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## New answers tagged lagrange-multiplier

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Assume you want to optimize the system $f(x_1,x_2)$ subject to $g(x_1,x_2)=c$. To assure that the constraing holds $$dg=\frac{\partial g}{\partial x_1}dx_1+\frac{\partial g}{\partial x_2}dx_2=0$$ $$\Rightarrow dx_2=-\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial x_2}}dx_1$$ By definition $f$ is optimized when $$df=\frac{\partial ... 2 The thing is like in the following picture. of wikipedia. For f=d you increment d until you touch g=c. In the moment of contact you take a minimum. If you go on, just before f=d leaves the contact, you take the maximum. Thinking it well it is like parking! Really, the idea is so productive that is the base of the Morse's theory. 1 The level curves of f represent single values of f that increase in a direction parallel to the gradient. This means that, given a level curve that does not represent a local maximum, there is another level curve nearby whose value for f is greater than the first curve. Imagine g as a curve that cuts through a level curve of f at a point p. ... 1 Parametrize the curve g(x) = 0 with c(t) s.t c(0) = p where p is the local extrema of f, c'(0) \not = 0. Then you know that f(c(t)) has local min/max when t = 0 i.e;$$\frac{d}{dt} f(c(t)) |_{t=0} = \nabla f(p) \cdot c'(0) = 0$$We also know that \nabla g(p) \cdot c'(0) = 0 and so there exists a non-zero scalar \lambda s.t;$$\nabla f(p) ...

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$(2/\sqrt{11})(3,1,-1)$ is the closest point and $(-2/\sqrt{11})(3,1,-1)$ is the most distant point. If you have to use Lagrange multipliiers.... minimize/maximize: $(x-3)^2 + (y-1)^2 + (z+1)^2$ (this is the distance squared from x,y,z to your point.) constrained by: $x^2 + y^2 + z^2 = 4$ $F(x,y,z,\lambda) = (x-3)^2 + (y-1)^2 + (z+1)^2 - \lambda(x^2 + ... 4 Do you have to use Lagrange multipliers? The sphere is centered at the origin, and the point lies outside it. If you find the line through$(3,1,-1)$and the origin, the closest and farthest points will be the two points of intersection between the line and the sphere. Seeing that the length of the vector from the origin to the point of interest is ... 1 What's wrong is that$x^2 + y^2 - 1 = 0$is not the intersection of your two constraints. In fact it has only two points in common with that intersection. 1 Here's a different way to see that Lagrange multipliers are "intuitively" the correct way to go. To first order, the change in$f$due to a change in$\vec r$is simply:$$\delta f = f(\vec r + \delta\vec r) - f(\vec r) \approx \nabla f\cdot\delta\vec r.$$This has the obvious interpretation for$f$above, "go in the direction of$\nabla f$to increase$f$," ... 3 Look at the tangent space to$S = g^{-1}(c)$at$x_0$, say$S_{x_0}$. Then$S_{x_0} = \left[\nabla g(x_0) \right]^{\perp}$. Therefore,$S_{x_0}^\perp$is the one-dimensional subspace spanned by$\nabla g(x_0). Now, \nabla f(x_0) = \lambda \nabla g(x_0)\text{ for some } \lambda \in \Bbb{R} \iff \nabla f(x_0) \in S_{x_0}^\perp \iff \nabla f(x_0).v = 0 ... 15 Here's some intuition. First suppose that x_0 is a local minimizer for the problem \begin{align} \tag{\spadesuit} \text{minimize} & \quad f(x) \\ \text{subject to} &\quad Ax = b. \end{align} If Au = 0, then the directional derivative D_u f(x_0) = \langle \nabla f(x_0),u \rangle is nonnegative. Otherwise, it would be possible to improve ... 4 Write \alpha_i^2=\alpha_i^{\frac{1}{2}}\alpha_i^{\frac{3}{2}}, and then use the Cauchy-Schwarz inequality. 0 We have that f(x,y,z) \ge 0 for all (x,y,z). Now f(0,r/{\sqrt 2}, r/{\sqrt 2}) = 0, hence the minimum of f is 0. I am not sure what you mean by x,y,z \neq 0, but if it is " x \neq 0 and y \neq 0 and z \neq 0", then it is incorrect. What we know is that at least one of them should be \neq 0, but this is not important. From the system you ... 0 First of all your transposing of the constraint is not right. 720=30x+y\Rightarrow 720-30x-y=0. Therefore the lagrange function is g(x,y,λ)=240\sqrt{x}+y+λ(720-30x-y) Then the partial derivatives come next. Pay attention on the corresponding symbols. {\frac{\partial f}{\partial x}} = 120{x^{-1/2}}-30\lambda=0 {\frac{\partial f}{\partial y}} = ... 0 For the plane we have x + z = 1 \iff \\ (1,0,1) \cdot (x,y,z) = 1 \iff \\ (1/\sqrt{2}) (1,0,1) \cdot (x,y,z) = 1/\sqrt{2} $$The points on the ellipse have the coordinates$$ (x, \pm \sqrt{1-x^2}, 1-x) \quad (x\in I=[-1,1]) $$and distance from the origin$$ d(x) = \sqrt{x^2 + 1 - x^2 + (1-x)^2} = \sqrt{x^2 - 2x + 2} $$Looking for local extrema$$ 0 = ... 0 What you need to look at are the Karush-Kuhn-Tucker conditions. EDIT: In your example, the non-negativity constraintsx_1 \ge 0$and$x_2 \ge 0$(I hope you did mean$\ge$, not$>$, otherwise you might not get any minimum) each will get a Lagrange multiplier, which is required to be nonnegative (while the Lagrange multiplier for an equality constraint ... 0 You have 5 equations and 5 unknowns and you have to find all solutions to the system. Some solutions will correspond to local maxima, other to local minima, and the remaining ones to local inflection points. Then, among the solutions that you found, you have to evaluate f(x) and see which one has highest value (or use a Hessian test which is not necessary ... 2$\cos^2 x+\sin^2 x=1is usually a powerful formula. Since \begin{align} (2\lambda x)^2+(2\lambda y)^2 &= 4\lambda^2\\ &=e^{2x}, \end{align} \begin{align} -e^x \sin y + 2\lambda y &= e^x (\pm y-\sin y). \end{align} Sincee^x>0$,$\pm y -\sin y =0$and$y=0$is the unique root. Therefore, given function has two extreme point,$(1,0)$and ... 0 I do not see strange results. Using the derivatives with respect to$x,y,z$, you then have $$x=-\frac{1}{2 \lambda }\qquad y=-\frac{1}{\lambda }\qquad z=-\frac{3}{2 \lambda }$$ So $$L_\lambda=\frac{7}{2 \lambda ^2}-1=0$$ which makes $$\lambda=\pm \sqrt{\frac{7}{2}}$$ Replacing, you then have $$x=\pm \frac{1}{\sqrt{14}}\qquad y=\pm\frac{2}{\sqrt{14}}\qquad ... 0 x+2y+3z\leq\sqrt {x^2+y^2+z^2}\sqrt {1+4+9} = \sqrt {14}, by Cauchy-Schwarz, so \max occurs at x = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{1}{\sqrt{14}} and \min occurs at x = \dfrac{y}{2} = \dfrac{z}{3} = -\dfrac{1}{\sqrt{14}}. You should get a same answer if you solve the Lagrange equations correctly. 1 The geometrical picture is the following: We are asked to find the local extrema of the distance from the point (0,b) on the y-axis to points on the parabola y=x^2. From looking at a figure we can guess the following: If b\gg1 there are two local minima high up, and a local maximum at (0,0). If 0<b\ll1 there is just one local minimum at ... 1 The critical point (0,b) should not be included as an extrema, since the constraint is y=x^2, which is the boundary of the region, whereas (0,b) in inside the region y=x^2, so it should not be included. The way we use to determine local max or local min is through Second Derivative Test. Suppose the second derivative of f(x,y) are continuous on a open ... 1 Hint: substitute the values of t and for (3) use the magnitude to find the distance. 1 Your first equation -2x=\lambda\cdot 2x implies that either x=0 or \lambda=-1 (or both). Your second equation 2y=\lambda\cdot 2y implies that either y=0 or \lambda=1 (or both). Combining those two, you have a total of four possibilities. Look at each possibility and find the possible points (if any) for that possibility. Then look at the ... 1 The Lagrange multiplier condition tells you that \nabla f(x_0) = \sum_{i=1}^m \lambda_i\nabla g_i(x_0) is normal (orthogonal) to the affine subspace M. This means that \nabla f(x_0)\cdot (y-x_0) = 0 for any y\in M. Thus, since f is convex, for any y\in M we have$$f(y)\ge f(x_0)+\nabla f(x_0)\cdot (y-x_0) = f(x_0).$$This means x_0 is a global ... 1 The theory of Lagrange multipliers says: Under the given circumstances there is a value \lambda\in{\mathbb R} such that at the conditionally stationary point x the equations$$2Lx-\lambda d=0,\qquad x^Td=k\tag{1}$$are simultaneously true. Now$(1)$is a system of$N+1$equations in$N+1$unknowns, one of them$\lambda$. If the situation is not ... 1 There is a second derivative test for constrained extrema. Here's the statement for the case of one constraint$g(x)=c$. Suppose$a$is a constrained critical point of$f$on the constraint set$g(x)=c$. Then we have$\nabla f(a)=\lambda\nabla g(a)$for some scalar$\lambda$. Consider the Hessian matrix$H$(matrix of second partial derivatives) of ... 0 I find it easier to understand the method of Lagrange multipliers by looking at$f$and$g$separately. The constraint is expressed as$g(x,y)=\text{const}$, i.e., as a level curve of$g$. If we look at how$f$changes along one of these curves, we see that at a stationary point, the directional derivative along a tangent to this curve vanishes, just as ... 0 Generally, a major part of such problems is establishing that a solution exists. The problem doesn't have a solution in the usual sense. However,$\inf \{ \sum_k p_k^3 | \sum_k p_k =1, p_k \ge 0 \} = 0$. To see this, note that the cost always non negative, and taking$p_1=\cdots = p_n = {1 \over n}$and$p_k = 0$for$k >n$, we have$\sum_k p_k^3 = {1 ...

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For those interested I have asked for help in a different place and was able to reach a solution. Both @Olorin and @Ido have found the same solution. The hint that allowed me to solve this was to try and maximize $\ln f$ instead of $f$. The complete solution is: Let $g\left(x,y,z\right)=x^{k}+y^{k}+z^{k}-1$. We will search for the maximum of $\ln f$ ...

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