Skip to main content
Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.5.60
Chapter 4, Problem 4.5.60

Business and Economics
60. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost.

Verified step by step guidance
1
Define the average cost (AC) as the total cost (C) divided by the quantity produced (Q), i.e., AC = C/Q.
Define the marginal cost (MC) as the derivative of the total cost with respect to quantity, i.e., MC = dC/dQ.
To find the production level where the average cost is smallest, take the derivative of the average cost with respect to quantity, d(AC)/dQ, and set it equal to zero to find critical points.
Use the quotient rule to differentiate AC = C/Q, which gives d(AC)/dQ = (Q * dC/dQ - C) / Q^2.
Set the derivative d(AC)/dQ equal to zero and solve for Q, which simplifies to Q * MC = C, or AC = MC, proving that the production level where average cost is smallest is where average cost equals marginal cost.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

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

Average Cost

Average cost is the total cost of production divided by the number of units produced. It is a measure of the cost per unit of output and is crucial for determining the efficiency of production. Understanding average cost helps in analyzing how costs behave as production levels change.
Recommended video:
06:37
Average Value of a Function

Marginal Cost

Marginal cost is the additional cost incurred by producing one more unit of a good or service. It is derived from the derivative of the total cost function with respect to quantity. Marginal cost is essential for decision-making, as it helps determine the optimal level of production where profits are maximized.
Recommended video:
07:32
Example 3: Maximizing Profit

Optimization in Calculus

Optimization involves finding the maximum or minimum value of a function. In the context of production, it is used to find the level of output that minimizes average cost. This often involves setting the derivative of the average cost function equal to zero and solving for the production level, which is where average cost equals marginal cost.
Recommended video:
10:13
Intro to Applied Optimization: Maximizing Area
Related Practice
Textbook Question

101. In Exercises 101 and 102, the graph of f' is given. Determine x-values corresponding to local minima, local maxima, and inflection points for the graph of f.

Textbook Question

99. In Exercises 99 and 100, the graph of f' is given. Determine x-values corresponding to inflection points for the graph of f.

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dr/dθ = −π sin (πθ), r(0) = 0

Textbook Question

Absolute Extrema on Finite Closed Intervals


In Exercises 37–40, find the function’s absolute maximum and minimum values and say where they occur.


g(θ) = θ³ᐟ⁵, −32 ≤ θ ≤ 1

Textbook Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.

7. y=sin|x|, -2π≤x≤2π

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.


∫x⁻¹ᐟ³ dx