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  •  » maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

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Topic review (newest first)

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.

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