## 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
$x=50$
/