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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

## #1 2012-10-29 01:43:12

engrymbiff
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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

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.

## #2 2012-10-29 01:52:16

bobbym

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### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

Hi engrymbiff;

How did you set it up?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #3 2012-10-30 04:51:48

engrymbiff
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### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

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.

## #4 2012-10-30 08:37:36

bobbym

Online

### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

I did not have much luck with it either, perhaps they want you to do it by an inequality.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #5 2012-11-04 07:24:50

engrymbiff
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### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

Help from someone?

engrymbiff
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## #7 2012-11-13 22:40:47

scientia
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### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

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.

Last edited by scientia (2012-11-13 22:51:01)

## #8 2012-11-14 07:31:34

bobbym

Online

### Re: maximize (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) s.t. -1 <= a, b, c, d <= 1

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.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
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