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December 7th, 2017, 03:12 PM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 371 Thanks: 10  Find the dimensions of the rectangular corral split into two pens ...................
Hello sorry if I already posted this one? I did check threads I have started but did not see it.... Find the dimensions of the rectangular corral split into two pens (sharing a fence) of the same size producing the greatest enclosed area given 630 feet of fencing. I know it's not really much that I have done or remember but I think this was my first step? Which was also confusing for my tutors and teachers to at first figure out until I was able to tell them that you have 3 widths now, and four lengths. (Even the word "pens" was confusing at first for the language barrier ) thanks for your help I will do my best to grasp what your're able to show me. Last edited by GIjoefan1976; December 7th, 2017 at 03:13 PM. Reason: words 
December 7th, 2017, 03:35 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,652 Thanks: 838 
$630 = 4L + 3W$ $L = \dfrac{6303W}{4}$ $A = 2LW = 2 \dfrac{6303W}{4}W$ $A = \dfrac 1 2(630W  3W^2) = \dfrac 3 2 (W^2210W)$ $A = \dfrac 3 2 \left((W105)^2 (105)^2\right)$ $A = \dfrac 3 2 (W105)^2 + \dfrac 3 2 (105)^2$ The left term is negative so making it zero will maximize the area. This clearly occurs at $W=105$ $L = \dfrac{630315}{4} = \dfrac{315}{4} =78.75$ thanks to Skipjack Last edited by romsek; December 7th, 2017 at 03:49 PM. 
December 7th, 2017, 03:43 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,241 Thanks: 1438 
You are given that 4L + 3W = 630 ft. To maximize 2LW, you need 4L = 3W, so W = 105 ft and L = (315/4) ft. (Note: romsek calculated (630  420)/4, but should have calculated (630  315)/4.) 

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corral, dimensions, find, pens, rectangular, split 
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