用一條線圍出一個面積

First the proof. I will approach that with calculus and polar coordinates.

Imagine a general closed-path r(θ), enclosing an area about the origin.
r being the radius at angle θ.

Rewording the question as finding the shape (i.e. type of function r(θ)) which has the smallest perimeter for the same area. For a small slice of this shape, this will mean finding the smallest arc length for the same sector area.

Consider the elemental area spread out by an elemental angle of dθ.

The elemental sector area is given by
dA = r(θ)^2 / 2 * dθ
(ref. http://en.wikipedia.org/wiki/Polar_coordinate_system)

The elemental arc length swept by this angle is
dS = 1 / square_root( r(θ)^2 + (d r(θ) / dθ)^2 ) * dθ
(ref. http://en.wikipedia.org/wiki/Arc_length)

For the same area dA, r(θ) is fixed. To minimize the arc length dS, the only variable left is dr(θ) / dt. Since it is squared, the minimum dS can only be obtained when dr(θ)/dt is zero.

If dr(θ)/dt = 0, then
r(θ) = C (i.e. an arbitary constant, not a function of angle θ).

This gives you the polar form of a circle!

Now the easy arithmetics:

Circumference of a circle = 2*pi*r = 100cm
r = 100 / (2*p) = 15.92cm

Area of a circle = pi * r^2 = pi*15.92^2 = 795.78cm^2

試想：
若圍成三角形，面積最大 = 1/2(100/3)^2 x sin60 = 481.125.......
若圍成四邊形，面積最大 = (100/4)^2 = 625
若圍成六邊形，面積最大 = 6x[1/2x(100/6)^2 x sin60] = 721.6878.....
可見圍的圖形邊數愈多，面積愈大。
亦因圖形邊數愈多愈接近圓形，所以面積最大應是圓形，面積為：
[100/(2pi)]^2 x pi = 795.7747....

用一條線可以做的圖形只有圓形，圖形的計算方法如下

半徑×半徑×圓周率=圓面積
即:R×R×π=圓面積

圓的半徑:
100/3.14
=31.847133757961783439490445859873
=31.85(取二位)
圓的面積:
31.85*31.85*3.14
=3185.28665(cm2)

我覺得系圓形!
圓:795.8cm^2
正:625cm^2
三:481.1^2

應該是圓形
面積為(100/2pi)2(pi)=796cm2

圓形,
設x是半徑,
2x3.14=100
x=16
面積是16x16x3.14=803cm2