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2L +W =180  (I )
 
A = WL    / Area that needed to be maximized
 
Substitute  W from in terms of L from (1)
 
     w = 180 -2L
 
   A = ( 180 - 2L) L
 
  A= f( L ) =  - 2L^2 + 180 L  
 
         f( L ) is maximized at : L = -180/ -4 = 45
 
          w = 180 - 2L = 180 -90 = 90
 
           A rectangle with dimensions of 45* 90 = 4050 yrdwill be the answer.    
 
     
 
 
 
 
 
We can assign  aa length and width such that 2x+y=180 and we try to maximize the area xy. Since we know y=180-2x, this becomes 180x-2x^2. To maximize the area we need todifferentiate it and make sure its roots are either max or min poin so we get the right ones. For our area to be max we willl need to examine between points. More on this later..
 
Differentiated..180-4x. When we set it to zero we get x=45. So the dimensions are 45 by 90 with a total area of 405.
 
Hope that helps,
Deanna

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