Proposition . For xo 属于P to be primal optimal it is sufficient that f(x) and the active components of g(x) at Xo are Type I functions with respect to a common rl(X ) at Xo and conditions ()-() hold at x o .for some Yo. We note that if r(x) is identically zero then () and () imply that xo is primal optimal. It follows that if a primal constraint qualification holds, then for xo属于P to be primal optimal it is necessary as well as sufficient that conditions ()-() hold at xo for some Y0 and that f(x) and g(x) are Type I functions with respect to a common r(x) at x0. However in the following theorem we see that for optimality, under an additional qualification on the constraints there must exist a vector r(x) which is not identically zero for each feasible x. Theorem . If x0 属于 P and the number of active constraints at Xo is k, where k < n, then for Xo to be optimal in the primal problem it is necessary that f(x) and g(x) are Type I functions with respect to 命题 。为XO笔记本电脑属于磷是原始最佳它是足够的F （ x ）和活性成分的G （十）在厂是I型职能方面的一个共同的北京（十）在XO和条件（ ） - （ ）保持在XO笔记本电脑。一些哟。 我们注意到，当R （ x ）是相同的零然后（ ）和（ ）意味着原始XO笔记本电脑是最佳的。因此，如果原持有资格限制，那么为XO笔记本电脑属于磷是原始最优这是必要的，也是充分的条件 （ ） - （ ）举行XO笔记本电脑的一些Y0和函数f （ x ）和g （ x ）的是I型职能方面的共同住宅（十）在x0 。然而在下列定理，我们看到的最优化，根据附加资格的限制，必须存在一个矢量住宅（十）这是不相同的每个可行零十定理 。如果x0属于P和积极的数目限制在XO笔记本电脑却钾，其中k “氮，然后厂为最优的原始问题是必要的函数f （ x ）和g （ x ）的是I型职能方面
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