The Fencing Problem.

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The Fencing Problem

A farmer has brought 1000 metres of fencing. With this fencing he wants to enclose an area of land. The farmer wants the fencing to enclose an area of the largest size. I will investigate different shapes the fencing can make to achieve the largest area.

Firstly I am going to investigate:

Squares

I am investigate the use of a square with a maximum area and a 1000m perimeter.

The general formula to work out the area for this square is:

AREA= xy

NOT TO SCALE

As the square has four equal sides there can only be one length of each side and one overall area. The length of this side must be:

000m ÷ 4 = 250m

NOT TO SCALE

Therefore the maximum area of the square is 62500m². As I know this, I do not need to display my results in a table or graph as there is only one possible result of a 250m length.

RECTANGLES

To begin with I am going to investigate numerous rectangles that all have a perimeter of 1000 meters, to find out if there is any relevant pattern in my results.

Firstly, the basic formula for quadrilateral shapes such as the rectangle and square is

AREA = LENGTH x WIDTH

In this case it is 498m x 2m = 994m²

NOT TO SCALE

The generalized formula to find out the area of this rectangle is:

2x+2y =1000

x +y=500

Area =xy

=x (500-x)

=500x -x²

NOT TO SCALE

Below is my table of figures for rectangles length, width and area:

Length (m)

x Width

Area (m2)

0

500

0

0

490

4900

20

480

9600

30

470

4100

40

460

8400

50

450

22500

00

400

40000

50

350

52500

200

300

60000
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250

250

62500

300

200

60000

350

50

52500

400

00

40000

450

50

22500

500

0

0

Looking at my results it is easy to see that's there an increase, a peak and a decrease in the measurements. Therefore I would expect my graph to show a parabola.

From my table and graph, it is easy to see that the maximum possible base for a rectangle is that with a base of 250m . This shape is ...

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