## The Fencing Problem

Many optimisation problems give rise naturally to quadratics.
100m of fence is used to enclose part of a field in the shape of a rectangle. One side of the enclosure is formed by a hedge.

If the sides of the enclose are
$x$
and
$y$
then the area is
$A=xy$
and since there is 100m of fencing,
$100=2x+y$
.
The second equation can be arranged to give
$y=100-2x$
. In terms of
$x$
the area is
$A=x(100-2x)=100x-2x^2$
.
We can find the maximum of this expression by completing the square.
\begin{aligned} A &= 100x-2x^2 \\ &= -2(x^2-50x) \\ &= -2((x-25)^2-25^2) \\ &= -2 \times -(25)^2-2(x-25)^2 \\ &= 1250-2(x-25)^2 \end{aligned}

The maximum area is
$1250m^2$
, occurring when
$x=25m$
.