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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

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.


0

I'm assuming x and y commute, right? So, if you add the equations you end up with: -x^2-2xy+3x=0 or -x-2y+3=0


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 $\gamma = -3/2$.


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.


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

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


0

The Lagrange multiplier is $$L=2(xy+xz+yz)+\lambda(a-(4x+4y+4z))$$ We can find the partial derivatives: $$\frac{\partial L}{\partial x}=2y+2z-4\lambda\\ \frac{\partial L}{\partial y}=2x+2z-4\lambda\\ \frac{\partial L}{\partial z}=2x+2y-4\lambda\\ \frac{\partial L}{\partial \lambda}=a-(4x+4y+4z)\\ $$ You are correct that the critical point is ...


0

This is connected to something proved in various places in MS (for instance Question related to Lagrange multipliers): the maximum of a quadratic form $Q$ on the unit sphere of a real euclidean space $E$ is the biggest eigenvalue of the matrix of $Q$ wrt any ortonormal basis of $E$. So we take $E\subset\mathbb R^n$ defined by $\langle ...


0

I found similar question in my textbook .hope you can follow on same lines


0

I claim that the point on the ellipsoid with the shortest distance to your plane will be such that the vector normal to the ellipsoid at that point will be parallel to the normal to the plane. The normal at (x, y, z) has the form (2x, 2y, 8z), and the normal to the plane is (1, 1, 1). Therefore, at the closest (and furthest) point on the ellipsoid, there is ...


0

Let $(r, s, t)$ be a point on the ellipsoid $x^2+y^2+4z^2=4$. Then, the signed distance from the point to the plane is $$d=\dfrac{r+s+t-6}{\sqrt{1^2 + 1^2 + 1^2}}.$$ Now we just need to minimize $d$ subject to $r^2+s^2+4t^2 - 4 =0$. Lagrange Multipliers should work here.


0

One writes that the gradient of $\alpha(x)=f(x,h(x))$ wrt x vanishes as you say. Using the chain rule to compute the derivatives wrt the $x_j$'s: $$ 0=\frac{\partial\alpha}{\partial x_j}=\Big(\sum_{i=1}^D\frac{\partial f}{\partial x_i}\frac{\partial x_i}{\partial x_j}\Big)+ \frac{\partial f}{\partial y}\dfrac{\partial h}{\partial x_i}=\frac{\partial ...


0

Your approach is reasonable. Let $f((x,y)) = |ax+by|$, $g((x,y)) = {1 \over 2}(x^2+y^2)$, solve $\max \{f((x,y))| g((x,y)) = {1 \over 2} \}$. The constraint $g((x,y)) = {1 \over 2} $ shows that the feasible set is compact, so we know a maximum exists. Note that if $a=b=0$, then the result is true, so suppose $(a,b) \neq 0$. If $(a,b) \neq 0$, choosing ...


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: $$L(x,\lambda) = f(x) - \langle \lambda, c - g(x)\rangle$$ ...


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 ...


0

can we do this. we will show that you can write $$ ax + by = \sqrt{a^2 + b^2}\cos(t - \phi)$$ which will give us the inequality $$ |ax + by| \le \sqrt{a^2 + b^2}$$ here is an outline: since $x^2 + y^2 = 1,$ there is $t$ such that $x = \cos t, y = \sin t$ and $ax + by = \sqrt{a^2 + b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}} \cos t + \dfrac{b}{\sqrt{a^2+b^2}} ...


0

For the other method, parametrize the circle by $x=\cos t$ and $y=\sin t$. Then $$ ax+by=a\cos t+ b\sin t=\sqrt{a^2+b^2}\sin(t+\phi) $$ for some $\phi$ (you should be able to check this). Maybe you can continue from here?


1

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 = ...


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

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

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) \\ ...


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 = ...


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.


0

Concerning the differentiability of $g_\rho(y)=\min_x L(x,y)_\rho$, I think that the result is a direct consequence of the Danskin's Theorem (see Proposition B.25 in [Bertsekas, "Nonlinear Programming", 2nd Edition]). More specifically, if $L(\cdot,y)_\rho$ is strictly convex and $L(x,\cdot)_\rho$ is differentiable, then $g_\rho(y)$ is differentiable with ...


0

As $f(x) = x^p $ is convex (for $p > 1$). We can use Jensen's inequality $$ \frac{x_1^p+x^2_p+\cdots+x_n^p}{n} \geq \left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^p$$ and hence it shows that the critical point you obtained using Lagrange multipliers is a minimum.


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


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 ...


0

Since there's a computer result posted now, I guess I'll describe my approach further. From the set of "Lagrange equations" you describe, we can solve each one for $ \ \lambda \ $ to establish $$ \lambda \ = \ ( \ 2 \ - \frac{\mu}{2} \ ) \ x \ = \ ( \ 2 \ - \frac{2\mu}{5} \ ) \ y \ = \ ( \ \frac{2\mu}{25} \ - 2 \ ) \ z \ \ . $$ We can solve the implied ...


0

Since I am lazy to do it by hands, I give you the sage code. $###lagrange method sage:x, y, z, lam , mu= var('x, y, z, lam, mu') sage:f=x^2+y^2+z^2 sage:g1 = x^2/4+y^2/5+z^2/25-1 sage:g2 = x +y -z sage:h = f - g1 * lam -g2 * mu sage:gradh = h.gradient([x, y, z, lam, mu]) sage:critical = solve([gradh[0] == 0, gradh[1] == 0, gradh[2] == 0, gradh[3] ...


1

Let´s say $x=x_1$ and $1-x=x_2$. Your intermediate result becomes $\frac{x_1}{x_2}=\frac{\gamma_1}{\gamma_2}$ Solving the equation for $x_1$: $x_1=x_2\frac{\gamma_1}{\gamma_2}$ The condition is, that $1=x_1+x_2$. Inserting the expression for $x_1$: $1=x_2\frac{\gamma_1}{\gamma_2}+x_2$ Now you can solve the equation for $x_2$. At the end you should ...


1

Found it: $ \frac{x}{1-x} = \frac{\gamma_1}{\gamma_2} $ $ \gamma_2 x = \gamma_1 - \gamma_1 x $ $ (\gamma_1 + \gamma_2) x = \gamma_1 $ $ x = \frac{\gamma_1}{\gamma_1 + \gamma_2} $ $ 1-x = 1-\frac{\gamma_1}{\gamma_1 + \gamma_2} = \frac{\gamma_1 + \gamma_2}{\gamma_1 + \gamma_2}-\frac{\gamma_1}{\gamma_1 + \gamma_2} = \frac{\gamma_2}{\gamma_1 + \gamma_2} $



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