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Help ASAP

Panamonium8730  Feb 24, 2017
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3x meters of fencing is used for the sides and the middle fence

 

So.....120 - 3x is left for the top and bottom. So....the length can be expressed as

 

[ 120 - 3x] / 2

 

So....the total area  A  =

 

W * L  =     x ( [120 - 3x]/2 )  =  (1/2)x [120 - 3x]  =  60x - (3/2)x^2 =

 

-(3/2)x^2  + 60x

 

And the width, x, that maximizes the area is forund as    60 / [-2(-3/2)]  =

 

  60 / [6/2]  =  60 / 3    = 20 m

 

And the length is    20[120 - 3(20) ] / 2   =  10 [ 120 - 60] = 10 * 60  =

 

600m^2

 

Here's a graph that shows the max area at (20, 600)

 

https://www.desmos.com/calculator/gzzu0hnyz1

 

 

 

cool cool cool

CPhill  Feb 24, 2017

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