Question 1207440
<pre>
Rectangle A has length 2.5, width 4/2.5 = 1.6, perimeter 2(2.5)+2(1.6)=8.2
{{{drawing(400, 1440/7, -1,6,-1,2.6,

line(0,0,2.5,0), line(2.5,0,2.5,1.6), line(2.5,1.6,0,1.6), line(0,1.6,0,0) )}}}

Rectangle B has length 3.2, width 4/3.2 = 1.25, perimeter 2(3.2)+2(1.25)=8.9
{{{drawing(400, 1440/7, -1,6,-1,2.6,

line(0,0,3.2,0), line(3.2,0,3.2,1.25), line(3.2,1.25,0,1.25), line(0,1.25,0,0) )}}}

Rectangle C has length 4, width 4/4 = 1, perimeter 2(4)+2(1)=10
{{{drawing(400, 1440/7, -1,6,-1,2.6,

line(0,0,4,0), line(4,0,4,1), line(4,1,0,1), line(0,1,0,0) )}}}

Rectangle D has length 5, width 4/5 = 0.8, perimeter 2(5)+2(0.8)=11.6
{{{drawing(400, 1440/7, -1,6,-1,2.6,

line(0,0,5,0), line(5,0,5,0.8), line(5,0.8,0,0.8), line(0,0.8,0,0) )}}}

So for given area 4, we might conjecture that rectangles with smaller perimeters
have their lengths closer to their widths.  And therefore, rectangles with
larger perimeters have their lengths much greater than their widths.
The nearer the rectangles are to a square, the smaller their perimeter.

So we might make a more general conjecture that of all rectangles that have a
given area, the one with the smallest perimeter possible is the one when the
rectangle is a square.

Edwin</pre>