Inequality-constrained quadratic program

Question

I am trying to find the optimal solution to this: $$\begin{aligned} \min_{} \quad & x^TAx\\ \textrm{s.t.} \quad & c^Tx \ge 1\\ \end{aligned}$$ where $A \in \mathbb{S}_{+}^{n}$ ($n \times n$ symmetric positive semidefinite matrix) and $c \in \mathbb{R}^n$ is a nonzero vector.

I know I can use the first order optimality condition and would get $2Ax(y-x) \ge 0$ for all feasibly $y$. $x$ is only optimal when this inequality is true for all feasible $y$ but I'm not sure how to get the optimal $x$ knowing the inequality.

Answer 1

assuming $A$ is symmetric positive definite, it has a Cholesky decompostions $A= W^T W.$ Taking $y= Wx,$ the objective becomes $y^T y$ which is just its squared length. The constraint is $(c^T W^{-1}) y \geq 1$ where the point nearest the origin is a multiple of the vector $(c^T W^{-1})$ as a row, or ${W^{-1}}^T c$ as a column. Something like that.

Answer 2

At optimality $c^Tx=1$ (if $c^Tx>1$, you can always rescale $x$ without making the objective function worse). Minimizing a quadratic function under a linear equality constraint is easy and boils down to solving a linear system.