kevin 寫到:Meowth你的解法跟算幾不等式有關係嗎?
L.H.S?
無關算幾不等式吧
[數學]證明
由 Raceleader 於 星期二 六月 24, 2003 11:47 pm
all values are not fixed, will the orientation of the rectangles change, then the total area will be larger?
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Raceleader - 訪客
由 --- 於 星期三 六月 25, 2003 2:57 pm
Meowth 寫到:三元對稱不等式, may try 圓柱座標變換
A=k-a
B=k-b
C=k-c
let x=a/k, y=b/k, z=c/k
1>= x,y,z >=0
let
x=t+r*sqrt(2/3)*cos(m)
y=t+r*sqrt(2/3)*cos(m+120')
z=t+r*sqrt(2/3)*cos(m-120')
1>=t>=0
r=d(L: x=y=z, (x,y,z))
f=Ab+Bc+Ca
=kk[x(1-y)+y(1-z)+z(1-x)]
=kk[(x+y+z)-xy-yz-zx]
=kk[3t- 3tt + rr/2]
=kk[3(1/4-(t-1/2)^2) +rr/2]
To Maximize f:
we should maximize r, that is we should choose the ridges of the cube.
(1) when 1/3>=t>=0
r=s*t
As t increased, 3(1/4-(t-1/2)^2) & rr all increasing.
so, f is maximized at t=1/3, where 3 points (0,0,1),(0,1,0)(1,0,0) get maximized f values. (f=kk)
(2) when 1>=t>=2/3
r=s*(1-t)
As t increased, 3(1/4-(t-1/2)^2) & rr all decreasing.
so, f is maximized at t=2/3, where 3 points (0,1,1),(1,1,0)(1,0,1) get maximized f values. (f=kk)
(3) when 2/3>=t >=1/3
we should maximize r, that is we should choose the ridges of the cube.
that are (1,0,q),(0,1,q),(1,q,0),(0,q,1),(q,1,0),(q,0,1).
==> f=kk
from (1)(2)(3), we get max(f)=kk, at (x,y,z)= (1,0,q),(0,1,q),(1,q,0),(0,q,1),(q,1,0),(q,0,1), that is, (a,b,c)=(k,0,qk),(0,k,qk,(k,qk0),(0,qk,k),(qk,k,0),(qk,0,k); 1>=q>=0.
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We can also get the minimum of f:
minimize 3t(1-t) & rr/2
==> t=0 or 1, r=0
==> (x,y,z)=(0,0,0) or (1,1,1)
==> (a,c,b)=(0,0,0) or (k,k,k)
==> min(f)=0
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Any body know what I write?
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--- - 訪客
