# Tag Info

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)]^2for 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 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 ... 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 ... 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 inequalitiesg_1(x)+s_1=-(x-3.0)^2+s_1=-1g_2(x)+s_2=-(x-5.3)^2+s_2=-1g_3(x)+s_3=-(x-7.0)^2+s_3=-1s_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)$$where ... 4 In general when optimizing f(x) subject to g(x)=0, you solve the problem \nabla f(x)=\lambda \nabla g(x) and the critical points can be checked by the bordered Hessian matrix:$$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} ...

4

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 ...

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

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

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

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

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

${\bf 1\ }$ 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

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 ...

2

Have a look for SDP (semidefinite programming). Your problem is \begin{align*} \text{Minimize} \quad &f(Q)\\ \text{such that} \quad & Q \succeq 0 \\ \text{and} \quad &\mathrm{trace}(Q \, A ) \le P\end{align*} You can associate the Lagrangian $$L(Q, R, \lambda) = f(Q) + \mathrm{trace}(Q \, R) + \lambda \, \mathrm{trace}(Q \, A ),$$ for $R ... 2$(xf(t)-1)^2\ge 0,x\in R\Rightarrow x^2f(t)^2-2xf(t)+1^2\ge0\displaystyle\Rightarrow \int_0^1(x^2f(t)^2-2xf(t)+1)dt\ge0\displaystyle\Rightarrow x^2\int_0^1f(t)^2dt-2x\int_0^1f(t)dt+1\int_0^1dt\ge0$This is a quadratic in x which is always greater than or equal to zero. So we must have its discriminant to be less than or equal to zero (otherwise ... 2 Well, your problem is one-dimensional, so you don't need a Lagrangian. The only possible extremal points are those where$y'(x)=0$, and the end-points$0$and$2\pi/3$. So you don't take any derivatives for the end-point, you simply compute the values. You would need a Lagrangian for the boundaries, but only in more dimensions, for example when optimizing a ... 2 Addendum: I've included the full set of solutions Mathematica finds for the case$n=3$at the end of this answer. The comments and answers so far don't show how Lagrange multipliers can solve the problem (because of the non-negativity condition). Here's how that technique can be used. To maximize a function$f(q_1,\dots, q_n)with the requirement that ... 2 You can learn the method of Lagrange multipliers here and here. Maybe, the second link is more useful for you because it is using a problem-solving approach. In your case, you have: $$f(x,y)=6x+\frac{96}{x}+\frac{4y}{x}+\frac{x}{y}$$ and $$g(x,y)=x+y=6.$$ Now, applying the method of Lagrange multipliers. $$\Lambda(x,y,\lambda)=f(x,y)-\lambda(g(x,y)-6). ... 2 You want to maximize the function f(x,y)=x-2y on the part of the parabola y=10-x^2 within the first quadrant. Substituting we have the function f(x,10-x^2)=2x^2+x-20, and we are only considering values of x from 0 to \sqrt{10}. You've already shown there is no local extrema within these values, so we only check the endpoints. f(0,10)=-20 and ... 2 Use Lagrange multiplier as you tried.$$ \begin{align*} f(a_1, \cdots, a_n) &= \sigma^2 \sum a_i^2 \\ g(a_1, \cdots, a_n) &= 1 - \sum a_i \\ F(a_1, \cdots, a_n; \lambda) &= f - \lambda g \end{align*} $$Partial derivatives are$$ \begin{align*} \frac{\partial F}{\partial a_j} &= 2 \sigma^2 a_j - \lambda \\ \frac{\partial F}{\partial ... 2 First, it's sort of crazy to apply Lagrange multipliers in dimension1$. But, yes, the hypothesis you're missing is that$\nabla g(x)\ne 0$(here$g'(x)\ne 0$) for all$x$in the constraint set. (In general, if there are$m$constraints$g_1(x)=c_1$,$\dots$,$g_m(x)=c_m$, the hypothesis is that$\nabla g_i(x)$,$i=1,\dots,m, are all linearly independent ... 2 This is straightforward: your Lagrangian is $$L(x,y,z,\lambda) = 2x+y-z -\lambda(4x^2 + 2y^2+z^2 - 40).$$ Now just write that all the partial derivatives are 0, and you are done: $$0 = \frac{\partial L}{\partial x} = 2-8\lambda x\implies \lambda = \frac 1{4x}; \\ 0 = \frac{\partial L}{\partial y} = 1-4\lambda y\implies \lambda = \frac 1{4y}; \\ 0 = ... 2 The volume of a pyramid (of any shaped base) is \frac13A_bh, where A_b is the area of the base and h is the height (perpendicular distance from the base to the opposing vertex). In this particular case, we're considering a triangular pyramid, with the right triangle OAB as a base and opposing vertex C. The area of the base is \frac12ab, and the ... 2 Define$$H(x,y,\lambda):=x^ae^{-x}y^be^{-y}+\lambda(x+y-1)\Longrightarrow\begin{align*}H'_x&=y^be^{-y}x^{a-1}e^{-x}\left(a-x\right)+\lambda=0\Longrightarrow \lambda=-y^be^{-y}x^{a-1}e^{-x}\left(a-x\right)\\H'_y&=x^ae^{-x}y^{b-1}e^{-y}\left(b-y\right)+\lambda=0\Longrightarrow ... 2 By using Lagrange multipliers or the KKT conditions, you transform an optimization problem ("minimize some quantity") into a system of equations and inequations -- it is no longer an optimization problem. The new problem can be easier to solve. It is also easier to check if a point is a solution. But there are also a few drawbacks: for instance, it only ... 2 (I guess you're missingdt$in the constraint integral? -Also, it has to be that$\zeta_t$is not the symbol of a function, but a separate entity multiplying the parnethesis that follows it). There are a million things that must be assumed as premises to do straightforward Lagrange optimization here, but let's say that they are indeed assumed. In that ... 2 This looks like a usual problem for Lagrange multipliers. You define a Lagrange function via $$L(x,y,\lambda)= x+2y + \lambda \cdot (x^2+y^2-80)$$ Now you look for critical points of the function$L$. But here is the lagrange part more explicit. A critical point is a point where the gradient of$L\$ is zero, as \begin{align*} \frac{\partial L}{\partial x} ...

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