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: 1,764 Thanks: 621 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: 2,299 Thanks: 1218  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: 2,299 Thanks: 1218  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.  

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