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- bobbym
- 2012-11-14 07:31:34
Hi;
I am getting the same results through numerical methods, so it looks like your method is fine. Very good!
There is a missing comma in your answer that might cause some confusion.
- scientia
- 2012-11-13 22:40:47
I don't know if this method is acceptable but I suppose we could argue like this.
First, we take so all the factors in the product will be positive. And as we want the product to be as big as possible, we take to be as big as possible and to be as small as possible; thus we have
As the expression is now antisymmetrical in and , we can let (so is as large as possible); thus we have
Now you can maximize using normal calculus methods.
- engrymbiff
- 2012-11-13 19:57:27
I'll give it one more bump. Please help me out here anyone
- engrymbiff
- 2012-11-04 07:24:50
Help from someone?
- bobbym
- 2012-10-30 08:37:36
I did not have much luck with it either, perhaps they want you to do it by an inequality.
- engrymbiff
- 2012-10-30 04:51:48
As
L(a,b,c,d,u1,u2,u3,u4,u5,u6,u7,u8) = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) - u1(1-a) - u2(1-b) - u3(1-c) - u4(1-d) - u5(1+a) - u6(1+b) - u7(1+c) - u8(1+d)
and then try to identify which a,b,c,d,u1,u2,u3,u4,u5,u6,u7,u8 that satifies dL/da = dL/db = dL/dc = dL/dd = dL/du1 = ... = 0.
I guess that I've made a misstake when I set up the inequality constraint in L as
dL/du1 = a-1 = 0 => a = 1
dL/du5 = -a-1 = 0 => a = -1
so dL/du1 and dL/du5 cannot be equal to zero at the same time.
- bobbym
- 2012-10-29 01:52:16
Hi engrymbiff;
How did you set it up?
- engrymbiff
- 2012-10-29 01:43:12
Hi,
Could anybody help me with this problem?
maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1
I've tried using Lagrange multipliers but without any luck.