Finding maximum and minimum in a region via Lagrange multipliers

Question

Consider the following function in n-dimensional space

$$f(x_{1},x_{2},...,x_{n})=x^{2}_{1}+x_{2}^{2}+...+x_{n}^{2}$$

What are the minimum and maximum values of $f$ in the region

$$x_{1}^{2}+2x_{2}^{2}+3x_{3}^{2}+...+nx_{n}^{2}\leq 1$$?

I think the minimum value is 0 when all $x_{1}=x_{2}=...=x_{n}=0$.

What is the maximum value?

I could use Lagrange multipliers but I would have to restrict myself to the boundary $$x_{1}^{2}+2x_{2}^{2}+3x_{3}^{2}+...+nx_{n}^{2}= 1$$.

Answer 1

You can start with the inequality

\begin{equation} \begin{split} x_1^2+2x_2^2+3x_3^2+ \ldots +nx_n^2 &\leq 1\\ \mbox{i.e., } x_1^2+x_2^2+x_3^2+ \ldots +x_n^2 &\leq 1-\sum_{k=2}^n(k-1)x_k^2 \end{split} \end{equation}

In other words, the value of $f\leq 1-\sum(\mbox{+ve terms})$. The max is attained when $x_2=x_3=\ldots x_n=0$ and $x_1=1$.

Answer 2

The maximum value occurs at some point such that $x_1^2+2x_2^2+\cdots +nx_n^2=1$. This is because the maximum clearly does not occur at the origin. And if it occurred at some $(a_1,\dots,a_n)$ with $0\lt a_1^2+2a_2^2+\cdots +na_n^2\lt 1$, we could by multiplication by a suitable $\tau$ obtain a larger sum of squares using a point on the boundary.

So the Lagrange multipliers method can be used. (The maximum will turn out to be $1$.)

Added: When we use Lagrange multipliers, it is easy to see that $\lambda\ne 0$, for $\lambda=0$ would force all the $x_i$ to be $0$. From the equations $x_i=\lambda i x_i$, we find that all but one of the $x_i$ must be $0$. And among these $n$ options, the best choice is $x_1=1$, and $x_i=0$ for $i\gt 1$.