# A farmer wants to enclose a 450,000m^2 rectangular field with fence. She then wants to sub-divide this field into three smaller fields by placing additional lengths of fence parallel to one of the side. How can she do this so that she minimizes the cost?

## My equations are 450,000=xy and P=6x+4y, where x is the width and y is the length. However my teachers equation was different for the perimeter, his was P=4x+2y

Dec 30, 2016

The minimum perimeter is given when $x \approx 474.3$ and $y \approx 948.7$, which gives a corresponding perimeter of $P \approx 3794.7$

#### Explanation:

Without loss of generality let us assume that the field is divided as shown:

The area enclosed is given as $450 , 000$ and so:

$x y = 450000 \implies y = \frac{450000}{x}$

The total perimeter of all fencing is given by:
$\setminus \setminus \setminus \setminus \setminus P = 4 \times \left(\text{vertical") + 2 xx ("horizontal}\right)$
$\therefore P = 4 x + 2 y$

As indicated wlog we could interchange $x$ and $y$. We can now eliminate one variable (in this case $y$) as we have two equation, so we get:

$P = 4 x + 2 \cdot \frac{450000}{x}$
$\setminus \setminus \setminus = 4 x + \frac{900000}{x}$

Differentiating wrt $x$ we get:

$\frac{\mathrm{dP}}{\mathrm{dx}} = 4 - \frac{900000}{x} ^ 2$

At a critical point we have $\frac{\mathrm{dP}}{\mathrm{dx}} = 0$:

$\implies 4 - \frac{900000}{x} ^ 2 = 0$
$\therefore \frac{900000}{x} ^ 2 = 4$
$\therefore {x}^{2} = \frac{900000}{4}$
$\therefore {x}^{2} = 225000$
$\therefore x = 150 \sqrt{10} \approx 474.3 \implies y \approx 948.7$

This gives us $P \approx 3794.7$

We can visually verify that this corresponds to a minimum by looking at the graph:
graph{4x + 900000/x [-10, 1000, -1000, 7830]}