Upper bound for Maximization problem I have an optimization problem of the form 
Max $x_1+x_2+x_3+\cdots+x_n$
subject to  $x_0^2+x_1^2+x_2^2+\cdots+x_n^2+x_{12}^2+x_{13}^2+x_{14}^2+ \cdots+x_{1n}^2+x_{23}^2 + \cdots +x_{2n}^2+ \cdots +x_{n n-1}^2+x_{123}^2+\cdots+x_{12..n}^2=1$,
$x_0+x_1+x_2+\cdots+x_n+x_{12}+x_{13}$+$x_{14}+ \cdots +x_{1n}+x_{23}+ \cdots +x_{2n}+ \cdots +x_{n n-1}+x_{123}+ \cdots +x_{12..n}=1$,
where  $x_0,x_1,x_2,\ldots,x_{123\ldots n}$ are unrestricted.
What is the upper bound for the cost function ..?
Thank you.
 A: It's unclear exactly how many total variables you have; call it $m$. Let $\textbf{1}_n$ be the vector with the first $n$ components 1, the rest 0; and $\mathbf1$ the vector with all components $1$.
Then the KKT conditions of your maximization problem are
$$\mathbf{1}_n + 2\lambda x + \mu \mathbf{1} = 0,$$
where $\lambda$ and $\mu$ are the Lagrange multipliers. Call $v$ the value of the objective at the critical point. Then, dotting the above by $x$ gives
$$v + 2\lambda + \mu = 0.$$
Dotting by $\mathbf{1}$ gives
$$n + 2\lambda + m\mu = 0,$$
and lastly, dotting by $\mathbf{1}_n$ gives
$$n + 2\lambda v + n\mu = 0.$$
Now we are left with a system of equations; they have only one nonlinear term so we can proceed by eliminating variables. Combining the first two equations gives
$$ 2\lambda(m-1) = n - mv,$$
and the first and second,
$$ 2\lambda (n - v)  = n - nv.$$
We can now eliminate $\lambda$:
$$(n-mv)(n-v) = (n-nv)(m-1)$$
$$n^2 -nv -nmv + mv^2 = n(m-1) - n(m-1)v$$
$$mv^2  -2nv + n(n-m+1) = 0.$$
So
$$v = \frac{n \pm \sqrt{n(m-1)(m-n)}}{m}.$$
I'll leave it to you to check that the positive sign choice indeed gives a maximum.
A: A thought: If you stack all variables inside a vector, say $x$, it should be straight forward to re-write the above problem as
\begin{align}
&\min_x b^Tx \\
\text{subject to}&x^Tx=1 \\
&c^Tx=1;
\end{align}
This is not a convex problem. I am not sure about a closed form solution though. KKT conditions should be able to do something. 
