2016AC-3-4 - Rhea

Solution

We form the lagrangian:
$ l(x,\lambda)=x_1x_2+\lambda_1(x_1+x_2+x_3-1)+\lambda_2(x_1+x_2-x_3) $
$ \begin{cases}\nabla_xL=\begin{bmatrix} x_2+\lambda_1+\lambda_2 \\ x_1+\lambda_1+\lambda_2 \\ \lambda_1+\lambda_2\end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ x_1+x_2+x_3-1=0 \\ x_1+x_2-x_3=0 \end{cases} $

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