# Thread: Vector calc minimum problem

1. ## Vector calc minimum problem

I am given that the summation from i = 1 to 3 = 1
I am to find x sub 1, xsub2, xsub3 so that the summation from i = 1 to 3 of x sub i ^2 is at a minimum. Then I am suppose to come up for a general formula based on my findings. If anyone is willing to give me a push in the right direction it would be much appreciated! Frostking

2. Originally Posted by Frostking
I am given that the summation from i = 1 to 3 = 1
I am to find x sub 1, xsub2, xsub3 so that the summation from i = 1 to 3 of x sub i ^2 is at a minimum. Then I am suppose to come up for a general formula based on my findings. If anyone is willing to give me a push in the right direction it would be much appreciated! Frostking

So what we know is that we are trying to minimize

$m=x_1^2+x_2^2+x_3^2$ subject to

$x_1+x_2+x_3=1 \iff F(x_1,x_2,x_3)=x_1+x_2+x_3-1$

with these we can use Lagrange multipliers

$\nabla{m}=\nabla \lambda F$

$(2x_1)\vec i +(2x_2) \vec j +(2x_3)\vec k =(\lambda)\vec i+(\lambda)\vec j +(\lambda) \vec k$

so $x_1=x_2=x_3=\frac{\lambda}{2}$

subbing this into the constraint equation
$\frac{\lambda}{2}+\frac{\lambda}{2}+\frac{\lambda} {2}=1 \iff \lambda =\frac{2}{3}$

This then implies that $x_1=x_2=x_3=\frac{1}{3}$

First of all, $F(x_1,x_2,x_3)$ doesn't have to equal zero, so you can drop the -1.
And you don't need to solve for $\lambda$.
your constraint to obtain what you really want, the $x_i$'s.