Textbook Question
Theory and Examples
Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.5.62
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 guidance1
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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Video duration:
3mKey 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.
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Guided course
06:37
Average Value of a Function
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.
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04:50Critical Points
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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06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(cscθ cotθ) / 2 dθ
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Textbook Question
56. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the accompanying figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax^3+bx^2+cx+d, where y(-L)= H and y(0)=0.
a. What is dy/dx at x = 0?
b. What is dy/dx at x = -L?
c. Use the values for dy/dx at x = 0 and x =- L together with y(0) = 0 and y(-L) = H to show that y(x)=H[2(x/L)^3+3(x/L)^2]
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Textbook Question
Absolute Extrema on Finite Closed Intervals
In Exercises 21–36, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
g(x) = √(4 − x²), −2 ≤ x ≤ 1
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Textbook Question
The 8-ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.
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Textbook Question
Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(−3csc²x)dx
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