Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer does not exist, enter DNE.) f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 36

The Lagrangian is[tex]L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)[/tex]with partial derivatives (set equal to 0)[tex]\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0[/tex][tex]\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0[/tex]for each [tex]1\le i\le n[/tex].Let [tex]\Lambda[/tex] be the sum of all the multipliers [tex]\lambda_i[/tex],[tex]\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n[/tex]We notice that[tex]x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0[/tex]so that[tex]\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0[/tex]We know that [tex]\sum\limits_{i=1}^n{x_i}^2=36[/tex], so[tex]\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda[/tex]Solving the first [tex]n[/tex] equations for [tex]x_i[/tex] gives[tex]1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}[/tex]and in particular[tex]\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}[/tex]It follows that[tex]-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}[/tex]which gives us[tex]x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}[/tex]That is, we've found two critical points,[tex]\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)[/tex]At the critical point with positive signs, [tex]f(x_1,\ldots,x_n)[/tex] attains a maximum value of[tex]\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n[/tex]and at the other, a minimum value of[tex]\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n[/tex]