Math Problem Statement
Solution
We are given a cost function , where represents the cost in dollars of producing units of a commodity.
Part (a): Finding the Average Rate of Change
The average rate of change of with respect to between two points and is calculated using the formula: We need to compute this for two values of : 105 and 101.
(i) For :
- Compute and :
- Calculate the average rate of change between and .
(ii) For :
- Similarly, compute :
- Calculate the average rate of change between and .
Part (b): Finding the Instantaneous Rate of Change
The instantaneous rate of change of with respect to at is given by the derivative .
To find the derivative of the cost function : At : So, the marginal cost (instantaneous rate of change) at 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 :
(ii) For :
Part (b) Instantaneous Rate of Change (Marginal Cost) at :
Do you need further details or explanations on any of the steps?
Here are 5 related questions:
- What is the significance of the average rate of change in this context?
- How does the marginal cost relate to production decisions?
- How do you interpret the difference between average and instantaneous rates of change?
- Could we calculate the marginal cost at other values of ?
- How would increasing the power of 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
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