Finding Maximum Area With Gien Length of Fence

Optimisation problems typically involve answering questions such as 'find the shape that maximises the enclosed area with a given length of fence'.
Suppose we have a 200m of fence to close of part of a field which backs onto a hedge. The hedge and fence are to form a rectangle. This is shown in the diagram, with the green rectangle being the hedge.

The length of fencing is  
\[2x+y\]
  so  
\[2x+y=200\]
.
The enclosed  
\[xy\]
.
The pronlem is now 'maximise  
\[xy\]
subject to  
\[2x+y=200\]
'.
We can substitute  
\[2x+y=200\]
  into  
\[A(x,y)\]
  to obtain an equation in terms of  
\[x\]
, which we can then maximise by completing the square.
 
\[2x+y=200 \rightarrow y=200-2x \rightarrow A(x)=x(200-2x)=200x-2x^2\]
.
Now complete the square.
\[100-2x^2=-2(x^2-100x)=-2((x-50)^2-50^2)=2 \times 50^2-2(x-50)^2=5000\]
.
The maximum value of  
\[A\]
  is 5000, when  
\[x=50\]
/