The Teachers Hair Ribbon

Let 'L' be the length of the rectangular paddock, and 'w' be the width. the perimeter is the constraint, and the area is to be maximised. Firstly, we have to find the function of area in terms of L.

p=2L+2w

A=Lw

A=L( $\frac{p}{2}$ -L)

taking the derivative of the area function, we get:

$\frac{p}{2}$ -2L

therefore at the maximum area,

$\frac{p}{2}$ -2L =0

$\frac{p}{2}$ =2L

p=4L

Substituting the value of p into w, we get

w=2L-L

therefore;

w=L

the rectangle MUST be a square to maximize its area

• Firstly, notice that if the perimeter is p then the side length for the square is p/4.* Also, the ratio is perimeter to area which affects this.

Suppose the perimeter is length one, it is possible to get the unit rate with all these ratios. In this case, a square would have area 1/16(the side length is a quarter of the perimeter and is squared), and an equilateral triangle would have side length 1/3 and therefore has area sqrt(3)/36-smaller then 1/3. The regular Hexagon would be six equilateral triangles(but with side length 1/6) and therefore each of the six equilateral triangles would have area sqrt(3)/144, except it would have six triangles, making the total area sqrt(3)/24(easily smaller than 1/3). *3D shapes have no perimeter and were tricks and irregular polygons are not close enough to a circle, but closer to an oval(less area)