# Functions 11 Word Problem?

## A rectangular part of a parking lot is to be fenced off to allow some repairs to be done. The workers have fourteen 3-m sections of pre-assembled fencing to use. They want to create the greatest possible area in which to work. How can the fencing be used to create as large an enclosed area as possible?

Mar 1, 2017

We have a maximum plot size of $7 \times 7$ panels , leading to a maximum area of $441$ (m)

#### Explanation:

We aim to fence off a corner plot of land, and we can only use complete panels.

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{No of panels used along width of plot" \\ y & "No of panels used along length of plot" \\ A & "Total Area enclosed by the panels (sq m)}\end{matrix}\right.$

Our aim is to find $A$, as a function of a single variable, and to maximize the total area, A, wrt that variable. ie we want a critical point of $A$.

Now, the total number of panels is set (constant) and so:

$x + y = 14$

And the total Area enclosed by the panels is given by:

$A = \left(3 x\right) \left(3 y\right)$
$\setminus \setminus \setminus = 9 x y$

And substitution of $y$ into the second result gives us:

$A = 9 x \left(14 - x\right)$
$\setminus \setminus \setminus = 126 x - 9 {x}^{2}$

We now have the Area, $A$, as a function of a single variable, so Differentiating wrt $x$ we get:

$\frac{\mathrm{dA}}{\mathrm{dx}} = 126 - 18 x$

At a critical point we have $\frac{\mathrm{dA}}{\mathrm{dx}} = 0 \implies$

$126 - 18 x = 0$
$\therefore x = 7$

We should check this corresponds to a maximum:

$\frac{{d}^{2} A}{{\mathrm{dx}}^{2}} = - 18 < 0$

So fortunately we have an integer result, and a maximum area. Substituting $x = 7$ into our earlier equation;

$7 + y = 14$
$y = 7$

Thus we have a maximum plot size when $x = y = 7$, leading to a maximum area of:

$A = 9 \left(7\right) \left(7\right)$
$\setminus \setminus \setminus = 441$ (m)