# Tag Info

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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 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)$$where ... 4 Suppose you want to maximize z=f(x,y) subject to the constraint g(x,y)=c. You've used the method of Lagrange multipliers to have found the maximum M and along the way have computed the Lagrange multiplier \lambda. Then \lambda={dM\over dc}, i.e. \lambda is the rate of change of the maximum value with respect to c. Said another way, you can ... 4 I had not seen this before, it is quite impressive. The argument as I understand it is as follows. Given a symmetric real matrix A, our first plan of action is to find an eigenvector v with a real eigenvalue. We do so as follows, consider the map f: S^{n-1} \to \mathbb{R} via f(x) = \langle Ax , x \rangle. This map is continuous (it is polynomial ... 4 In general when optimizing f(x) subject to g(x)=0, you solve the problem \nabla f(x)=\lambda \nabla g(x) and the critical points can be checked by the bordered Hessian matrix:$$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} ...

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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 ...

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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.

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