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.

the fencing problem

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.
\[\begin{equation} \begin{aligned} A &= 100x-2x^2 \\ &= -2(x^2-50x) \\ &= -2((x-25)^2-25^2) \\ &= -2 \times -(25)^2-2(x-25)^2 \\ &= 1250-2(x-25)^2 \end{aligned} \end{equation}\]

The maximum area is  
\[1250m^2\]
, occurring when  
\[x=25m\]
.

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