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Given $n$ positive values. $x_i \ge 1 (1\le i \le n)$. Their sum is $k$. $ \sum_{i=1}^{n}x_i = k $ Define the following value: $ \sum_{i=1}^{n}x_i(x_i-1) $

Now use Lagrange multipliers to find when this value is minimum.

Define $J = \sum_{i=1}^{n}x_i(x_i-1) + \lambda ((\sum_{i=1}^{n}x_i)-k)$. Then $\frac{\partial J}{\partial x_i}=0$ implies that $x_{i} = -\frac{(\lambda-1)}{2}$. Substituting this back into the constraint give $-\frac{\lambda -1}{2} = k/n$. Thus we have $x_i=k/n$.

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    Yes, your usage of the Lagrange multipliers rule is correct.2011-10-27

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