My Math Forum Find the dimensions of the rectangular corral split into two pens ...................

 Algebra Pre-Algebra and Basic Algebra Math Forum

December 7th, 2017, 02:12 PM   #1
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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.
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Last edited by GIjoefan1976; December 7th, 2017 at 02:13 PM. Reason: words

 December 7th, 2017, 02:35 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,495 Thanks: 1370 $630 = 4L + 3W$ $L = \dfrac{630-3W}{4}$ $A = 2LW = 2 \dfrac{630-3W}{4}W$ $A = \dfrac 1 2(630W - 3W^2) = -\dfrac 3 2 (W^2-210W)$ $A = -\dfrac 3 2 \left((W-105)^2 -(105)^2\right)$ $A = -\dfrac 3 2 (W-105)^2 + \dfrac 3 2 (105)^2$ The left term is negative so making it zero will maximize the area. This clearly occurs at $W=105$ $L = \dfrac{630-315}{4} = \dfrac{315}{4} =78.75$ thanks to Skipjack Last edited by romsek; December 7th, 2017 at 02:49 PM.
 December 7th, 2017, 02:43 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,810 Thanks: 2151 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.) Thanks from romsek

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