February 26th, 2017, 10:05 PM  #1 
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  Confusing word problem
A metal barrel is to be manufactured out of two different types of metal, one type for the top and bottom and one for the curved side. The metal for the curved side piece costs \$2.50 per square meter and the metal for the top and bottom costs \$4.50 per square meter. The top and bottom circles must be cut out of a square piece of metal whose side length is the diameter of the circle and the rest of the square is wasted (so contributes to the cost). If the volume is to be 9 cubic meters, find the dimensions of the barrel that minimizes the cost. Identify the interval you are minimized over and show your solution is the minimum. Last edited by skipjack; February 27th, 2017 at 12:24 AM. 
February 26th, 2017, 10:09 PM  #2  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,849 Thanks: 661 Math Focus: Yet to find out.  Quote:
Last edited by skipjack; February 27th, 2017 at 12:29 AM.  
February 26th, 2017, 11:59 PM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 2,649 Thanks: 1476  Quote:
$h = \dfrac{V}{\pi r^2}$ $area_{tb} = (2r)^2$ $area_{side} = 2\pi r h = 2\pi r \dfrac{V}{\pi r^2} = \dfrac{2 V}{r}$ $Cost = area_{tb}(4.50) + area_{side}(2.50)$ $Cost = (2r)^2(4.50) + \dfrac{2 V}{r}(2.50)$ Now, find the solution of $\left .\dfrac{dCost}{dr}\right_{r=r_{min}}= 0$ check that $r$ is the minimum by ensuring that $\left . \dfrac{d^2 Cost}{dr^2}\right_{r=r_{min}} > 0$ I get $r_{min}= \sqrt[3]{\dfrac{45}{36}}$ Last edited by skipjack; February 27th, 2017 at 12:30 AM.  
February 27th, 2017, 06:05 AM  #4  
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  Quote:
am I solving for r or v? also for the last part am I using a derivative test?  
February 27th, 2017, 06:34 AM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,649 Thanks: 1476  Quote:
$V=9$ I just like to wait as long as possible before plugging numbers in. Yes, this whole method is basically the derivative test for extreme points and then the second derivative test to ensure that the point(s) found are minima.  

Tags 
confusing, problem, word 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Confusing word problem, finding the probability of the mean  CreepyPastaMansion  Probability and Statistics  1  June 23rd, 2014 06:49 AM 
Confusing Word Problem  YoungMoustache  Elementary Math  1  September 20th, 2012 07:01 PM 
This Distance Word Problem Is Confusing  kensclark15  Algebra  9  September 17th, 2012 05:22 PM 
Confusing algebra word problem  matt93  Algebra  3  November 3rd, 2011 09:53 AM 
Another confusing word problem  cherryperry  Elementary Math  1  May 29th, 2010 02:29 PM 