SOLUTION: A gardener has 200m of fencing to enclose two adjacent rectangular plots.what dimensions will produce a maximum enclosed area?

  

  

The Figure on the right shows the condition.      


In the Figure, L is the length and W is the width.

So, we have 3 pieces of fencing of the length L each and 4 pieces 
of fencing of the length W each.

Then we have this equation 

3L + 4W = 200,

from which we have W = .
 
  




        Figure. 

 
 
Next, the combined area of the two corals is 

A = L*2W =  =  = ,

and we have to find the length L in a way to maximize the area A, i.e. maximize the quadratic function

A = .


  

  
  Now let me remind you that, if you have a quadratic function f(x) =  of the general form,   
    then it reaches the maximum/minimum at x = .
  

      See the lessons
          - HOW TO complete the square to find the minimum/maximum of a quadratic function
          - Briefly on finding the minimum/maximum of a quadratic function
      in this site.


For our situation,  a =   and  b = 100.

Therefore, the maximum is at L = -  =  = 33 .

Thus the area get a maximum at L = 33  feet.

Then W =  =  = 25 feet.

Answer.  The area is maximal at L = 33  feet and 2W = 2*25 feet = 50 feet.