Math Problem Statement

For a company that manufactures tennis rackets, the average cost per racket is given by C(x) = 360/x + 5 + (1/3)x for x ≥ 1, where x is the number of rackets produced per hour. Use the differential to approximate the change in average cost per racket if production is increased from 25 per hour to 31 per hour and from 35 per hour to 41 per hour.

Solution

The problem displayed in the image states:

"For a company that manufactures tennis rackets, the average cost per racket is found to be:

C(x)=360x+5+13xC(x) = \frac{360}{x} + 5 + \frac{1}{3}x

for x1x \geq 1, where xx 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:

  1. Take the derivative C(x)C'(x) of the cost function C(x)C(x) to find how the cost changes as production changes: C(x)=360x+5+13xC(x) = \frac{360}{x} + 5 + \frac{1}{3}x C(x)=360x2+13C'(x) = -\frac{360}{x^2} + \frac{1}{3}

  2. Use the derivative C(x)C'(x) to estimate the change in cost dCdC when xx changes using the formula: dCC(x)dxdC \approx C'(x) \cdot dx where dxdx 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:

  • x=25x = 25, and dx=3125=6dx = 31 - 25 = 6
  • Approximate the change in cost dCdC using C(25)C'(25).

Part (B) – Production increase from 35 to 41 per hour:

  • x=35x = 35, and dx=4135=6dx = 41 - 35 = 6
  • Approximate the change in cost dCdC using C(35)C'(35).

Let me calculate the values for these now.Here are the results:

  1. The derivative of the cost function is: C(x)=360x2+13C'(x) = -\frac{360}{x^2} + \frac{1}{3}

Part (A) – Production increase from 25 to 31 per hour:

  • C(25)0.243C'(25) \approx -0.243
  • Change in production: dx=6dx = 6
  • Approximate change in cost: dCC(25)6=1.456dC \approx C'(25) \cdot 6 = -1.456 Therefore, the average cost per racket will decrease by approximately $1.46.

Part (B) – Production increase from 35 to 41 per hour:

  • C(35)0.0395C'(35) \approx 0.0395
  • Change in production: dx=6dx = 6
  • Approximate change in cost: dCC(35)6=0.237dC \approx C'(35) \cdot 6 = 0.237 Therefore, the average cost per racket will increase by approximately $0.24.

Would you like more details or have any further questions?

Related questions:

  1. How do differentials provide approximations for changes in cost?
  2. How does the curvature of the cost function affect the results?
  3. What is the significance of the production level where the derivative is zero?
  4. How could one refine this approximation using integration instead?
  5. 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)