1
   

Application of derivative

 
 
Reply Thu 30 Jun, 2005 07:33 pm
Suppose the average costs of a mining operation depend on the number of machinges used, and average cost given by

(there is a line over the C) C(x) = 2900x+1,278,900/x, x>0

Where x, is the number of machines used.
a. Find the critical values for (there is a line over the C) C(x) that lie in the domain of the problem.
b. Over what interval in the domain do average cost decrease?
c. Over what interval in the domain do average cost increase?
d. How many machines give minimum average cost?
e. What are the minimum average cost?
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Discussion • Score: 1 • Views: 575 • Replies: 2
No top replies

 
raprap
 
  1  
Reply Thu 30 Jun, 2005 08:29 pm
C=ax+b*x^(-1/2)
dC/dx=a-1/2b*x^(-3/2)=0
x^(3/2)=b/(2a)
x^3=[b/(2a)]^2
x={[b/(2a)]^2}^(1/3)
if a=2900
b=1278900
x=34.548
check if minimum
d^2C/dx^2=(-1/2)(-3/2)b*x^(-5/2)
if x is positive d^2C/dx^2 is positive
so this is a minimum on C
but x=34.548
& you can't have a fraction of a machine so compare x=34 and 35 to see which is least using
C=ax+b*x^(-1/2)

Rap
0 Replies
 
greatwhiteshark
 
  1  
Reply Fri 1 Jul, 2005 02:52 pm
Rap
Thank you for your help and service.

Janet
0 Replies
 
 

Related Topics

Evolution 101 - Discussion by gungasnake
Typing Equations on a PC - Discussion by Brandon9000
The Future of Artificial Intelligence - Discussion by Brandon9000
The well known Mind vs Brain. - Discussion by crayon851
Scientists Offer Proof of 'Dark Matter' - Discussion by oralloy
Blue Saturn - Discussion by oralloy
Bald Eagle-DDT Myth Still Flying High - Discussion by gungasnake
DDT: A Weapon of Mass Survival - Discussion by gungasnake
 
  1. Forums
  2. » Application of derivative
Copyright © 2025 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.04 seconds on 01/22/2025 at 12:55:41