## anonymous 4 years ago 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

1. anonymous

aren't u supposed to use lagrangian Constrained Optimization ?

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

yeah, but why there is n+1 equations?

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

L=f+g入

11. anonymous

then dL/dx1 =1+2x1入 but not zero?

12. anonymous

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

13. anonymous

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!

14. anonymous

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?

15. anonymous

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

16. anonymous

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

17. anonymous

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$x_1=x_2=...=x_n=-\frac{1}{\sqrt{n}}$

18. anonymous

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

19. anonymous

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

20. anonymous

i hope its clear

21. anonymous

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

22. anonymous

no problem :)