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In the objective function of quadratic programming, the vector x and its transposed are existing. In the constraints also vector x exists. If there is a way that both in the cost function and in the constraints x be a function of x? Not x itself?

Simin Hesami
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    Are you talking about finding the minimum of $\frac12 x^T Qx + c^T x$ subject to $Ax \preccurlyeq b$? If so, what do you mean by "$x$ be a function of $x$"? – Joe May 17 '22 at 00:24

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