New answers tagged

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Bounds on the variables should be dealt with in the same way as other constraints. Even though only one variable is out-of-bounds, forcing that variable to be in-bounds may cause other variables to hit their bounds. So although there's nothing wrong with trying (2) (if you get an optimial solution of (2) where all variables happen to be in-bounds, it's an ...


3

The $x$ in the Lagrangian can be any value. However the "subject to" constraints tell us that the domain over which you are to minimize $L$ are only those that satisfy $c^Tx = 0$ and $x^Tx = 1$, i.e. all $x$ in the intersection of the unit $n$-sphere and the plane $c^Tx = 0$. After you have derived the Lagrangian equations $$\frac{\partial L}{\partial \mu} ...


1

You have three equations from the first order conditions: $$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$ Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This gives us infinitely many solutions with ...


-1

You didn't consider the case $x = y = \lambda = 0, z = 1$ which satisfies the nabla equations, and gives the minimum value for $f$ of $0$.


0

The above approaches are correct, but just to make the link with your solution, here is another way of writing things: $$ \begin{cases} \nabla_x \mathcal{L} = 0\\ \nabla_{\lambda} \mathcal{L} = 0\\ \end{cases} \quad \Rightarrow\quad \begin{cases} Ax = \lambda x\\ x^{T}x=1\\ \end{cases} $$ In other words, unitary eigenvectors of $A$ satisfy the Lagrangian ...


1

$$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \lambda_{\min} (\mathrm A) > 0$$ because $\|\mathrm x\|_2 = 1$ and $\mathrm A \succ \mathrm O$. The minimum is attained at the intersection of the eigenspace of $\lambda_{\min} (\mathrm A)$ with the unit Euclidean sphere.


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Sketch: Since $A$ is positive-definite, it is symmetric and hence has an orthonormal eigenbasis. Let $v_1,\cdots,v_n$ be the eigenvectors with corresponding eigenvalues $\lambda_1,\cdots,\lambda_n$. Therefore we can write $x=\sum a_iv_i$ where $\sum a_i^2=1$. The sum condition comes from the fact that $\langle\sum a_iv_i,\sum a_iv_i\rangle=\sum a_ia_j\...


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After having several thoughts, understood that the vector and bias can be adjusted so that the product can be fit to 1. Not 100% sure this is the correct understanding. Appreciate feedback.


1

Say you have: $$\left< 3z, 6, 3x \right> = \lambda \left< 2x, 4y, 2z\right>$$ Clearly this is equal to: $$\left< z, 2, x \right> = \frac{2\lambda}{3} \left< x, 2y, z\right>$$ You can let $\lambda^* = 2\lambda/3$ and proceed as usual. As you said, the direction is the most important part of Lagrange Multipliers, for the most part $\...


1

In the 1D case, the second derivative test is inconclusive when the second derivative vanishes at the critical point. You can see this by looking at $f(x)=x^3,g(x)=x^4,x_0=0$; the second derivative test gives the same information but the behavior is different. The analogous thing happens in higher dimensions, including in constrained problems. tl;dr: You ...


0

We want to solve $$\begin{array}{lc} \text{maximize} & x^\alpha y^\beta\\ \text{subject to} & p x + q y = r\end{array}$$ We define the Lagrangian $$\mathcal{L} (x,y,\lambda) := x^\alpha y^\beta - \lambda (p x + q y - r)$$ Taking the partial derivatives and finding where they vanish, $$\alpha x^{\alpha-1} y^\beta = \lambda p \qquad \qquad \beta x^...


1

If $I=px+qy$, then $y = (I-px)/q$, so $x^ay^b =x^a((I-px)/q)^b =x^a(I-px)^b/q^b $. Differentiating, we want $\begin{array}\\ 0 &=(x^a(I-px)^b)'\\ &=ax^{a-1}(I-px)^b-x^apb(I-px)^{b-1}\\ &=x^{a-1}(I-px)^{b-1}(a(I-px)-xpb)\\ &=x^{a-1}(I-px)^{b-1}(aI-apx-xpb)\\ &=x^{a-1}(I-px)^{b-1}(aI-xp(a+b))\\ \text{so}\\ x &=\dfrac{aI}{p(a+b)}\\ \...


3

The solution The answer can be been found on the internet in any number of places. The function $U$ is a Cobb-Douglas utility function. The Cobb-Douglas function is one of the most commonly used utility functions in economics. The demand functions you should get are: $$x(p,I)=\frac{\alpha I}{(\alpha+\beta)p}\qquad y(p,I)=\frac{\beta I}{(\alpha+\beta)q}$$ ...


1

You can certainly simplify that system! $$\lambda = \frac{\alpha x^{\alpha - 1}y^\beta} p$$ $$\lambda = \frac{\beta y^{\beta - 1}x^\alpha} q$$ Thus $$\frac{\alpha x^{\alpha - 1}y^\beta} p = \frac{\beta y^{\beta - 1}x^\alpha} q$$ and $$q\left(\alpha x^{\alpha - 1}y^\beta\right) = p\left(\beta y^{\beta - 1}x^\alpha\right)$$ You can reduce powers: $$q\...


0

observe that $$f_x=y$$ and $$f_y=x-2y$$ thus we get the solution $$x=0,y=0$$ from the system $$f_x=0$$ and $$f_y=0$$


0

The sum of a convex function and a strongly convex function is again strongly-convex. This is a direct application of the definition of strong-convexity. On the other hand, it should be clear that $u \mapsto u^TP_iu$ is strongly convex if $P_i$ is positive definite.


1

The Hessian of a function is a generalization of the 2nd derivative of a single-variable function $f(x)$. If $x=a$ is a critical point of $f(x)$, a sufficient criterion for $a$ be a maximum is that the second derivative of $f$ at $a$ be negative: $f''(a) < 0$ (https://en.wikipedia.org/wiki/Derivative_test#Second_derivative_test). For multivariable ...


1

Unless you restrict the problem to $n \geq 1$ the statement is false. Consider $x=1, y=2, n=\frac12$: $$ \sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}\right)^n $$


0

Set $$f(x,y,z)=(x-5)^2+y^2+(z-1)^2$$ $$g(x,y,z)=z-x^2-3y^2=0$$ By application of Lagrange method, we have $$\nabla f=\lambda \nabla g$$ $$(2x-10\,,\,2y\,,\,2z-2)=\lambda(-2x\,,\,-6y\,,\,1)$$ therefore $$\left\{ \begin{align} & y=0\,\,\,\,\quad\Rightarrow \,\,\,z={{x}^{2}} \\ & \lambda =-\frac{1}{3}\,\,\,\Rightarrow \,\,\,x=\frac {15} 2\,\,\,\,,\,...


0

I think I can provide a simpler method for this specific problem that does not use Lagrange multipliers. Take a cross section of the 3-dimensional solid $z=x^2+3y^2$ at the plane $y=0$. We then have a parabola on the $xz$ plane: $z=x^2, y=0$. Note that point $P$ is also on this plane. To find the smallest distance between the original surface and the point,...


0

This was a hard question A common technique is to find the x,y restrictions for equations 1 and 2 to be equal to later be used in equation 3 to find actual values of x and y. A great way to do this is by manipulating dg(x) and dg(y) so that dg(x) = dg(y) and so then df(x) = df(y) and from there find restrictions. $$dg(x) = 2\lambda x+2\lambda$$ $$df(x) = ...


0

We have a homogeneous linear system in $x \in \mathbb R^n$ $$A x = 0_m$$ where $A \in \mathbb R^{m \times n}$. We also have constraints on the last $n_2$ entries of $x$. Hence, we write $$\begin{bmatrix} A_1 & A_2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} = 0_m$$ where $A_1 \in \mathbb R^{m \times n_1}$ and $A_2 \in \mathbb R^{m \times n_2}$...


1

Have you tried a software like Mathematica? There are four solutions to your system: $x=y=Sqrt[5/2]$ and the negative of that root (both with $λ=5/8$), and $x=-y=Sqrt[10]$ and the negative of that root (both with $λ=5/2$).


3

Imagine that you're aiming to cover as much of the $\sum_i v_i$ square as possible: The bigger the largest inner square, the closer it gets to covering more of the background square. The maximum sum of squares is reached when all but one of the $v_i$ is at the specified minimum - in this case, $1$.


3

If there are two numbers with $1\lt v_j\le v_k$, you can decrease $v_j$ and increase $v_k$ to increase the sum of squares. Thus at most one of the $v_i$ is greater than $1$.


0

(At the moment, this is not a complete answer. It is only my attempt to prove something I suggested in the comments.) Theorem 1. If $\vec{x}=(x_1,\dots,x_n)$ is a minimum, then at least one of $x_i$ is zero or the entries of $\vec{x}$ take at most two distinct values. The result of the Theorem will follow directly from the following lemma. Lemma 2. If $...


0

In general, if you can show that your optimization problem has continuous and differentiable objective function and constraint functions and satisfies a special technical hypothesis called a constraint qualification (there are many constraint qualifications in the literature) then the Lagrange multiplier condition is a necessary condition for a point $x^{*}$ ...



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