certainly,but we have to know if x+y+z=n ,when x=y=z,

=n/3,we get x*y*z is the Max.
V=4x*y^2=4xb^2/a^2(a^2-x^2)=4b^2/a^2(a^2x-x^3)
only a^2x-x^3 useful
a^2x-x^3=x(a-x)(a+x)=[(1-c)x(a-x)(ca+cx)]/c(1-c)=[
(1-c)x+(a-x)+(ca+cx)=a+ca=n
so,x=(a+ca)/3(1-c),a-x=(a+c)/3,cancel c we get x=(squat3/3)a.
Thank your question.

所有跟帖： 

• This is a wonderful method. -皆兄弟也- ♂ (149 bytes) () 09/13/2010 postreply 15:42:19

• Thanks. Parameter Method. -jinjing- ♀ (233 bytes) () 09/13/2010 postreply 17:29:16

• Sorry,First term is not right. -jinjing- ♀ (224 bytes) () 09/13/2010 postreply 17:47:58

• 此壇真乃從龍臥虎之處也。佩服，佩服。 -NaCl- ♂ (0 bytes) () 09/13/2010 postreply 21:09:47

• Thanks,I'm not strong for some open questions. -jinjing- ♀ (59 bytes) () 09/14/2010 postreply 10:16:19

• can you transit（a1+b1x）（a2+b2x）...（an+bnx）step by step? -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 07:59:03

• Sorry,I could say clearly and have some carelessness. -jinjing- ♀ (774 bytes) () 09/14/2010 postreply 10:00:53

• I really appreciate your effort. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:13:56

• For n,it seems (n-1)th equition should be delt with. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:22:32

• really wonderful. there are no (n-1)th equition . -皆兄弟也- ♂ (259 bytes) () 09/14/2010 postreply 10:43:52

• I got this transition. but,can you show what n equations are? -皆兄弟也- ♂ (333 bytes) () 09/14/2010 postreply 08:41:16

• Thank you,you correct me the +,-.some where.Up is answer. -jinjing- ♀ (0 bytes) () 09/14/2010 postreply 10:09:35

• I got n-1 independent equations.why do you use b2+c3b3+...+cnbn? -皆兄弟也- ♂ (213 bytes) () 09/14/2010 postreply 10:11:26

• n=2,your first Q is wrong. In fact t, you can choice any (n-1) -jinjing- ♀ (113 bytes) () 09/14/2010 postreply 10:41:32

• I have realized that it is wrong. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:53:35

• The sum of all factors should be constant(no x). -jinjing- ♀ (150 bytes) () 09/15/2010 postreply 10:20:17