Question

An enclosed area is designed using one existing wall, and 72m of fence material. What is the maximum area of the enclosure?

Answer 1

Answer:

The maximum area possible is 648 squared meters.

Step-by-step explanation:

Let the length of the existing wall be $\ell$.

And let the width of the fence be $w$.

The area of the enclosure will be given by:

$A=w\ell$

Since the area is bounded by one existing wall, the perimeter (the 72 meters of fencing material) will only be:

$72=2w+\ell$

We want to maximize the area.

From the perimeter, we can subtract 2w from both sides to obtain:

$\ell=72-2w$

Substituting this for our area formula, we acquire:

$A=w(72-2w)$

This is now a quadratic. Recall that the maximum value of a quadratic always occurs at its vertex.

We can distribute:

$A=-2w^2+72w$

Find the vertex of the quadratic. Using the vertex formula, we acquire that:

$\displaystyle w=-\frac{b}{2a}=-\frac{(72)}{2(-2)}=18$

So, the maximum area is:

$A=-2(18)^2+72(18)=648\text{ meters}^2$