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=(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=(1)/(√(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,(1)/(√(n)))`.

So, finally,
`a+b+c=0+1+(1)/(n)=(n+1)/(n).`