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-   -   Find the dimensions of the rectangular corral split into two pens ................... (http://mymathforum.com/algebra/343028-find-dimensions-rectangular-corral-split-into-two-pens.html)

GIjoefan1976 December 7th, 2017 03:12 PM

Find the dimensions of the rectangular corral split into two pens ...................
 
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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.

romsek December 7th, 2017 03:35 PM

$630 = 4L + 3W$

$L = \dfrac{630-3W}{4}$

$A = 2LW = 2 \dfrac{630-3W}{4}W$

$A = \dfrac 1 2(630W - 3W^2) = -\dfrac 3 2 (W^2-210W)$

$A = -\dfrac 3 2 \left((W-105)^2 -(105)^2\right)$

$A = -\dfrac 3 2 (W-105)^2 + \dfrac 3 2 (105)^2$

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

skipjack December 7th, 2017 03:43 PM

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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