I shall draw a graph and find the value of w for which A is greatest.
The graph below shows me that a rectangle with a perimeter of 1000 m has a maximum area when it becomes a square, having sides of 250 m. The area is then 62 500 m2. I am now going to do the same with a triangle. I will find out the greatest area for a triangle.
Area of Triangle = ½ Base * Height
a= ½ 200 * 300 = 30 000 m
This is much less than the area of a square. Perhaps if I make the base and the height equal, the result will improve.
Pythagoras gives y = x + x.
y = 2x = 2x
P = x + x + 2x
2x + 2x = 1000
x ( 2 + 2 ) = 1000
x = 1000 / 2 + 2 ~ 292.9 M
A = 1/2 * 292.9 * 292.9
= 42, 893.2 m
Would an equilateral triangle be better than a right angled triangle, since it has 3 equal sides?
Area = ½ a b sin C
= ½ 333.33333333*333.33333333* sin 60
= 48,112.521 m 2, which is the best result I have found for a triangle
From the results for triangles and rectangles, I predict that a regular pentagon will give better results.
tan 36 = 100/h.
h= 100 / tan 36
= 137.63819 m
Area of pentagon
= 5 * ½ * 200 * 137.63819 m
= 68,819.095 m2.
This is easily the best area so far. So I will try a hexagon which , I predict will, give a greater area.
Area of regular hexagon.
= 6 * ½ * 166.666 * 166.666 sin 60
= 72, 168.784 m 2.
This is a big improvement from the area of a pentagon
I will now try to find out if the decagon has a greater area than a hexagon.
I predict that the decagon with perimeter of 1000 m will have a greater area than a hexagon’s. 50 50
Tan 18 = 50/h
H = 50/ tan 18
= 153.88418
AREA OF DECAGON =
10* ½ *100 * 153.88418
= 76942.1 m
My prediction is confirmed.
Next, I will try to make a formula for a polygon with n sides, since the method is exactly the same as used above for the decagon.
Reg Polygon of “n” sides
1000/h = length of side
180 / ntan (180/n) = 500/n/h
h = 500 /n / tan(180/n)
Area of reg. polygon with n sides
= n * 500 / n * 500 / n / tan (180 / n ) = 250,000/ n TAN (180 / n )
I shall use a table to test my formula for various values of n.
These three results confirm that my general formula is true, because the values agree with the previous areas.
A circle is very close to a regular polygon having a very large number of sides. What is the area of a circle with perimeter of 1000 m ? I will calculate this value.
P = 1000 = 2πr
R = 1000/2π = 500/π
= 159.1549431 m
Area = πr2.
Π(159.1549431)2
= 79577.47155 m2
Conclusion
For a given length of fencing (such as 1000 m ) the greatest possibleb area which the farmer could enclose would be given by a circle pf radius 159.1549431 m,; the biggest shape it could be is a circle.
