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Any hints on how to solve this problem? \begin{align}\label{eq:opt} \max &~~3x_1 + 5x_2 \\ {{\rm s}.{\rm t}.} &~~x_1 + x_2 = 1~\text{and}~x_1^2 + 2x_2^2 - 4x_1x_2 \leq 1. \end{align}

From the first order conditions, $\bigtriangledown f(x) + \lambda \bigtriangledown h(x) + \mu \bigtriangledown g(x) = 0$, and the constraint, $\mu \cdot g(x) = 0$, I have that

\begin{align*} 3 + \lambda + \mu (2x_1 - 4x_2) = 0 \\ 5 + \lambda + \mu (4x_2 - 4x_1) = 0 \\ \mu(x_1^2 + 2x_2^2 - 4x_1x_2 - 1)= 0, \end{align*}

However, I do not see how I am to find $\lambda$, $\mu$, $x_1$ and $x_2$ from only these three. We can solve for $\lambda$ using the first two, but that doesn't end up leading to the answer, which should be $(x_1, x_2) = \left(\frac{1}{7}, \frac{6}{7}\right).$

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  • $\begingroup$ The system of equations should also include the equality constaint $x_1+x_2=1$. Also, it follows from the first two equations that it must hold $\mu \neq 0$, therefore, the third equation should be simplified to $x_1^2+2 x_2^2 - 4x_1 x_2 -1 = 0$ $\endgroup$
    – Stelios
    May 5, 2017 at 11:54
  • $\begingroup$ Ah, yes adding in $x_1 + x_2 = 1$ as an equality constraint helps. Thank you. $\endgroup$
    – jj8989
    May 6, 2017 at 2:11

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