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3

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 ...


3

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 ...


3

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 $$xy+2xz+3yz\ge6\times3=18$$ ...


3

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 $\gamma = -3/2$.


3

It doesn't matter. All that changes is the sign of $\lambda^*$, where $(x^*,y^*,\lambda^*)$ is the critical point. Dealing with maximization doesn't change it either. You can see this because regardless of how you formulate the method, you still have $$\nabla L(x,y,\lambda) = \begin{bmatrix} f_x(x,y) \pm \lambda g_x(x,y) \\ f_y(x,y) \pm \lambda g_y(x,y) \\ ...


2

I guess that $\mu_i\geq 0$ for all $i$. You can reformulate the problem as $$ \max \sum_{i=1}^n \log( y_i )\\ s.t.\\ y_i = x_i\cdot \mu_i \quad i=1,\ldots,n\\ \sum_{j=1}^k \mu_i = 1\\ \mu_i \ge 0 \quad i=1,\ldots,n $$ Assuming that $x_i$ are such that $y_i\geq 0$ for all $i$, this is a convex problem you can solve with a standard optimizer.


2

Let $H$ be a Hilbert space and define $M$ by $$M=\{u\in H:\ F(u)=0\},$$ where $F:H\to\mathbb{R}$ is a $C^1(H)$ function. Theorem: Suppose that for all $u\in M$, $F'(u)\neq 0$. Then, $M$ is a $C^1$ Hilbert Manifold of $H$. To prove it, fix $u\in H$. Remember that $F':H\to H^\star$, so $F'(u)\neq 0$ means that the linear function $F'(u)$ has non trivial ...


1

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1

There is a closed form solution for your problem under the following assumption: $x_i > 0, \mu_i > 0 $, for all $i$. You can rewrite $\sum_{i=1}^N \log(x_i \mu_i) = \log(\prod_{i=1}^N x_i \mu_i)$. Using this identity, it is clear that your problem is equivalent to $$ \max c \prod_{i=1}^N \mu_i~~~{\rm s.t.} ~~ \sum_{j=1}^N \mu_i = 1,$$ where $c = ...


1

Note that $xy \cdot yz = \gamma^2$. If $\gamma \neq 0$ then dividing by $xz$ gives $y^2 = \gamma$. Similarly for $x,z$. If $\gamma = 0$, then exactly two of $x,y,z$ are zero and the other is 3.


1

(0,0) satisfies both of your equations , Check its nature at that point .Also factoring first equation you get $x=y$ and $x+y=3$ .use in second equation


1

If $y=0$ then we cannot cancel it. Moreover, $x^2+y^2=1$ implies $x= \pm 1$ when $y=0$ hence the mysterious extra points.


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Here is the basic definition of lagrange multipliers: $$ \nabla f = \lambda \nabla g$$ With respect to: $$ g(x,y,z)=xyz-6=0$$ Which turns into: $$\nabla (xy+2xz+3yz) = <y+2z,x+3z,2x+3y>$$ $$\nabla (xyz-6) = <yz,xz,xy>$$ Therefore, separating into components gives the following equations: $$ \vec i:y+2z=\lambda yz \rightarrow \lambda = ...


1

$f(x,y,z)=\sqrt{x^2+y^2+z^2}$ 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 ...


1

You should solve the problem \begin{equation} \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} \end{equation} 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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