Little help with min/max problem w/ n-variables

  • Thread starter Thread starter TheMadCapBeta
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on optimizing an n-dimensional linear function defined as f(x1,x2,...,xn) = a1*x1 + a2*x2 + ... + an*xn under the constraint x1^2 + x2^2 + ... + xn^2 = 1 using Lagrange multipliers. The critical point is derived, leading to the conclusion that L = sqrt(a1^2 + a2^2 + ... + an^2)/2. The maximum value of the function approaches infinity as n increases, while the minimum value occurs at k when n = 1, where k is a constant representing the values of a1, a2, ..., an.

PREREQUISITES
  • Understanding of Lagrange multipliers
  • Familiarity with n-dimensional functions
  • Knowledge of optimization techniques
  • Basic algebra and calculus skills
NEXT STEPS
  • Study the application of Lagrange multipliers in constrained optimization problems
  • Explore the properties of n-dimensional linear functions
  • Learn about the implications of constraints in optimization
  • Investigate the behavior of functions as dimensions approach infinity
USEFUL FOR

Mathematicians, optimization specialists, and students studying advanced calculus or linear algebra who are interested in the application of Lagrange multipliers and n-dimensional function optimization.

TheMadCapBeta
I'm given an n-dimensional function such that

f(x1,x2,...,xn) = a1*x1 + a2*x2 +...+ an*xn

where a1, a2, ...,an are all positive numbers. This is with the restraint that

x1^2 + x2^2 + ... + xn^2 = 1

Using Lagrange multipliers (I'll use 'L' for Lambda):

a1 = L(2*x1)
a2 = L(2*x2)
...
an = L(2*xn)

Solving for L,

L = a1/(2*x1) = a2/(2*x2) = ... = an(2*xn) [1]

So, generally, xn = an/(2*L). Plugging into the constraint function, I get:

(a1/2*L)^2 + (a2/2*L)^2 + ... + (an/2*L)^2 = 1
(1/4*l^2)*(a1^2 + a2^2 + ... + an^2) = 1
L = sqrt(a1^2 + a2^2 + ... + an^2)/2

So, for each x, generally,

xn = an/sqrt(a1^2 + a2^2 + ... an^2) [2]

Which implies that a1 = a2 = ... = an from [1]

a1/(2*L) = a2/(2*L) = ... = an/(2*L)

So [2] is basically my critical point.

Plugging back into f(...) I get

f(x1,x2,...,xn) = a1*(a1/sqrt(a1^2 + a2^2 + ... + an^2)) ...etc

<=> ... (a1^2 + a2^2 + ... + an^2)/sqrt(a1^2 + a2^2 + ... + an^2)

Furthermore...

<=> ... sqrt(a1^2 + a2^2 + ... + an^2)

Since a1 = a2 = ... = an, then we can say (using an arbitrary (a))

<=> ... sqrt(n*a^2) <=> a*(sqrt(n)).

So, to my question (heh), for the minimum value it's obvious that it would be when n = 0. But I'm a little confused as to what the maximum would be.
Say a constant k = a, then would it be k*(sqrt(n))?? Just a simple question I guess, as it comes down to it. Thanks.

Actually, wouldn't the min be when n = 1, since there mustb e at least one variable. So the min would be k.
 
Last edited by a moderator:
Physics news on Phys.org
I'm confused as to what your question really is!

You start out by saying "I'm given an n-dimensional function" but then you start saying that " for the minimum value it's obvious that it would be when n = 0".

No, n is a given constant. The problem is to find x1, x2,..., xn so as to minimize the function.

Also you say
"So, for each x, generally,

xn = an/sqrt(a1^2 + a2^2 + ... an^2) [2]

Which implies that a1 = a2 = ... = an from [1]"

No, a1, a2, etc. are GIVEN. You can't place any constraints on them.
[1] implies a1= a2=...= an ONLY if you assume x1= x2=...= xn which is NOT in general true.
 
As for the max, it would be when n approaches infinity, since the function would approach infinity as well. So the max would be k*(sqrt(n)).
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 97 ·
4
Replies
97
Views
18K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
Replies
28
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K