Answer:
The maximum area possible is 648 squared meters.
Step-by-step explanation:
Let the length of the existing wall be [tex]\ell[/tex].
And let the width of the fence be [tex]w[/tex].
The area of the enclosure will be given by:
[tex]A=w\ell[/tex]
Since the area is bounded by one existing wall, the perimeter (the 72 meters of fencing material) will only be:
[tex]72=2w+\ell[/tex]
We want to maximize the area.
From the perimeter, we can subtract 2w from both sides to obtain:
[tex]\ell=72-2w[/tex]
Substituting this for our area formula, we acquire:
[tex]A=w(72-2w)[/tex]
This is now a quadratic. Recall that the maximum value of a quadratic always occurs at its vertex.
We can distribute:
[tex]A=-2w^2+72w[/tex]
Find the vertex of the quadratic. Using the vertex formula, we acquire that:
[tex]\displaystyle w=-\frac{b}{2a}=-\frac{(72)}{2(-2)}=18[/tex]
So, the maximum area is:
[tex]A=-2(18)^2+72(18)=648\text{ meters}^2[/tex]
