Question

Let a and b, respectively, be the absolute minimum and maximum values of the function f(x1,x2,...,xn)=x21+x22+...+x2n within the region x21+2x22+3x23+...+nx2n≤1. Let c be the absolute minimum value of f(x1,x2,...,xn) on just the boundary of the region.What is a + b + c ?

Answer 1

Answer: $a+b+c=\frac{n+1}{n}.$

Step-by-step explanation: The function $f(x_1,x_2,\ldots)=x_1^2+x_2^2+\ldots$ is always positive except at the origin where it is equal to zero. This means that the absolute minumum of this function must be $a=0$. Absolute maximum is when all of the variables are equal to zero except $x_1$ which is equal to 1 (f evaluated at this point is equal to 1 do b=1). The function itself is then equal to 1. This is because when $f(\cdots)=x_1^2+x_2^2+\ldots\leq x_1^2+2x_2^2+3x_3^2+\ldots\leq1$ so it is at most equal to 1 and this happens exactly at the point $(x_1,x_2,x_3,\ldots)=(1,0,0,\ldots).$

The absolute minimum at the boundary of this function happens when all the variables are equal to 0 except $x_n=\frac{1}{\sqrt{n}}$ and this minimum is equal to c=1/n. To see this notice that

$nf=nx_1^2+nx_2^2+\cdots nx_n^2\geq x_1^2+2x_2^2+\cdots nx_n^2=1$

(the equality sign is because now we are on the boundary). We notice that nf is greater than or equal to 1 and the minimum of nf=1 (this implies the minimum for f to be 1/n) is attained exactly when $(x_1,x_2,\ldots,x_n)=(0,0,\ldots,\frac{1}{\sqrt{n}})$.

So, finally, $a+b+c=0+1+\frac{1}{n}=\frac{n+1}{n}.$