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Satifying Kuhn-Tucker conditions.

Given $\lambda$ is a row, and $\dfrac{\partial L}{\partial \lambda}$ is a column,

why does $\lambda$ $\dfrac{\partial L}{\partial \lambda}=0$ ?

1 Answers1

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I worked it out. I didn't realise it's just a condition that must be satisfied.

Either $\lambda=0$

or $\dfrac{\partial L}{\partial \lambda}=0$

Fair enough.