Challenging optimization

正如maximal_ideal_space所講，可以用AM>=GM黎做如下：

Let the length, width, height be x, y, z respectively.

First we know that
A = 2[xy+yz+zx] and V = xyz
we want to maximise V when A is fixed.

Thus applying the AM>=GM, we get

A / 2
= xy+yz+zx
>= 3 [xy*yz*zx]^{1/3}
= 3 [xyz]^{2/3}
= 3 V^{2/3}

Thus we have
V <= (A / 6)^{3/2}

Equality holds iff xy=yz=zx iff x=y=z.

正如maximal_ideal_space所講，可以用AM>=GM黎做如下：

Let the length, width, height be x, y, z respectively.

First we know that
A = 2[xy+yz+zx] and V = xyz
we want to maximise V when A is fixed.

Thus applying the AM>=GM, we get

A / 2
= xy+yz+zx
>= 3 [xy*yz*zx]^{1/3}
= 3 [xyz]^{2/3}
= 3 V^{2/3}

Thus we have
V <= (A / 6)^{3/2}

Equality holds iff xy=yz=zx iff x=y=z.

Remark. When you are doing such problems with fixed sum or fixed product, and want to maximise the product or minimise the sum, then you may simply use AM>=GM to solve the optimization problems.


2009-01-31 17:32:10 補充：
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請大家唔好投#003！！！

其實簡單的用AM-GM inequality也可以。

Let use Lagrange Multiplier Method to solve this problem
Let length x, width y, height z
Then total surface area = 2(xy+yz+zx)=A
V=xyz
So we want to maximize xyz subject to 2(xy+yz+zx)-A=0
Using Lagrange Multiplier Method
Consider f(x,y,z)=xyz-λ[2(xy+yz+zx)-A]
∂f/∂x=yz-λ(2y+2z)
∂f/∂y=xz-λ(2x+2z)
∂f/∂z=xy-λ(2x+2y)
∂f/∂λ= 2(xy+yz+zx)-A
Set them all equal to zero
We have
yz/(2y+2z)=xz/(2x+2z)=xy/(2x+2y)
yz/(y+z)=xz/(x+z)=xy/(x+y)
It can be eqaily shown that x=y=z
So 2(xy+yz+zx)-A=0 =>6x^2=A
x=SQRT(A/6)
V=xyz=x^3=(A/6)^(3/2)


2009-01-16 18:34:30 補充：
正如maximal_ideal_space大大所知﹐這裡面有好高深的數理哲學呢。

正如周界相同﹐圓的面積最大。

http://hk.knowledge.yahoo.com/question/question?qid=7009011202123

您應該展示下你的實力嘛

2009-01-16 23:32:34 補充：
經貓朋提點﹐我知道答案了