## Finding Maximum Area With Gien 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 \[x=50\]

/