We observe that $F \equiv n(2n-1) + n(n-2)+L(x_{2},x_{3} \cdots x_{n-2})$ where the form is obtained by utilizing constraint 1 and then the constraint obtained by subtracting 2 by 1. Where $L$ is linear in each variable and has just pluses in it.So $L \geq 0$ Also we observe that $x=(n-1,0,0,\cdots 0,1)$ is a solution of the linear equations.This will imply $L(x)=0$ Thus $F$ is minimum at $(n-1,0,0 \cdots 0,1)$ the value being $3n(n-1)$

See that $F(x_1,x_2,...,x_{n-1})=2n\displaystyle\sum_{k=1}^{n-1}kx_k-\displaystyle\sum_{k=1}^{n-1}k^2x_k=4n(n-1)-\displaystyle\sum_{k=1}^{n-1}k^2x_k$, so we should find $\max \displaystyle\sum_{k=1}^{n-1}k^2x_k$. After checking small cases we guess that our maximum is achieved when $(x_1,x_2,...,x_{n-1})=(n-1,0,...,1).$ Indeed, we can easily prove now that $\displaystyle\sum_{k=1}^{n-1}k^2x_k=\displaystyle\sum_{k=1}^{n-1}x_k +\displaystyle\sum_{k=2}^{n-1}(k-1)(k+1)x_k\le n+n\displaystyle\sum_{k=2}^{n-1} (k-1)x_k=n(n-1)$.