New answers tagged

3

Imagine that you're aiming to cover as much of the $\sum_i v_i$ square as possible: The bigger the largest inner square, the closer it gets to covering more of the background square. The maximum sum of squares is reached when all but one of the $v_i$ is at the specified minimum - in this case, $1$.


3

If there are two numbers with $1\lt v_j\le v_k$, you can decrease $v_j$ and increase $v_k$ to increase the sum of squares. Thus at most one of the $v_i$ is greater than $1$.


0

(At the moment, this is not a complete answer. It is only my attempt to prove something I suggested in the comments.) Theorem 1. If $\vec{x}=(x_1,\dots,x_n)$ is a minimum, then at least one of $x_i$ is zero or the entries of $\vec{x}$ take at most two distinct values. The result of the Theorem will follow directly from the following lemma. Lemma 2. If $...


0

In general, if you can show that your optimization problem has continuous and differentiable objective function and constraint functions and satisfies a special technical hypothesis called a constraint qualification (there are many constraint qualifications in the literature) then the Lagrange multiplier condition is a necessary condition for a point $x^{*}$ ...


2

According to the constraints and objective function, it seems that the objective function can get the maximum value on the boundary of feasible solutions, and since it is an open set, I think the objective function does not have any maximum in this case. Now, let replace the last three constraints as given below. $x_2\geq 0$, $x_3\geq 0$, and $x_4\geq 0$...


1

I am adding this just to demonstrate another way to use Lagrange multipliers. Because of the symmetry of the constraint disk $ \ x^2 \ + \ y^2 \ \le \ 1 \ $ , we can also apply the Lagrange-multiplier method through the interior of the disk* by considering the set of concentric circles $ \ x^2 \ + \ y^2 \ = \ c^2 \ $ , with $ \ 0 \ \le \ c^2 \ \le \ 1 \ $ , ...


1

Okay, so let's split this up into two problems: Extrema in $x^2+y^2<1$: $$\nabla f(x,y)=(4x^3-2x,2y)=0$$ So $x=\pm\frac{\sqrt{2}}{2}$ or $x=0$, and $y=0$ Extrema on $x^2+y^2=1$: $$x=\cos\theta$$ $$y=\sin\theta$$ $$\frac{d}{d\theta}\left[\cos^4\theta-\cos^2\theta+\sin^2\theta\right]=4\sin\theta\left(\cos\theta-\cos^3\theta\right)$$ So $\theta \...


5

First we look for candidates in the interior of the disk. Setting the partials equal to $0$, we find that the candidates are $x=0,y=0$ and $x=\pm \frac{1}{\sqrt{2}}$, $y=0$. Next we look for candidates on the boundary of the disk. Lagrange multipliers are fine, or else use the fact that $y^2=1-x^2$ to note that we are maximizing/minimizing $(x^2-1)^2$ on ...


4

Why use Lagrange multipliers for this problem? Putting $x=y+\frac{\pi}{4}$, we have essentially a function of single variable $$\cos(y+\pi/4)^2+\cos^2y=1+\frac{1}{\sqrt2}\cos(2y+\frac{\pi}{4})$$ which has maximum value $1+\frac{1}{\sqrt2}$ and minimum value $1-\frac{1}{\sqrt2}$.


0

OK, first off, I believe you should end up with: $$ \sin(2x) = -\sin(2y) \\ \sin(2x) = \sin(-2y) $$ You need to be very careful with sine because it's not simply true that $2x = -2y \rightarrow y = -x$. First off, the sines are equal not only when the two arguments are equal but also when they differ by $2\pi n$, so we would really have: $2x = -2y+2\pi n \...


0

There's no need to use a Lagrange multiplier. The ellipse $$\{ (x,y) \in \mathbb{R}^2 \mid 2 x^2 + y^2 = 1 \}$$ is parametrized as follows $$x (\theta) = \frac{\sqrt 2}{2} \, \cos (\theta) \qquad \qquad \qquad y (\theta) = \sin (\theta)$$ Hence, $$g (\theta) := f (x (\theta), y (\theta)) = \frac{\sqrt 2}{2} \, \cos (\theta) + \sin^2 (\theta)$$ ...


3

You correctly got the three equations for the Lagrange method. The 2nd equation tells you that $\lambda=1$ or $y=0$. $\lambda=1$ gives you the stationary point you identified, which has $x=\frac{1}{4},y^2=\frac{7}{8}$ and hence $f(x,y)=\frac{9}{8}$. $y=0$ gives $x^2=\frac{1}{2}$ and hence $x=\pm\frac{1}{\sqrt2}$ and $f(\pm\frac{1}{\sqrt2},0)=\pm\frac{1}{\...


1

To find the extreme values, you need to also check the conditions where $f_x=0$ and $f_y=0$, which is the way to find local extremas before using Lagrange. In this case you can find through $f_y=0$, that $y=0$


0

For this problem, you may not use Lagrange multiplier. The objective function is a cone and the constraint is, as you said, circle with center (1,2) and radius $\sqrt5$. So the max-min is on the curve which is the intersection between the cone and the cylinder. You can vision that there are two points on the curve corresponding to min (close to the origin) ...


0

The constraint defines the circle $$(x-1)^2+(y-2)^2=5$$ going through the origin, and you are told to find the two points on this circle which are at maximal, resp., minimal distance from the origin. The nearest point is $(0,0)$, of course, and the farthest point is the other end of the diameter through $(0,0)$, i.e., the point $(2,4)$. It follows that $\...


2

You can divide one equation by the other. For this purpose we assume, for the moment, that the denominators are unequal $0$. $\frac{x}{y}=\frac{x-1}{y-2}$ Multipliying both sides by $y$ and $y-2$ $x(y-2)=y(x-1)$ $xy-2x=yx-y$ Substracting xy on both sides. $-2x=-y$ $2x=y$ Plugging in the term for y into the constraint. $x^2-2x+(2x)^2-4\cdot 2x=0$ $...


1

Let's do it in a few steps. Firstly, there is a solution: you minimize a continuous function over a nonempty compact domain, so a solution exists by the Extreme Value Theorem. Secondly, it is unique by looking at the convexity properties of the goal function and domain. Thirdly, it must satisfy your KKT (or Fritz John) conditions by a standard application ...


1

If you accept without proof that such a curve has a regular parametric representation $$t\mapsto{\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(t\in{\mathbb R}/{\mathbb Z})$$ of period $1$ the claim can be proven as follows: Consider the pullback $$\phi(t):=f\bigl({\bf z}(t)\bigr)\ .$$ Convince yourself by means of the chain rule that $\phi'(t_0)=0$ iff ${\bf z}(t_0)...


1

I think I can give you an intuitive argument. Say you parametrize $\gamma$ by a function $\vec r(t)$ and then consider $f(\vec r(t))$. If the Lagrange method gives you three stationary points, then you have: $$\left.\frac{\partial f(\vec r(t))}{\partial t}\right|_{t=t_i} = 0$$for $i=1,2,3$; i.e. $f$ has stationary points on $\gamma$ in the points ...


0

The Lagrangian for the problem is $$L(x,\lambda) = - \sum_{i=1}^n \ln (a_i + x_i) - \lambda \left(1- \sum_{i=1}^n x_i\right) = - \lambda + \sum_{i=1}^n \left( \lambda x_i - \ln(a_i + x_i) \right).$$ According to the Lagrangian sufficiency theorem (see Theorem 2.1 in [1]), if there exist $\lambda^{\ast}$ and $x^{\ast}$ such that $x^{\ast}_i \geq 0$ $L(x^{\...


4

This is the standard water-filling problem (As an example, see http://www.comm.utoronto.ca/~weiyu/loading_icc.pdf ). You would see this problem in any communications textbook including https://people.eecs.berkeley.edu/~dtse/book.html - Chapter 5, (5.39)-(5.40). Rather than repeating steps, I would direct you to http://www.net.in.tum.de/fileadmin/TUM/NET/...


3

Let $\mathrm{x} := (x_1, x_2, x_3)$ and $\mathrm{y} := (y_1, y_2, y_3)$ be two free points in $\mathbb{R}^3$. The Euclidean distance between them is $\|\mathrm{x} - \mathrm{y}\|_2$. If we impose the constraint that each of these two points lie on each of the given lines, then we have the linear equality constraints $$\begin{bmatrix} 1 & -1 & 0\\ 1 &...


1

I think your calculation of $(Dh)_{x_0}$ is wrong, but I'm having trouble following it so I'm not quite sure. Notice that ${|x-a|}^2$, which I assume is what you call $h_a$, can be written as: $$\langle x, x \rangle -2 \langle x, a \rangle + \langle a, a \rangle=-2 \langle x, a \rangle + \left( 1 +\langle a, a \rangle\right)$$ So that $\frac{\partial}{\...


0

Of course if $x_{i} \in \{1,k\}$, or equivalently $(x_i - 1)(x_i - k) = 0$, is part of the constraints then you won't get any simpler in the KKT system. Theoretically speaking, depending on the other equations of the system, one may obtain a solution to the problem (which is specific) without doing brute force. However, for discrete optimization problems, ...


0

The objective function can be written in the form $$\frac{1}{2} \begin{bmatrix} \mathrm{w}\\ \theta\end{bmatrix}^T \begin{bmatrix} \mathrm{I}_n & 0_n\\ 0_n^T & 0\end{bmatrix} \begin{bmatrix} \mathrm{w}\\ \theta\end{bmatrix} - \begin{bmatrix} \mathrm{w}_t\\ 0\end{bmatrix}^T \begin{bmatrix} \mathrm{w}\\ \theta\end{bmatrix} + \frac{1}{2} \|\mathrm{w}_t\...


0

Solve the problem for fixed $\theta$ as you described; then optimise with respect to $\theta$ without constraints.


0

Let $$f(x) := - \sum_{i=1}^n \ln(\alpha_i +x_i) = - \ln\left(\prod_{i=1}^n \alpha_i + x_i\right)$$ be the objective function. Using the AM-GM inequality, $$\frac{1}{n}\sum_{i=1}^n \alpha_i + x_i = \frac{1}{n}\sum_{i=1}^n \alpha_i + \frac{1}{n}\underbrace{\sum_{i=1}^n x_i}_{=1} \geq \left(\prod_{i=1}^n \alpha_i + x_i\right)^{\frac{1}{n}}$$ Thus, $$\left[ ...


0

You only have equality for equality-constraints. The inequality constraints should be satisfied by, well, inequality. Consider the problem $$\min\limits_{x\in \mathbb{R}^n} f(x)$$ Subject to \begin{align} c_i(x) &= 0, & &i\in\mathcal{E}\\ c_i(x) &\geq 0, & &i\in\mathcal{I}. \end{align} Define the Lagrangian $$\mathcal{L}(...


0

The partial derivatives in the KKT system are with respect to the primal variables (i.e. $x$) and not to the dual variables. So if $\lambda_j$ and $\mu_j$ are the dual variables (or multipliers) respectively corresponding to the constraints $-x_j \le 0$ and $x_j - 1 \le 0$, you will have $-\lambda_j x_j + \mu_j (x_j -1)$ in the Lagrangian, and thus the ...


3

From $f(x,y):=3x^2+y$ we obtain $\nabla f(x,y)=(6x,1)$, hence $|\nabla f(x,y)|=\sqrt{36x^2+1}$. It is obvious that the points on the closed unit disk where the largest value $|\nabla f(x,y)|$ occurs are the two points $(\pm1,0)$, with $\nabla f(\pm1,0)=(\pm6,1)$ and $|\nabla f(\pm1,0)|=\sqrt{37}$. Now at each point the direction of maximal increase of $...


1

We have a quadratic program $$\begin{array}{ll} \text{minimize} & \|\mathrm{x} - \mathrm{k}\|_2^2\\ \text{subject to} & 1_n^T\mathrm{x} = c\end{array}$$ where $c = -1 + \displaystyle\prod_{i=1}^n (k_i+1)$. Using a Lagrange multiplier, $$\mathcal{L} (x,\lambda) := \frac{1}{2}\|\mathrm{x} - \mathrm{k}\|_2^2 + \lambda (1_n^T \mathrm{x} - c)$$ Taking ...


-2

You can rewrite the constraint as: $$0= 1-B +\sum_{i=1}^nx_i$$ and then turn it into a cost function: $$\lambda\left\|1-B+\sum_{i=1}^{n}x_i\right\|_2$$ If we want to we can rewrite this with matrices, $\bf 1$ being the column vector full of ones: $$\min_{\bf x} \left\{ \|{\bf x}-{\bf k}\|_2 + \lambda\|1-B+{\bf 1}^T{\bf x}\|_2\right\}$$ You will ...


0

So after the procedure at end of the day formula to be remembered is: $$ \dfrac{\partial f/ \partial x } {\partial f/ \partial y } =\dfrac{\partial g/ \partial x } {\partial g/ \partial y } = -\lambda. \tag{1} $$ EDIT1: Let us take it further to the last logical step. After all we have to finally express maximum/minimum f in terms of g... Just as, in ...


1

From what you obtained $$\displaystyle \nabla_{x,y,\lambda}\mathcal L(x,y,\lambda) = 0 \Leftrightarrow \begin{cases} x^2 + 2\lambda x & = & 0 \\ 1+2\lambda y & = & 0 \\ x^2 + y^2 - 1 & = & 0 \end{cases}$$ we have, from the first equation, either $x=0$ or $x=-2\lambda$. From the second $y=-\frac{1}{2 \lambda }$. So, replacing in the ...


2

Do the following substitution, $$x_1=\frac1x$$ $$y_1=\frac1y$$ $$z_1=\frac1z$$ Then the problem is transformed to, $$f(x_1,y_1,z_1)=x_1^2+y_1^2+z_1^2$$ s.t. $x_1+y_1+z_1=1$ The constraint is a plane passing (0,0,1), (0,1,0), and (1,0,0). The objective function is a sphere with radius r. So the minimum r is when the sphere tangents to the plane or the ...



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