smallest volumn

Let the plane be x/a+y/b+z/c=1, then
(1) 1/a+2/b+3/c =1
(2) the indicated volume V=abc/6

f(a,b,c, u)=abc/6+u(1/a+2/b+3/c -1)
∂f/∂a = bc/6 - u/a² =0, u=a²bc/6
∂f/∂b =ac/6-2u/b²=0, u=ab²c/12
∂f/∂c = ab/6 -3u/c² =0, u=abc²/18
so, a/6=b/12=c/18, (a,b,c)=(t, 2t, 3t)
∂f/∂u = 1/a+2/b+3/c -1 =0, then
1/t+2/(2t)+3/(3t) =1, t= 3, (a, b, c)=(3, 6, 9), u=81, V=27

(method 2)
1/a+ 2/b+ 3/c =1
AP>=GP, then 1/27 >= 6/abc, so V=abc/6 >= 27 (minimum)

2012-05-04 14:29:40 補充：
(intercept form) If a plane passes (a,0,0), (0,b,0), (0,0,c) then the plane is
x/a+ y/b + z/c = 1

a(x-1) + b(y-2) + c(z-3) = 0
x-intercept = (a + 2b + 3c)/a
y-intercept = (a + 2b + 3c)/b
z-intercept = (a + 2b + 3c)/c

Vol = (a + 2b + 3c)^3/(6abc)

一個有 3 個 variable 的 vol. expression

2012-05-04 10:13:36 補充：
求 x-intercept, 代 y = 0, z = 0, 得出:
a(x - 1) -2b - 3c = 0
ax = a + 2b + 3c
x = (a + 2b + 3c)/a
求 y- z-intercepts 用類似手法.

a(x-1) + b(y-2) + c(z-3) = 0

明顯經過 (1, 2, 3) 的平面。

你要做的是求取 a, b, c，使得此平面與第一象限所圍之體積最小。

a:對x偏微 b:對x偏微

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