Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 30

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.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi engrymbiff;

How did you set it up?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 30

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.

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 30

Help from someone?

Offline

**engrymbiff****Member**- Registered: 2010-06-14
- Posts: 30

I'll give it one more bump. Please help me out here anyone

Offline

**scientia****Member**- Registered: 2009-11-13
- Posts: 224

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 haveNow you can maximize

using normal calculus methods.

*Last edited by scientia (2012-11-12 23:51:01)*

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline