Tag Info

Hot answers tagged

10

Let $\displaystyle g(x,y,z)=\frac{x}{y^3+54}+\frac{y}{z^3+54}+\frac{z}{x^3+54}$. If we consider $g(a,b,1-(a+b))$ we can use $\partial_ag=0$ and $\partial_bg=0$ to numerically find the following critical points (up to cyclic permutation): $$\begin{array}{lll|l} \text{a} & \text{b} & \text{c} & \text{g(a,b,c)} \\ \hline 1.20836 & -0.608416 ...


7

Hint: Test for Extrema Let $f$ be a function of two variables that has continuous second partial derivatives on a rectangular region $Q$ and let: $$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$$ for every $(x,y)$ in $Q$. If $(a,b)$ is in $Q$ and $f_x(a,b) = 0, f_y(a,b) = 0$, then: (i) $f(a, b)$ is a local maximum of $f$ if $g(a,b) \gt ...


5

Consider a point $p$ in the common domain $\Omega\subset{\mathbb R}^n$ of $f$ and the constraints $$g_k(x)=0\qquad(1\leq k\leq r)\ .\tag{1}$$ The gradients $\nabla g_k(p)$ define a subspace $U$ of allowed directions when walking away from $p$. In fact a direction $X$ is allowed only if it belongs to the tangent planes of all level surfaces $(1)$. This means ...


5

actually it is same,because we can consider signs as a alternatives of maximize or minimize,so you can use it without any problem. http://en.wikipedia.org/wiki/Lagrange_multiplier maybe also author's definition plays some role as well.so i think there is not big difference between $+$ sign ad $-$ sign in this case i found you case ...


5

I think you have some problems, because you use an incorrect notation. Let me rewrite your original problem: \begin{align*} \text{Minimize}\quad & J(y) = \int_{x_0}^{x_1} F(x, y(x), y'(x)) \, \mathrm{d} x \\ \text{such that}\quad & G(x, y(x), y'(x)) = 0 \quad\text{for all } x \in [x_0,x_1]. \end{align*} Here, $F : \mathbb{R} \times \mathbb R \times ...


4

Maximize $g$ ignoring the constraint. If the solution fulfills the constraint, you're done. If not, there's no maximum, since it would have to lie on the boundary, but the boundary is excluded by the constraint.


4

Here is a complete answer. The computations are rather long but every step is natural. Perhaps someone else can simplify the computational part of the proof. We will show that the maximum is $M=\frac{3}{2}-\sqrt{2}$ independently of $n$, just as claimed in Macavity's comments. Let $\phi(x)=1-\sqrt{x}$ for $x\in [0,1]$. The inequality to be shown can then ...


4

First of all just delete y. Modify the constraints and use dummy variables to get rid of functional inequalities $$g_1(x)+s_1=-(x-3.0)^2+s_1=-1$$ $$g_2(x)+s_2=-(x-5.3)^2+s_2=-1$$ $$g_3(x)+s_3=-(x-7.0)^2+s_3=-1$$ $$s_1,s_2,s_3\ge 0$$ Construct your Lagrangian $$Z=L(x)+\lambda_1(r_1-g_1(x)-s_1)+\lambda_2(r_2-g_2(x)-s_2)+\lambda_3(r_3-g_3(x)-s_3)$$ where ...


4

Suppose you want to maximize $z=f(x,y)$ subject to the constraint $g(x,y)=c$. You've used the method of Lagrange multipliers to have found the maximum $M$ and along the way have computed the Lagrange multiplier $\lambda$. Then $\lambda={dM\over dc}$, i.e. $\lambda$ is the rate of change of the maximum value with respect to $c$. Said another way, you can ...


4

I had not seen this before, it is quite impressive. The argument as I understand it is as follows. Given a symmetric real matrix $A$, our first plan of action is to find an eigenvector $v$ with a real eigenvalue. We do so as follows, consider the map $f: S^{n-1} \to \mathbb{R}$ via $f(x) = \langle Ax , x \rangle$. This map is continuous (it is polynomial ...


4

In general when optimizing $f(x)$ subject to $g(x)=0$, you solve the problem $\nabla f(x)=\lambda \nabla g(x)$ and the critical points can be checked by the bordered Hessian matrix: $$H=\begin{pmatrix} 0 & g_x & g_y\\ g_x & f_{xx}+\lambda g_{xx} & f_{xy}+\lambda g_{xy}\\ g_y & f_{yx}+\lambda g_{yx} & f_{yy}+\lambda g_{yy} ...


3

The circle can be parametrized as $$ x = 2 + \frac{4}{\sqrt 5} \sin \theta, \; y = 2 + 2 \cos \theta, \; z = 1 + \frac{2}{\sqrt 5} \sin \theta. $$ The squared distance of such a point from the origin is $$ f(\theta) = 13 + 4 \sqrt 5 \sin \theta + 8 \cos \theta. $$ Derivative is $$ f'(\theta ) = 4 \sqrt 5 \cos \theta - 8 \sin \theta. $$ So, the two ...


3

No one can be zero. So, in that surface, $z^2=2/xy$. Now, by AM-GM $$x^2+y^2+z^2=x^2+y^2+\frac{2}{xy}\geq2xy +\frac{2}{xy}\geq 2\sqrt{4}=4.$$ Study the conditions for the equality to happen and show that they actually happen for the points you already suspect.


3

$\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{\expo}[1]{\,{\rm e}^{#1}\,} ...


3

It is usually understood that $f_0(x)=+\infty$ for $x\notin D$. This would make both definitions equivalent.


3

First three equations lead to: $$\lambda = \frac{x-1}{x} = \frac{y-2}{y} = \frac{z-2}{z}.$$ Now subtracting 1 on every term implies $$\lambda-1 = -\frac{1}{x} = -\frac{2}{y} = -\frac{2}{z}.$$ Now you found a relation of $x,y$ and $z$: $$ 2x = y = z. $$ I believe you can take it from here.


3

Your description of Lagrange multipliers, and of the alternative quadratic penalty method, is a bit off. Rather than jointly minimizing over $(x,y,\lambda) \in \mathbb{R}^3$, we typically want to fix $\lambda\in\mathbb{R}$ and then minimize over $(x,y) \in \mathbb{R}^2$. The solution will be parameterized by $\lambda$, so we then appropriately choose ...


3

I don't have time to work out all the details in an answer, but here's a quick starter. The key idea behind Lagrange multipliers is that when two surfaces are tangent to each other, their normal vectors at that point are parallel. In this case, you want to find when the surface $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ is closest to the origin, that is, ...


3

Since $M$ is unitary you can write $$\Vert MA-B\Vert_F^2=\Vert MA\Vert_F^2-2\langle MA,B\rangle_F+\Vert B\Vert_F^2\\ = \Vert A\Vert_F^2-2\mathrm{Re}\langle M,BA^*\rangle_F+\Vert B\Vert_F^2.$$ Therefore, your optimization reduces to maximizing $\mathrm{Re}\langle M,BA^*\rangle_F$. Using H$\ddot{\text{o}}$lder inequality and SVD of $BA^*=USV^*$, it's ...


3

Since you said non-negative, I am going to assume that zero is allowed in which case I found a minimum value of 0.0184826 achieved at $a=0, b=0.748545, c=0.251455$ up to their cyclic permutations. After posting this I see that Oleg567 pointed to this solution already. Furthermore, if you want all $a,b,c$ to be strictly positive then it looks like the ...


3

$x^2+y^2 = 1$ so we can say $x = \sin \theta, y = \cos \theta$. Then we need to maximize and minimize $\frac{1}{3 \sin^3 \theta} + \cos \theta$ which can be easily done with calculus.


3

Hint: You want to minimize $$ f(x, y, z) = x^2 + y^2 + z^2 $$ while under the constraint $$ g(x,y,z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 $$ So you should set $$ \nabla f = \lambda \nabla g $$ i.e. $$ (2x, 2y, 2z) = \lambda \left(-\frac{1}{x^2}, -\frac{1}{y^2}, -\frac{1}{z^2} \right). $$


3

It doesn't matter which one you use, the $\lambda$ you get when solving the system of equations will just be the negative of the other method.


3

By the Implicit Function Theorem, near $P$ you can represent your level surface as a graph, say $z=\phi(x,y)$, where $\phi$ is continuously differentiable. If $P=\phi(a,b)$, take any line through $(a,b)$ and you get a nice curve.


3

You found $x=-\lambda/2$ and $y=-\lambda/3$. Now plug this into the constraint and obtain $$24=2x+3y=-\lambda-3\lambda\ .$$ It follows that necessarily $\lambda=-6$, so that you obtain $x=3$, $\>y=2$. So there is a unique conditionally stationary point $P=(3,2)$. In order to show that $f(P)=36$ is indeed the global maximum of $f$ for the given ...


3

Without loss of generality, we can assume that $A$ is symmetric since $f$ is a homogeneous quadratic form (if $A$ is not symmetric, note that $f(u)=\frac 12u^t\cdot (A+A^t)\cdot u$). Therefore $\nabla f=2 A\cdot u$. Let $V$ denote the variety $u_1+\dots+u_n=u_1^2+\dots+u_n^2-1=0$. Note that $V$ is smooth and compact, hence the minimum of $f$ is reached at ...


3

${\bf 1\ }$ We begin with the following two-dimensional problem: Maximize $$f(q_1,q_2):=(1-q_1)(1-q_2)$$ under the constraints $$q_1^2+q_2^2=r^2, \quad q_1\geq0, \ q_2\geq0\ .$$ Here $r$ is a parameter, $0<r\leq1$. Analyzing the graphs of the functions $$g_r(t):=(1-r\cos t)(1-r\sin t)\qquad(0\leq t\leq{\pi\over2})$$ we find the following (see the ...


3

Start with the equations that you have derived: \begin{eqnarray*} ye^{xy}&=&3\lambda x^2,\\ xe^{xy}&=&3\lambda y^2,\\ x^3+y^3&=&16. \end{eqnarray*} As a first step, show that none of $\lambda, x$ or $y$ can be zero. (If one of them is zero, then the first two equations show that all three must be zero, contradicting the third ...


3

Let $f(x,y):=x^3+y$ and $g(x,y):=y$. Then $\nabla f(0,0)=\nabla g(0,0)=(0,1)$, but $f(x,0)-f(0,0)=x^3$ assumes both signs in the immediate neighborhood of $(0,0)$.


2

It's not $f$ that has a local extremum at $x_{1}$, but rather $f|_{S}$. Consider, for example, $f(x,y)=xy$ restricted to the line $y=1-x$. This restricted function has a local maximum at $(\frac{1}{2},\frac{1}{2})$, but the full function $f$ is not at an extremum at $(\frac{1}{2},\frac{1}{2})$.



Only top voted, non community-wiki answers of a minimum length are eligible