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Regarding your equivalence, The direction $\impliedby$ is easy. Suppose $a$ and $b$ are parallel with $a=cb$ and $a'd<0$, where $c$ is a scalar and $a,b,d$ are vectors. Then $a'd=cb'd=0$, but this contradicts $a'd<0$, so there is no $d$ satisfying the conditions. The direction $\implies$ can by using the fact that $a'd = ||a||\: ||d||\cos\theta$ ...

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hint: $f(x,y)=\dfrac{(\frac{1}{2}+y)x+(y+1)+1}{(2+y)x+(y+1)}=1+\dfrac{1-\frac{3}{2}x}{(2+y)x+(y+1)}$ so it is a mono decreasing function for $x \implies f_{min}=f(y(y+1),y)=f_1(y)$ now use same method to check if $f_1(y)$ is mono decreasing function or not.thenthe answer is easy to get.

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i think first you should obtain grad of f then grad f=0 if you find x,y that Applies to domain its minimum else using lagrangian method.

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I think I might have arrived to at least part of the answer, but comments will still be highly appreciated. Consider the following LP: $$\max cx\\ a_ix\leq b_i,\quad i=1,\ldots,m\\ x\geq 0$$ The Lagrangian relaxation problem consists of finding, for $\lambda\geq 0$: $$ZLR(\lambda)=\max_x ... 0 You have to minimize: \mathcal L(K|L=1)= w\cdot 1+rK+ \lambda \left(\overline Y - C \cdot K^{\alpha} \cdot 1^{1-\alpha} \right )=w+rK+ \lambda \left(\overline Y - C \cdot K^{\alpha} \right ) \overline Y is the given output (output constraint). Now calulate \frac{\partial \mathcal L}{\partial K}=0,\frac{\partial \mathcal L}{\partial \lambda}=0 ... 1 If N and P are not too large, and @Macavity is correct about the solution minimizing the largest deviation of v_i+\alpha_i^2, the problem appears to be rather easily solved using mixed-integer linear programming. However, I do not always get the correct solution. I've developed a test below in the MATLAB Toolbox YALMIP (disclaimer, developed by me) ... -1 With the lagrange multipliers, I get : \forall i \ \ v_i=\frac{(P+\sum\alpha_i^2)}{N}- \alpha_i^2. What do you think ? 0 Let g(x,y,z)=x^2+y^2+z^2 and h(x,y,z)=x-y+z. Setting f_x=\lambda g_x+\mu h_x, \;\;f_y=\lambda g_y+\mu h_y, \;\;f_z=\lambda g_z+\mu h_z \;\;gives \;\;6x=\lambda(2x)+\mu(1)=2\lambda x+\mu \;\;2y=\lambda(2y)+\mu(-1)=2\lambda y-\mu \;\;6z=\lambda(2z)+\mu(1)=2\lambda z+\mu. Subtracting the 3rd equation from the 1st gives 6(x-z)=2\lambda(x-z), ... 0 We have three functions: f=3x^2+y^2+3z^2 (function to be maximized/minimized) g=x^2+y^2+z^2-6 (constraint 1) h=x-y+z (constraint 2) Take the partial derivatives of the three functions, and make a system of equations: \begin{cases}f_x=g_x\lambda+h_x\mu \\ f_y=g_y\lambda+h_y\mu \\ f_z=g_z\lambda+h_z\mu \end{cases} Solve this system for x, y ... 1 If \lambda isn't negative, then for large x we have f(x)\sim x^r with r=\frac{\alpha-1}{2}> 0. So f(x) doesn't vanish at infinity and as such the PDF cannot be normalized to have total probability 1. If \lambda is negative, then f(x) vanishes when x^2=-\lambda/\mu. Since probability densities must be nonnegative, the definition of ... 1 You have to be careful when solving the system$$2x(\lambda+1)=0,\quad y-b=8\lambda,\quad x^2-16y=0\ .\tag{1}$$According to the first equation x=0 or \lambda=-1. From x=0 we get y=0 and then \lambda=-{b\over8}. It follows that (0,0) is a conditionally stationary point of f in any case. When \lambda=-1 we get y=b-8 and then x=\pm ... 1 Compute the various distances, You will find that when b\gt 8 there are two nearest points, symmetric about the y-axis. When b\le 8, the nearest point is the origin. 2 You can replace g(x,y,z) with h(x,y)=g(x,y,1-x-y). Now you need to find the maximum of h on$$S=[0,1]\times[0,1] Which is a little easier. You know that if $h$ has a maximum in the interior of $S$, its partial derivatives will equal zero at that points. Then you simply check for possible maximums on the edges of $S$ (which is slightly annoying, since ...

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Well, this is not a finished solution. Your function is problematic, due to the absolute value, and I'm not sure whether there is something to be done regarding it. If the function is not differentiable, you probably won't be able to make it so. However, I do believe that the problem can be reduced from the 12 variables you have now, to a single one. From ...

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I apologize if this answer is not detailed enough as I do not have a thorough solution to your problem myself. But I do believe that you leave out one critical point because the way you create the square is somewhat arbitrary. For example, you could have written $3x^2 + 4xy + 3y^2 = 20 = 5(x^2 + y^2) - 2(x-y)^2=5D^2 - 2(x-y)^2$ so then you need to ...

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If $(x, y)$ is unconditional optimum then it is necessary the solution to $\nabla f = 0$, but in your case when minimizing $f(x, y)$ you still have constrainted problem and $\nabla f = 0$ is not the necessary condition for the optimum in such kind of problem. Also note that the solution $x = -y$ of $\nabla f = 0$ is the global maximum for $f(x, y)$. So ...

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As littleO explained above, the rationale behind introducing Lagrangian can be explained via its connection with proximal point methods. Despite the fact that many researchers recommend to refer to D. Bertsekas "Constrained Optimization and Lagrange Multiplier Methods", I personally find it quite unreadable. This paper provides a fairly good explanation ...

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I'll assume $f$ is convex. (I should probably also assume $f$ is closed and proper.) As you said, the Lagrangian is $L(x,y) = f(x) + \langle y, Ax - b \rangle$. The dual function is \begin{align*} g(y) &= \inf_x \, L(x,y) \\ &= \inf_x \, f(x) - \langle -A^Ty, x \rangle - \langle y, b \rangle \\ &= - \sup_x \, \langle -A^Ty, x \rangle - f(x) + ...

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Since the function was apparently altered after at least some of the answers were posted, I'll address a peculiarity of this problem that may have raised Ilda's original concern. It should be remarked upon that while the "level curves" of the function $\ f(x, \ y) \ = \ x^2 \ + \ 4xy \ + \ 4y^2 \$ may give the impression of being (rotated) ellipses, this ...

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Lets formulate the optimization problem you are trying to solve: $\min\limits_{c_i} \Sigma_i(c_i^2\sigma_i^2)$ s.t. $\sum_i c_i = 1$ (since unbiasedness requries $E[\sum_i c_iX_i] = \mu\sum_i c_i = \mu$) Lets vectorize this formulation. Let $c^2=(c_1^2,c_2^2...c_n^2)$ and $\sigma^2=(\sigma_1^2,\sigma_2^2....\sigma_n^2)$ Then we get a formulation that ...

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