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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): $$\begin{array}{lll|l} \text{a} & \text{b} & \text{c} & \text{g(a,b,c)} \\ \hline 1.20836 & -0.608416 ... 6 Hint: Test for Extrema Let f be a function of two variables that has continuous second partial derivatives on a rectangular region Q and let:$$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2$$for every (x,y) in Q. If (a,b) is in Q and f_x(a,b) = 0, f_y(a,b) = 0, then: (i) f(a, b) is a local maximum of f if g(a,b) \gt ... 5 actually it is same,because we can consider signs as a alternatives of maximize or minimize,so you can use it without any problem. http://en.wikipedia.org/wiki/Lagrange_multiplier maybe also author's definition plays some role as well.so i think there is not big difference between + sign ad - sign in this case i found you case ... 4 In general when optimizing f(x) subject to g(x)=0, you solve the problem \nabla f(x)=\lambda \nabla g(x) and the critical points can be checked by the bordered Hessian matrix:$$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 ...

4

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, $F : \mathbb{R} \times \mathbb R \times ... 3 I don't have time to work out all the details in an answer, but here's a quick starter. The key idea behind Lagrange multipliers is that when two surfaces are tangent to each other, their normal vectors at that point are parallel. In this case, you want to find when the surface$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$is closest to the origin, that is, ... 3 Start with the equations that you have derived: \begin{eqnarray*} ye^{xy}&=&3\lambda x^2,\\ xe^{xy}&=&3\lambda y^2,\\ x^3+y^3&=&16. \end{eqnarray*} As a first step, show that none of$\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 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 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 ... 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 Another example to supplement that of Sharkos:$f(x,y)=x+y$and$g(x,y)=\sqrt{x}+\sqrt{y}-1.$Then the only critical point found by LaGrange is at$(1/4,1/4)$where$f=1/2$, but the max of$f=1$occurs at both points$(1,0),(0,1)$. The constraint region here is the inwardly bent curve$y=(1-\sqrt{x})^2$for$0 \le x \le 1$and has$(1,0),(0,1)$as its ... 2 (I guess you're missing$dt$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$Lis zero, as \begin{align*} \frac{\partial L}{\partial x} ... 2 The functionf$is continuous on the disk$D=\{(x,y)\in\mathbb R^2| x^2+y^2\leq 1\}$, which is a compact subset of$\mathbb R^2$. By the theorem of Weiserstrass,$f$has absolute maximum and minimum on$D$. So you begin searching for stationary points, i.e. solutions of$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}=0$inside the disk, i.e. ... 2 As noted in several comments, the second equation yields$x=y^{3}$. Hence, by the third equation, $$y^{12}=3y^{4} \implies y^{8}=3$$ Assuming$y$is real, we get$y=\pm 3^{1/8}$and$x=\pm3^{3/8}$. By the first equation, then, (taking the positive roots): $$1+4\lambda3^{9/8}-4\lambda3^{1/8}=0 \implies 8\lambda3^{1/8}=-1$$ So$\lambda=\frac{-1}{8}3^{-1/8}$. ... 2 For (b), suppose that$x,y\in\Bbb R_{>0}$, and let$k=xy$; from (a) you know that$x+y\ge 2\sqrt{k}$and hence that$\frac{x+y}2\ge\sqrt{xy}$. For (a) you don’t really need anything as fancy as Lagrange multipliers. You have$y=\frac{k}x$, so$x+y=x+\frac{k}x$; its derivative with respect to$x$is$1-\frac{k}{x^2}$, which is increasing on$x>0$and ... 2 You've done the hard part. Now check the values of$f(x,y)$at the two points$(x,y) = (\pm 2/\sqrt{5}, \pm 1/\sqrt{5})$. Since $$f(\pm 2/\sqrt{5}, \pm 1/\sqrt{5}) = 2(\pm2/\sqrt5) \pm1/\sqrt5 = \pm\sqrt{5},$$ you can conclude that$f$is maximized at$(2/\sqrt5,1/\sqrt5)$and minimized at$(-2/\sqrt5,-1/\sqrt5)$. Edit: To clarify, the reason this is enough ... 2 Your setup is fine. This sort of problem will not (usually) have an analytic solution. You have a two-dimensional non-linear minimization problem. There are many numeric routines that can solve this in libraries, and they are discussed in any numerical analysis text. They really consist of informed trial and error, where the informed part comes from ... 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 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 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 You want to maximize the functionf(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 When you solve for$\lambda$in the first FOC, you should arrive at$\lambda=q_2/p_1$. You made a mistake there. Substituting into the second equation gives $$q1-q_2/p_1*p_2=0\Leftrightarrow q_1=p_2q_2/p_1.$$ Substitute this into the third equation to get $$I-q_2p_2-p_2q_2=0\Leftrightarrow q_2=I/(2p_2).$$ This is the optimal demand for good 2. Now substitute ... 2 Your last equation doesn't follow from anything you'd written above; certainly not from the equation above it, which is a scalar equation and thus can only determine at most one degree of freedom of$\hat{\mathbf w}$. If you want to get a vector equation such as the last equation, you have to solve the equation$\partial J(\mathbf w)/\partial\mathbf ...

2

It is not difficult to verify that the largest box in the unit sphere is the cube with sides $2/\sqrt3$, whence for such a sphere you get the volume $8/3\sqrt3$. Now notice that $f:\mathbb R^3\to\mathbb R^3$, $(x,y,z)\mapsto(ax,by,cz)$ maps unit sphere to your ellipsoid, and $V(f(X))=abc V(X)$ where $V$ is the volume and $X$ is any set. Moreover, $f$ brings ...

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 $(\frac{1}{2},\frac{1}{2})$.

2

A quick and cheap way to do this problem is to exploit the symmetry. Using Lagrange's method will yield a system of equations that is symmetric with respect to any permutation of $x,y,$ and $z$, so we can assume that $x=y=z$. It follows that they are all equal to $3$. To actually use the method, setting $\nabla(d^2)=\lambda\nabla g$ gives: ...

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