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  
  then the area is  
  and since there is 100m of fencing,  
The second equation can be arranged to give  
. In terms of  
  the area is  
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  
, occurring when  

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