Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

5. Graphical Applications of Derivatives

Applied Optimization

Calculus

5. Graphical Applications of Derivatives

Applied Optimization

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3:11 minutes

Problem 4.5.62

Textbook Question

Business and Economics
62. Production level Suppose that c(x)=x^3-20x^2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.

Verified step by step guidance

First, understand that the average cost function is given by dividing the total cost function c(x) by the number of items x. So, the average cost function A(x) is A(x) = c(x)/x = (x^3 - 20x^2 + 20000x)/x.

Simplify the average cost function A(x) by dividing each term in the numerator by x. This results in A(x) = x^2 - 20x + 20000.

To find the production level that minimizes the average cost, take the derivative of the average cost function A(x) with respect to x. This gives A'(x) = 2x - 20.

Set the derivative A'(x) equal to zero to find the critical points. Solve the equation 2x - 20 = 0 to find the value of x that minimizes the average cost.

Verify that the critical point found is indeed a minimum by using the second derivative test. Compute the second derivative A''(x) = 2, which is positive, indicating that the function is concave up at the critical point, confirming a minimum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Cost Function

The average cost function is derived by dividing the total cost function by the number of items produced, x. It represents the cost per unit of production and is crucial for determining the production level that minimizes costs. In this problem, the average cost function is c(x)/x, which simplifies to x^2 - 20x + 20,000.

Critical Points and Optimization

To find the production level that minimizes the average cost, we need to identify critical points by taking the derivative of the average cost function and setting it to zero. This process helps locate points where the function's slope is zero, indicating potential minima or maxima. Solving for these points allows us to determine the optimal production level.

Second Derivative Test

The second derivative test is used to confirm whether a critical point is a minimum or maximum. By evaluating the second derivative at the critical points, we can determine the concavity of the function. If the second derivative is positive, the function is concave up, indicating a local minimum, which is essential for ensuring the production level minimizes average costs.

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Business and Economics62. Production level Suppose that c(x)=x^3-20x^2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.

Key Concepts

Average Cost Function

Critical Points and Optimization

Second Derivative Test

Watch next

Business and Economics
62. Production level Suppose that c(x)=x^3-20x^2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.