Linear objective with quadratic constraints

Question

I have the problem $$ \text{maximize } f= c^Tx \\ \text{subject to } x^T Q x \leq 1 \\ x,c \in \mathbb{R}^n \text{ , } Q \in \mathbb{R}^{n \times n} $$ and $ Q $ is additionally symmetric positive definite. Using a Lagrange multiplier I get the answer $$ f = +\sqrt{c^T Q^{-1} c} $$ but I can't prove that it's correct. How would this be proven correct or incorrect?

Answer 1

With the Lagrangian $$ L(x,\lambda) = -c^Tx + \lambda \frac12 (x^TQx -1) $$ one finds the KKT conditions $$ -c + \lambda Qx = 0, \ \lambda\ge0 , \ \lambda(x^TQx -1)=0, \ x^TQx\le 1. $$ If $x^TQx<1$ then $\lambda=0$, and necessarily $c=0$. Hence $\lambda>0$, $x^TQx=1$, $$ x = \frac1\lambda Q^{-1}c, \ x^TQx = \frac1{\lambda^2}c^TQ^{-1}c = 1, $$ this implies $\lambda^2 = c^TQ^{-1}c$. And since $\lambda\ge0$, we get $\lambda = \sqrt{c^TQ^{-1}c}$, and $$ c^Tx = \lambda x^TQx = \sqrt{c^TQ^{-1}c}. $$ Since the function to be maximized is concave and the feasible set is convex, the global maximum is realized at a KKT point. The only KKT point is $x= \frac1{\sqrt{c^TQ^{-1}c}} Q^{-1}c$ with $c^Tx=\sqrt{c^TQ^{-1}c}$. Hence this value is indeed the maximum.