I have to following problem $$\max_{x}x^TAx+b^Tx\quad \mathrm{s.t.}\quad x^Tx\leq c,$$ where $A$ is real, symmetric and positive semi-definite.

Firstly I tried to solve the problem with the KKT, but that didn't lead to any good results for arbitrary matrices.

So now I argue in the following way

Since $x^TAx \leq \lambda_{max}x^Tx$, I get $x^TAx+b^Tx\leq \lambda_{max}x^Tx + b^Tx$ and with the constraint $x^Tx\leq c$ this leads to $$x^TAx+b^Tx\leq \lambda_{max}x^Tx + b^Tx \leq \lambda_{max}c+b^Tx$$ So my solution would be $$x=\sqrt{c}v_{max},$$ where $v_{max}$ is the unit-eigenvector corresponding to the largest eigenvalue $\lambda_{max}$ of $A$. The thing that confuses me about this solution is that the solution does not depend on $b$ at all, so I am sure it is not correct...

  • $\begingroup$ @Rahul Thanks for the hint to this questions. I would still like to know, why my solution is not correct then. $\endgroup$ – RastaP Apr 25 '16 at 11:33
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    $\begingroup$ From $x^TAx+b^Tx\le \lambda_{\max}c+b^Tx$ how do you conclude that the optimum is $x=\sqrt cv_{\max}$? $\endgroup$ – user856 Apr 25 '16 at 12:02
  • $\begingroup$ To get an equality instead of inequality, but I see your point. So I really have to go with the KKT conditions.. Thanks though! $\endgroup$ – RastaP Apr 25 '16 at 12:11