二條函數的問題

1.
唔明條題目講乜，係咪打錯字呀？

2.
設用來圍成正方形的鉄絲長為x m，用來圍成圓形的鉄絲長為(L – x) m

兩者面积之和 = s = (x/4)² + [(L – x)/(2π)]² π
　　　　　　　　= x²/16 + (L – x)²/(4π)
　　　　　　　　= x²/16 + L²/(4π) – Lx/(2π) + x²/(4π)

ds/dx = 0
ds/dx = x/8 + x/(2π) – L/(2π)

∴x/8 + x/(2π) – L/(2π) = 0
　　　　　　(π + 4)x = 4L
　　　　　　　　　x = 4L/(π + 4)

d²s/dx²│4L/(π + 4) = 1/8 + 1/(2π) > 0

所以，當我們將一條長為L m的鉄絲截成(4L/(π + 4)) m和(πL/(π + 4)) m兩段，並將那段長度為(4L/(π + 4)) m的鉄絲圍成正方形，那段長度為(4L/(π + 4)) m的鉄絲圍成圓形的時候，才使兩者面积之和s最少。

first of all, we let the length of the wire forming a square is x
so, the length of the wire forming a circle is y = (L - x)

total area = s = [(x / 4) ^ 2] * {[y / (2 * pi)] ^ 2} * pi

s = (x ^ 2) * (y ^ 2) / (64 * pi)
s = [(x ^ 4) - 2 * L * (x ^ 3) + (L ^ 2) * (x ^ 2)] / (64 * pi)
ds/dx = [2 * (x ^ 3) - 3 * L * (x ^ 2) + (L ^ 2) * x] / (32 * pi)
= x * [x - (L / 2)] * (x - L) / (32 * pi)
= 0
so, x = 0, 0.5L or L

However,
d2s/dx2 = [6 * (x ^ 2) - 6 * L * x + (L ^ 2)] / (32 * pi)
= {6 * x - [3 - sqrt(3)] * L} * {6 * x - [3 + sqrt(3)] * L} / (32 * pi) > 0
= [(x - 0.211 * L) * (x - 0.789 * L)] / (32 * pi) > 0
so, x < 0.211 * L or x > 0.789 * L

Hence, x can be only either 0 or L