x=105x = 105.

(ii) For x=101x = 101:

  1. Similarly, compute C(101)C(101): C(101)=4000+12(101)+0.05(101)2C(101) = 4000 + 12(101) + 0.05(101)^2
  2. Calculate the average rate of change between x=100x = 100 and x=101x = 101.

Part (b): Finding the Instantaneous Rate of Change

The instantaneous rate of change of CC with respect to xx at x=100x = 100 is given by the derivative C(x)C'(x).

To find the derivative of the cost function C(x)C(x): C(x)=ddx(4000+12x+0.05x2)=12+0.1xC'(x) = \frac{d}{dx} (4000 + 12x + 0.05x^2) = 12 + 0.1x At x=100x = 100: C(100)=12+0.1(100)=12+10=22C'(100) = 12 + 0.1(100) = 12 + 10 = 22 So, the marginal cost (instantaneous rate of change) at x=100x = 100 is $22 per unit.

Let me calculate the missing values for the average rate of change.Here are the computed values:

Part (a) Average Rate of Change:

(i) For x=105x = 105: Average rate of change=22.25dollars per unit\text{Average rate of change} = 22.25 \, \text{dollars per unit}

(ii) For x=101x = 101: Average rate of change=22.05dollars per unit\text{Average rate of change} = 22.05 \, \text{dollars per unit}

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

Marginal cost=22dollars per unit\text{Marginal cost} = 22 \, \text{dollars per unit}

Do you need further details or explanations on any of the steps?

Here are 5 related questions:

  1. What is the significance of the average rate of change in this context?
  2. How does the marginal cost relate to production decisions?
  3. How do you interpret the difference between average and instantaneous rates of change?
  4. Could we calculate the marginal cost at other values of xx?
  5. How would increasing the power of xx in the cost function affect the rate of change?

Tip: The marginal cost represents the cost of producing one additional unit of the commodity and is key for optimizing production levels.

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Math Problem Analysis

Mathematical Concepts

Average rate of change
Instantaneous rate of change
Derivatives
Cost function

Formulas

C(x) = 4000 + 12x + 0.05x^2
Average rate of change: (C(a) - C(b)) / (a - b)
Derivative of a cost function: C'(x) = dC/dx

Theorems

Derivative rules
Quadratic functions

Suitable Grade Level

College level or Advanced High School Calculus