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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

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/...


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 $1-\frac{1}{\sqrt2}$.


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}{\...


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 $...


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 &...


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 ...


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. $x^2-2x+(2x)^2-4\cdot 2x=0$ $...


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 \...


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

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 $y=0$


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 ...


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}{\...


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 ...


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 ...



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