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## Hot answers tagged lagrange-multiplier

6

We have $$\frac{x^2}4+\frac{y^2}9=1$$ so, any point can be written as $(2\cos t,3\sin t)$ If the distance if $d,$ $$d^2=(2\cos t-5)^2+(3\sin t-5)^2$$ Now use Second derivative test

4

Beginning with the equation for the Lagrange Multipliers, it is a matter of (tedious) algebraic manipulation with a goal to systematically eliminate parameters. We have $x+y-z=0$ and $\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$, $f(x,y,z)=x^2+y^2+z^2$ $$\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} + ... 3 Lagrange multipliers finds local extrema. On a domain with boundary, global extrema might be on the boundary instead. On an unbounded domain, the objective function might not be bounded above (resp. below), even on the surface where the constraint holds, in which case the maximum (resp. minimum) could be at infinity. In this case, assuming the problem ... 2 You have proved that \lambda=1 was impossible. So from (2):$$ 0 = (\lambda-1)y \implies y = 0; \\ x - 2 = \pm 1 $$which gives the maximizer and minimizer. 2 It's true that some of the x_i will tend to \infty as P does, however, it's not always true that all of them will. Intuitively, because the functions are increasing, the solutions must always lie on the top right-hand face of the simplex$$\Sigma_P:=\left\{x:\sum_i x_i=P\right\}.$$A quick argument for this goes like suppose that a solution y does ... 1 I didn't really see any error in your computation. However you shouldn't drop one constraint. Instead you should keep the two simpler constraints$$g_2(x,y,z)=x^2+y^2-1\\ g(x,y,z)=xy+z^2$$Use two parameters for Lagrange multiplier:$$L=f(x,y,z)+\lambda g_2(x,y,z)+\mu g(x,y,z)$$Then set up:$$\nabla f+\lambda \nabla g_2 +\nabla g=0$$together with the ... 1 The usual Fermat theorem gives you that Df(x)=0 simply because you can test the function in any direction around x. Here you can move only along directions that belong to the tangent space at x to S \cap g^{-1}(0), and you will not conclude that Df(x) must vanish. So to speak, in constrained optimization you lack too many direction to deduce that ... 1 x is an extremum along g^{-1}\{0\}\cap S, not an extremum on S. That means that the gradient of f doesn't necessarily have to be 0, it just has to be perpendicular to g^{-1}\{0\} at that point. What is also perpendicular to g^{-1}\{0\} is the gradient of g. That means the gradient of f and the gradient of g are parallel in the extremal ... 1 You have two equations in three unknowns, so just have to choose one variable to maximize over. When you eliminate z you have to do it from the second constraint as well as the objective function. Your problem becomes to maximize 2x^2+2y^2+2xy subject to x^2/4+y^2/5+(x+y)^2/25=1=\frac {29}{100}x^2+\frac 6{25}y^2+\frac 2{25}xy Now solve the second ... 1 First thing first: you can simplify that r out of the problem since it is homogeneous (but I would keep the 5 since it does simplify the problem). Second thing: f(x) = \ln x + \ln y + 3 \ln z = \ln (xyz^3), and \ln is monotonic, so you should discard it. As Winther said, this is the only stationary point. Your octant is border-less and therefore ... 1 The distance from (x,y,z) to the origin is \sqrt{x^2+y^2+z^2}. We therefore want to minimize \sqrt{x^2+y^2+z^2}, or equivalently x^2+y^2+z^2, subject to the condition x^2+2y^2-z^2-1=0. So x^2+y^2+z^2 should be your f. Now the usual process should work well. 1 How about (x,y,z)=(\cos t,\sin t,-\sin t\cos t) Check the derivative is never the zero vector. 1 You have to make sure when solving the equations that you are not dividing by something that is zero. Having done your substitution for \lambda, you are left with$$ y-x^2=0 \\ 2x=2xy, $$and if x=0, the second equation is satisfied, and then the first equation gives y=0. 1 Note that x^T A^T A x = (Ax)^T (Ax) = \lVert Ax \rVert^2, so this is expressible as a norm. We expect the derivative to be something like 2A^T Ax, which is what you have in the line$$ \frac{\partial L}{\partial x_l} = \delta_{\alpha l} + \lambda a_{jl}(a_{ji} x_i + a_{jk}x_k): $$changing the indices, this is$$ \frac{\partial L}{\partial x_l} = ...

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An alternative approach, if $A$ is invertible: First consider the case when $A=I$. Then the problem is to maximize $u^Tx$ subject to $x^T x\le r^2$, where $u$ is a fixed vector ($e_\alpha$ in your case). By Cauchy-Schwarz, the solution is $x=ru/|u|$. For general invertible $A$, the problem is to maximize $u^Tx$ subject to $x^TA^TAx\le r^2$. Let $y=Ax$. ...

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Heuristic answer (the sort you would see in, e.g., a thermodynamics class): Let $G : \mathbb{R}^2 \to \mathbb{R}$. Consider the surface $G(x,y)=0$. Then to move from a point $(x,y)$ to an infinitely close point $(x+dx,x+dy)$ on the surface, we must have $$dG=\frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy = 0.$$ So we get an implicit ...

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The optimization problem is $$\max_{x,y} 1-rx^2-y^2$$ subject to $$x+y=m$$. So the decision varibles are $\mathbf{x}=(x,y)$ and the parameters are $\mathbf{r}=(r,m)$. That's how we connect the symbols in the definition to the symbols in the problem. It will be useful to define the objective $f(\mathbf{x},\mathbf{r})=1-rx^2-y^2$ and constraint function ...

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