Math Problem Statement
Solution
The problem displayed in the image states:
"For a company that manufactures tennis rackets, the average cost per racket is found to be:
for , where is the number of rackets produced per hour.
Use the differential to approximate the change in average cost per racket if production is:
- (A) increased from 25 per hour to 31 per hour, and
- (B) increased from 35 per hour to 41 per hour."
To approach this:
-
Take the derivative of the cost function to find how the cost changes as production changes:
-
Use the derivative to estimate the change in cost when changes using the formula: where is the change in production.
Let's now apply this for parts (A) and (B):
Part (A) – Production increase from 25 to 31 per hour:
- , and
- Approximate the change in cost using .
Part (B) – Production increase from 35 to 41 per hour:
- , and
- Approximate the change in cost using .
Let me calculate the values for these now.Here are the results:
- The derivative of the cost function is:
Part (A) – Production increase from 25 to 31 per hour:
- Change in production:
- Approximate change in cost: Therefore, the average cost per racket will decrease by approximately $1.46.
Part (B) – Production increase from 35 to 41 per hour:
- Change in production:
- Approximate change in cost: Therefore, the average cost per racket will increase by approximately $0.24.
Would you like more details or have any further questions?
Related questions:
- How do differentials provide approximations for changes in cost?
- How does the curvature of the cost function affect the results?
- What is the significance of the production level where the derivative is zero?
- How could one refine this approximation using integration instead?
- What other factors might influence the real-world cost of manufacturing beyond this model?
Tip:
Always check the sign of the derivative before interpreting the result, as it tells whether the cost increases or decreases!
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Math Problem Analysis
Mathematical Concepts
Differentials
Cost Functions
Derivatives
Formulas
C(x) = 360/x + 5 + (1/3)x
C'(x) = -360/x^2 + 1/3
dC ≈ C'(x) * dx
Theorems
Differentiation and its Applications
Linear Approximation with Differentials
Suitable Grade Level
Grades 10-12 (Precalculus/Calculus)
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