I want to solve the following problem in $\boldsymbol{x} \in \mathbb R^{n}$

$$\begin{array}{ll} \text{maximize} & \boldsymbol{x}^T \boldsymbol{A} \boldsymbol{x}\\ \text{subject to} & \boldsymbol{q}^T \boldsymbol{x} = 1\\ & x_i \geq 0\end{array}$$

where matrix $\boldsymbol{A}$ is positive definite matrix and $x_i$ denotes the $i$-th entry of $\boldsymbol{x}$.

Actually, I have tried to use Lagrangian multiplier. I directly transformed the objective function to $-\boldsymbol{x}^T \boldsymbol{A} \boldsymbol{x} + \lambda ( \boldsymbol{q}^T \boldsymbol{x} - 1 )$ and take its first derivative and set that to zero.

However, the solution obtained did not maximize the objective function, it just makes $\boldsymbol{x}^T\boldsymbol{A} \boldsymbol{x}$ smaller and smaller. Then I found that the solution of $\min_{\boldsymbol{x}} \boldsymbol{x}^T \boldsymbol{A} \boldsymbol{x}$ with the same constraints is the same with that of $\max_{\boldsymbol{x}} \boldsymbol{x}^T \boldsymbol{A} \boldsymbol{x}$.

Any comments would be appreciated!

Update As comments suggested, I changed the situation to $x_i \geq 0, \forall i$. Thus for example, when $\boldsymbol{A}= \left[\begin{matrix} {2 \; 0\\ 0 \;1 }\end{matrix} \right]$ and $\boldsymbol{q} = [1,1]^T$. The problem has a solution $\boldsymbol{x} = [1 ,0]^T$ that maximize the objective function. Can this extend to more general case?

  • 4
    $\begingroup$ Geometric intuition should suggest that in a noticeable number of cases $\sup\{x^TAx\,:\, q^Tx=1\land \forall i, x_i> 0\}=\infty$ (specifically, cases where $\neg(q>0)$) $\endgroup$ – Gae. S. Mar 25 at 16:47
  • $\begingroup$ If you are looking for a numerical solution, cplex/gurobi can solve this. $\endgroup$ – LinAlg Mar 26 at 1:08
  • $\begingroup$ Thanks for your kind comment. I still want to figure it out analytically. $\endgroup$ – Kris Prokins Mar 26 at 1:13
  • $\begingroup$ Thanks for you comment, but when $x_i \geq 0$, it seems the question have a maximum at the boundary. $\endgroup$ – Kris Prokins Mar 26 at 2:58
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    $\begingroup$ With the constraint $x_i>0$, I don't think there is any closed-form solution. $\endgroup$ – user1551 Mar 26 at 7:25

Since $\mathbf q > \mathbf 0$, the feasible region $\{\mathbf x \in \mathbb R^n : \mathbf q^{\mathsf T} \mathbf x = 1, \mathbf x \ge \mathbf 0\}$ is bounded (we have $x_i \in [0, \frac1{q_i}]$ for each $i$). It's also closed, so the maximum of $f(\mathbf x) = \mathbf x^{\mathsf T} \!A \mathbf x$ must be achieved somewhere in the feasible region.

Because $f(\mathbf x)$ is convex, this maximum must be at an extreme point, and this feasible region has only $n$ extreme points: for each $i$, we can get one of them by setting $x_i = \frac1{q_i}$ and all other entries to $0$. This point has objective value $f(\mathbf x) = \frac{A_{ii}}{q_i^2}$. Now just compare the values $\frac{A_{11}}{q_1^2}, \dots, \frac{A_{nn}}{q_n^2}$ and pick the largest.

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  • $\begingroup$ Nice solution. The maximum of a convex function over the polyhedron is achieved at one of its vertices. $\endgroup$ – River Li Mar 27 at 0:15

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