Lagrange Multipliers: Absolute minimum of $f(x_1,x_2,...,x_n)$ on the boundary of the region $x^2_1+2x^2_2+3x^2_3+...+nx^2_n≤1$

Question

I'm trying to use Lagrange Multipliers to solve the following problem

Let $f(x_1,x_2,...,x_n)=x_1^2+x_2^2+...+x_n^2$

What is the absolute minimum of $f(x_1,x_2,...,x_n)$ on the boundary $x^2_1+2x^2_2+3x^2_3+...+nx^2_n = 1$?

I've tried solving for $\lambda$ and got $\lambda=1/n$ but I'm not quite sure how to use that to come up with the generalization for the minimum.

I've also tried playing around with the equations a bit, and ended up with the equation $2 n^2 \lambda \Sigma x_n^2 = 1$, but I'm also not sure how to proceed from here.

Answer 1

Transforming to variables $y_i=x_i^2$, this becomes a linear programming problem - we wish to minimize the linear function $y_1+y_2+\cdots+y_n$ on the region $y_1+2y_2+3y_3+\cdots+ny_n=1$, $y_1\ge 0$, $y_2\ge 0$, ..., $y_n\ge 0$.

By standard theory, the minimum must come at one of the vertices. The $n$ vertices, in this case, have $y_i=\frac1i$ and $y_j=0$ for $j\neq i$. The values at those points are $\frac1i$, so the minimum $\frac1n$ comes at the $n$th vertex, where $y_n=\frac1n$ and the other $y_i$ are zero. In terms of the $x_n$, that's $(x_1,x_2,\dots,x_n)=\pm(0,0,\dots,\frac1{\sqrt{n}})$.

The choices are a) $(−1)/+1$ b) $(−1)/\sqrt{n}$ c) $(n^2+1)/((n+2)^2)$ d) $1/(n^2)$ and e) $(n+1)/n$

These choices are wrong, at least for the problem presented. There's a mistake either in your source or in your transcription.

I've tried solving for $\lambda$ and got $\lambda = 1/n$ but I'm not quite sure how to use that to come up with the generalization for the minimum.

Lagrange multipliers as I'd run it:
We're looking to minimize $f(x_1,x_2,\dots,x_n)=x_1^2+x_2^2+\cdots+x_n^2$ on the region $g(x_1,x_2,\dots,x_n)=x_1^2+2x_2^2+\cdots+nx_n^2=1$. The two gradients are $\nabla f = 2(x_1,x_2,\dots,x_n)$ and $\nabla g = 2(x_1,2x_2,\dots,nx_n)$. If two or more of the $x_i$ are nonzero, those coordinates have different ratios in $\nabla f$ and $\nabla g$, so the critical points come when all but one of the $x_i$ are zero. There are $2n$ such points, with $x_i=\pm\frac1{\sqrt{i}}$ and each other $x_j=0$. Testing each point, we find that the minimum comes at the last pair of points, where $x_n$ is the nonzero one.
We should also look for any places where the region has edges, corners, or any sort of lower-dimensional boundary - such places break the assumptions behind Lagrange multipliers and we must repeat the process in lower dimensions. In this case, there aren't any. The ellipsoid we're working with is smooth.

I'd like to see your process on this one. We indeed have the ratio $\lambda = \frac1n$ at the minimum, but what did you do with the other critical points?

Answer 2

For $\sum\limits_{k=1}^nkx_k^2=1$ we obtain: $$\sum_{k=1}^nx_k^2\geq\frac{1}{n}\sum_{k=1}^nkx_k^2=\frac{1}{n}.$$ The equality occurs for $x_1=x_2=...=x_{n-1}=0$ and $x_n=1,$

which says that $\frac{1}{n}$ is a minimal value.

For $\sum\limits_{k=1}^nkx_k^2\leq1$ we obtain: $$\sum_{k=1}^nx_k^2\geq0,$$ where the equality occurs for $x_1=x_2=...=x_n=0,$ which says that $0$ is a minimal value.