# Tag Info

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Convenient conditions for checking that a certain stationary point is optimal generally require that the objective function is convex. Nonconvex optimization is a much more difficult subject with a lot of special cases for different problems. Two of the many reasons for this are that there can be stationary points that are not local minima and that there can ...

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First of all, there is a fairly easy way to show this inequality: $$|ax + by| = |(a, b)\cdot (x, y)| \leq \|(a, b)\|\|(x, y)\|\cos\theta \leq \|(a, b)\|\|(x, y)\| = \|(a, b)\| = \sqrt{a^2 + b^2}.$$ As for the Lagrange multipliers method, you're almost there. As $2\lambda x = a$ and $2\lambda y = b$, $x = \frac{a}{2\lambda}$ and $y = \frac{b}{2\lambda}$. To ...

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Warning: This method is cute, but does not use Lagrange multipliers. Let $x=3w$ and $y=2u$. Then we have $xyz=6\implies uwz=1$ and we want to minimize $$xy+2xz+3yz=6uw+6wz+6uz=6(uw+wz+uz)$$ Now by the AM-GM inequality, we have $$uw+wz+uz\ge3\sqrt[3]{uwz}=3,$$ with equality when $uw=wz=uz$ which implies $u=w=z=1$, so we have $$xy+2xz+3yz\ge6\times3=18$$ ...

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From $4y = 2y\gamma$, if $y = 0$ then $\gamma$ isn't specified. In which case from $x^2 + y^2 = 1$, we have $x = \pm 1$. I.e., the points $(1,0)$ and $(-1,0)$, which correspond respectively to $\gamma = 1/2$ and $\gamma = -3/2$.

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$f(x,y,z)=\sqrt{x^2+y^2+z^2}$ assuming that you are talking about the Euclidean minimum distance Here you have two constraints, that means two Lagrangian multipliers namely $\lambda_1$ and $\lambda_2$ for the constraints $x+y-z+2=0$ and $x^2+y^2-z^2=0$. One can write them all as ...

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You should solve the problem $$\begin{array}{c} \min \hspace{3mm} x^2 +y^2 +z^2 \\ s.t. \hspace{3mm} x+y -z +2 =0 \\ \hspace{8mm} x^2+y^2-z^2 = 0. \\ \end{array}$$ The lagrangian function is given by $$\mathcal{L} = x^2 +y^2 +z^2 + \lambda (x+y -z +2)+\mu (x^2+y^2-z^2).$$ Hence, $$\frac{\partial \mathcal{L}}{\partial x} = 2x ... 1 Here's the TL;DR version, for your specific example. The Lagrangian is$$L(X,Z) = f(X) - \langle Z, K - XX^T \rangle$$where the inner product is the simple elementwise inner product, and the Lagrange multiplier Z is positive semidefinite. A more general discussion: the Lagrangian looks like this:$$L(x,\lambda) = f(x) - \langle \lambda, c - g(x)\rangle ...

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