Question

find the maximum and minumum values of the n variable function:X1+X2+...Xn subject to the constraint X1^2+X2^2+...+Xn^2=1

Answer 1

aren't u supposed to use lagrangian Constrained Optimization ?

Answer 2

so u want to find extremums of \[f(x_1,x_2,...,x_n)=x_1+x_2+...+x_n\]subject to this constraint\[x_1^2+x_2^2+...+x_n^2=1\]

Answer 3

set up ur lagrangian\[L=f-\lambda g\]and consequencly ur equations\[x_1^2+x_2^2+...+x_n^2=1\]\[\frac{\partial L}{\partial x_1}=0\]\[\frac{\partial L}{\partial x_2}=0\]...\[\frac{\partial L}{\partial x_n}=0\]these n+1 equation with n+1 unknown (degree of freedom is 0) will give u the values of \[x_1,x_2,...,x_n,\lambda\]for which \(f\) is max or min

Answer 4

still one missed point\[g(x_1,x_2,...,x_n)=x_1^2+x_2^2+...+x_n^2-1\]

plz let me know if u got it from here

Answer 5

yeah, but why there is n+1 equations?

Answer 6

we also have dL/d入=x1^2+x2^2+...xn^2-1 right?

Answer 7

n partial derivatives and the constrained itself is one of the equations so u have n+1 equation

Answer 8

and we also have dL/d入=-g right? @mukushla

Answer 9

then i got x1^2+x2^2+....xn^2=1

Answer 10

L=f+g入

Answer 11

then dL/dx1 =1+2x1入 but not zero?

Answer 12

one of equations for example\[\frac{\partial L}{\partial x_1}=0\]\[\frac{\partial f}{\partial x_1}-\lambda\frac{\partial g}{\partial x_1}=0\]\[1-2\lambda x_1=0\]set up other equations like this anf find \(\lambda\) first

Answer 13

ok, so what I did is:L=g+fλ,
and then x1=x2=x3=...xn=-1/(2λ)
for the constraint function, we can get nX1^2=1, then x=(1/n)^0.5,
then I plug this in to L again,
the function L can write as L=nX1+(-1/2λ)(nX1^2-1)
after plugging into x=(1/n)^0.5
i FINALLY get L=n^0.5
so there is only one answer, but I have to find maximum and min, So I really get confused, can you give me more hint? appreciates a lot!

Answer 14

maybe i made a mistake, x=(+or -) n^0.5 right, so there r 2 answer for L, n^0.5 and -n^0.5? so one is max and the other one is min, am I RIGHT?

Answer 15

emm...what i know as standard form for L is f-gλ

Answer 16

f+gλ or f-gλ doesnt matter, cause the answer is same. if gλ, then g=0, if -gλ then -g=0, g also =0, so this isnt the point

Answer 17

ok u r right it doesn't matter

and it gives\[x_1=x_2=...=x_n=\frac{1}{2\lambda}\]and plugging this in\[x_1^2+x_2^2+...+x_n^2=1\]gives\[n\frac{1}{4\lambda^2}=1\]and\[\lambda=\pm \frac{\sqrt{n}}{2}\]so the max of \(f\) is when\[x_1=x_2=...=x_n=\frac{1}{\sqrt{n}}\]and the min is when\[x_1=x_2=...=x_n=-\frac{1}{\sqrt{n}}\]

Answer 18

and note that u want to find max and min of f not L

Answer 19

\[\text{max} \ (f)=\frac{n}{\sqrt{n}}=\sqrt{n}\]and\[\text{min} \ (f)=-\frac{n}{\sqrt{n}}=-\sqrt{n}\]

Answer 20

i hope its clear

Answer 21

@mukushla yes, we got the same answer, thx a lot!!!!

Answer 22

no problem :)