Math Problem Statement
The cost function for a particular product is given by C(x)=0.0001x3−0.018x2+12x+110 dollars, where 0≤x≤110. Find the minimum marginal cost of the product, rounded to the nearest cent.
Solution
To find the minimum marginal cost, we need to first understand that the marginal cost is the derivative of the cost function, . We’ll follow these steps:
- Differentiate with respect to to find , which represents the marginal cost.
- Find the critical points by setting .
- Evaluate the marginal cost at these critical points and the boundaries and to determine the minimum marginal cost.
Step 1: Differentiate
Given:
Differentiating with respect to :
Step 2: Set to find critical points
Solving:
This is a quadratic equation, which we can solve using the quadratic formula:
Simplifying:
Calculating the values of from this will provide the critical points.
Step 3: Evaluate at critical points and endpoints
Once we have the critical points, we will evaluate at these points and also at and .
Let me calculate this for you.The real values we obtained for the marginal cost at the endpoints are:
- At :
- At :
The critical points yielded complex numbers, which are not within the domain of interest for this problem.
Conclusion
The minimum marginal cost of the product, rounded to the nearest cent, is $11.67 at .
Would you like a further breakdown of any of the steps, or do you have additional questions?
- What is the significance of a minimum marginal cost in production?
- How would the cost function change if an additional fixed cost was added?
- What does it mean for a function to have complex critical points in this context?
- How would we interpret marginal cost in an economic or business context?
- Could there be a scenario where the minimum marginal cost occurs at ?
Tip: When differentiating polynomials, each term's exponent decreases by one, making it a straightforward process to find marginal costs in cost functions.
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Math Problem Analysis
Mathematical Concepts
Differentiation
Marginal Cost
Quadratic Equations
Formulas
C'(x) = 0.0003x^2 - 0.036x + 12
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
Theorems
Derivative of a function
Critical points
Quadratic formula
Suitable Grade Level
Grades 11-12