F.6數學微分法極大極小問題

一條長60cm的鋼線被截成兩段.一段折成一正方形,另一段則繞成一圓形.設該正方形的邊長為x cm.
Sol
圓形周長=60-4x
圓形半徑=(60-4x)/(2pi)=9.5493046-0.6366203x
總面積=x^2+3.14159*(9.5493046-0.6366203x)^2
=x^2+286.47913-38.197216x+1.2732405x^2
=2.2732405x^2-38.197216x+286.47913
f(x)= 2.2732405x^2-38.197216x+286.47913
=2.2732405(x^2-16.80298x)+286.47913
=2.2732405(x^2-16.80298x+8.40149^2)+286.47913-2.2732405*70.585034
=2.2732405(x-8.40149)^2+126.02238
0<=4x=60
0<=x<=15
-8.40149<x-8.40149<=6.59851
0<=(x-8.40149)^2<=70.585034
0<=2.2732405(x-8.40149)^2<=160.45675
126.02238<=2.2732405(x-8.40149)^2+126.02238<=286.9135
126.02238<=f(x)<=286.9135
(a)問最小的總面積是多少?
min=126.02238
(b)求使總面積為最大的x值
-8.40149<x-8.40149<=6.59851
0<=(x-8.40149)^2<=70.585034
當f(x)=286.9135 x=0

設總面積為 A cm2。
正方形邊長為 x cm，所以 x >= 0。
繞成一圓形一段長為 (60 - 4x) cm，所以 60 - 4x >= 0，即 x <= 15。
即 0 <= x <= 15。
總面積 = 正方形面積 + 圓形面積
A = x2 + [(60 - 4x)/2π]2 π
A = x2 + (30 - 2x)2 /π
dA/dx = 2x + 2(30 - 2x)*(-2) /π
dA/dx = 2x + 8(x - 15) /π
當 dA/dx = 0，
2x + 8(x - 15) /π = 0
x + 4(x - 15) /π = 0
(π + 4)x - 60 = 0
x = 60/(π + 4)
當 x = 0，A = 900 /π (~~286)
當 x = 60/(π + 4)，A = 900 /(π + 4) (~~126)
當 x = 15，A = 225


(a)最小的總面積是 900 /(π + 4) cm2。

(b)使總面積為最大的 x 值為 0。

Find the extreme values of the area (denoted by A) of sum of the circle and square:
A = pi*r^2 + x^2
2pi^r + 4x = 60
=> x = 15 - 1/2*pi*r
A(r) = pi*r^2+(15-1/2*pi*r)^2, 0 <= r <= 30/pi
Note that the endpoints of the domain r implie to the instances when the string is not cut,
i.e. r = 0 when no circle is formed, r= 30/pi when no square is formed.
A'(r) = 2pi*r - pi(15-1/2*pi*r)
A'(r) = 0 when r = 30/(4+pi)
Since A''(r) = 2pi + pi^2/2 > 0 on (0, 30/pi), the absolute minimum of A occurs when r = 30/(4+pi)
and the absolute max of A occurs at one of the endpoints, so check: A(0) =225 , A(30/pi)= 900/pi > 225

a) 代 r= (30/4+pi) 入圓的面積
A = 2pi*r = 60pi/(4+pi) = 26.39
所以最小的總面積是大約26.39

b) 因為用所有的鋼線繞成圓形就會得到最大面積, 所以x=0.

b part有冇model ans?