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What will the dimensions of the resulting cardboard box be if the company wants to maximize the volume and they start with a flat piece of square cardboard 20 feet per side, and then cut smaller squares out of each corner and fold up the sides to create the box?

I'm trying to solve the kind of linear equation below such that the sum of the unknowns is maximised, but have been unable to find the solution.
$\frac{10}{y_1}+\frac{12}{y_2}+\frac{15}{y_3}=50$
It may just be I am searching using the wrong terminology, if so direction of where to look would be greatly appreciated too.EDIT: To add some more information, the initial problem involved four equations.
$y_1=K_1/x_1$
$y_2=K_2/x_2$
$y_3=K_3/x_3$
$x_1+x_2+x_3=X$
Where K and X values are given constants, and x and y values are unknown. I am trying to find a solution that maximises the sum of the y values.

It is from this problem that I simplified it to
$\frac{K_1}{y_1}+\frac{K_2}{y_2}+\frac{K_3}{y_3}=X$
With the first equation in the initial question just being arbitrary values.

Suppose that there is a positive definite matrix $\mathbf{A}\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$, and a vector $\mathbf{b}\in <!--\; \in \; -->{\mathbb{R}}^n$, then minimization of quadratic functions with linear terms can be done in closed form as
$\arg <!--\; \; -->\underset{\mathbf{x}\in <!--\; \in \; -->{\mathbb{R}}_n}{min}(\frac{1}{2}{\mathbf{x}}^{\mathsf{T}}\mathbf{A}\mathbf{x}-<!--\; -\; -->{\mathbf{b}}^{\mathsf{T}}\mathbf{x})={\mathbf{A}}^{-<!--\; -\; -->1}\mathbf{b}$

I met this in a machine learning book. However, the book didn't provide a proof. I wonder why this can be well-formed. Hope that someone can help me with it. I find that many machine learning books like to skip all of the proofs, which made me uncomfortable.

Suppose there is a set of n linear constraints $\{a_i^{T}x+b_i\le <!--\; \le \; -->0{\}}_{i=1}^n$ with $a_i\in <!--\; \in \; -->{\mathbb{R}}^d$, $b_i\in <!--\; \in \; -->\mathbb{R}$, $x\in <!--\; \in \; -->{\mathbb{R}}^d$. How can I find $x^{\ast <!--\; \ast \; -->}$ that maximizes $|\{i\in <!--\; \in \; -->[n]\mid <!--\; \mid \; -->a_i^T{x}^{\ast <!--\; \ast \; -->}+b_i\le <!--\; \le \; -->0\}|$? Is this problem solvable or can be approximated within polynomial time?
Some reference materials would also be nice. Thanks!

What is the Proximal Operator ($\mathrm{Prox}$) of the Pseudo $L_0$ Norm?
Namely:
${\mathrm{Prox}}_{\mathrm{\lambda <!--\; \lambda \; -->}{\Vert \cdot <!--\; \cdot \; -->\Vert }_0}<!--\; \; -->\left(\mathit{y}\right)=\arg <!--\; \; -->\underset{\mathit{x}}{min}\frac{1}{2}{\Vert \mathit{x}-<!--\; -\; -->\mathit{y}\Vert }_2^2+\mathrm{\lambda <!--\; \lambda \; -->}{\Vert \mathit{x}\Vert }_0$
Where ${\Vert \mathit{x}\Vert }_0=\mathrm{n}\mathrm{n}\mathrm{z}(x)$, namely teh number of non zeros elements in the vector $\mathit{x}$.

This might be trivial but I am struggling to justify the following simplification.from:
$h(x^{\ast <!--\; \ast \; -->}+d)-<!--\; -\; -->h(x^{\ast <!--\; \ast \; -->})\ge <!--\; \ge \; -->\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}_{j}f(x^{\ast <!--\; \ast \; -->})\ast <!--\; \ast \; -->d_j+\mathrm{\gamma <!--\; \gamma \; -->}\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}|d_j|$
to:
$h(x^{\ast <!--\; \ast \; -->}+d)-<!--\; -\; -->h(x^{\ast <!--\; \ast \; -->})\ge <!--\; \ge \; -->-<!--\; -\; -->\underset{j\text{⧸}\in <!--\; \in \; -->G}{max}|{\mathrm{\nabla <!--\; \nabla \; -->}}_{j}f(x^{\ast <!--\; \ast \; -->})|\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}|d_j|+\mathrm{\gamma <!--\; \gamma \; -->}\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}|d_j|$

Specifically, why is there a negative in front of the maximization?

Note:
I can get behind the fact that
$\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}_{j}f(x^{\ast <!--\; \ast \; -->})\ast <!--\; \ast \; -->d_j\ge <!--\; \ge \; -->\underset{j\text{⧸}\in <!--\; \in \; -->G}{max}|{\mathrm{\nabla <!--\; \nabla \; -->}}_{j}f(x^{\ast <!--\; \ast \; -->})|\underset{j\text{⧸}\in <!--\; \in \; -->G}{\sum <!--\; \sum \; -->}|d_j|$
provided the jacobian is nonnegative element-wise. But then why add the negative sign?

Soo, fix $0\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}\le <!--\; \le \; -->1$. Given $\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->1,x_i\ge <!--\; \ge \; -->0$, I know that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i\ge <!--\; \ge \; -->\mathrm{\lambda <!--\; \lambda \; -->}n$. I also know that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i^2\le <!--\; \le \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}^{2}n+\mathrm{ϵ<!--\; ϵ\; -->}$. I am trying to prove a multiplicative error on each $x_i$, mainly something along the lines of
$|x_i-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}|\le <!--\; \le \; -->f(\mathrm{ϵ<!--\; ϵ\; -->})\mathrm{\lambda <!--\; \lambda \; -->}$
Where $f(\mathrm{ϵ<!--\; ϵ\; -->})$ is some function of $\mathrm{ϵ<!--\; ϵ\; -->}$, say $f(\mathrm{ϵ<!--\; ϵ\; -->})=2\mathrm{ϵ<!--\; ϵ\; -->}$. Is there any inequality that would bound that distance?

Let $\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->\Vert <!--\; \Vert \; -->$ be a norm invariant under unitary trasformations.
Is it true that
$\Vert \left(\begin{array}{c}\stackrel{^<!--\; ^\; -->}{L}-<!--\; -\; -->L\\ G\end{array}\right)\Vert$
is minimized when $\stackrel{^<!--\; ^\; -->}{L}=L$ ($\stackrel{^<!--\; ^\; -->}{L}$ and $G$ are fixed, $L$ is the only variable quantity), i.e $\Vert \left(\begin{array}{c}\stackrel{^<!--\; ^\; -->}{L}-<!--\; -\; -->L\\ G\end{array}\right)\Vert =\Vert \left(\begin{array}{c}0\\ G\end{array}\right)\Vert$ ?
I'm unable to prove using just elementar inequalities such as triangle inequality or norm properties, any help would be appreciated.

Suppose that $p_1,\dots <!--\; \dots \; -->,p_n$ are nonnegative real numbers such that $p_1+\cdots <!--\; \cdots \; -->+p_n=1$; denote the corresponding set of vectors by ${\mathrm{\Delta <!--\; \Delta \; -->}}_n$.

I am interested in the following function, $f:<!--\; :\; -->{\mathrm{\Delta <!--\; \Delta \; -->}}_n\to <!--\; \to \; -->{\mathbb{R}}_{+}$, given by
$f(p)=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{p_k}{\underset{j=k}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_j}.$
We always have $f(p)\le <!--\; \le \; -->n$, using the lower bound $\underset{j\ge <!--\; \ge \; -->k}{\sum <!--\; \sum \; -->}{p}_j\ge <!--\; \ge \; -->p_k$. However I feel this must be a loose bound on the quantity
$\underset{p\in <!--\; \in \; -->{\mathrm{\Delta <!--\; \Delta \; -->}}_n}{sup}f(p),$
since it requires that $\underset{j>k}{\sum <!--\; \sum \; -->}{p}_j=0$ for all $k$ to be met with equality. Hence, I am wondering what the largest $f(p)$ can be when evaluated over the simplex?

Suppose we have an (not necessarily convex) optimization problem :
$\begin{array}{r}\underset{x}{min}{f}_0(x)\\ f_1(x)\le <!--\; \le \; -->0.\end{array}$
Let $L(x,\mathrm{\lambda <!--\; \lambda \; -->})=f_0(x)+\mathrm{\lambda <!--\; \lambda \; -->}(f_1(x))$. Then the above problem can be equivalently written as:
$\underset{x}{min}\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}L(x,\mathrm{\lambda <!--\; \lambda \; -->}).$
The dual of the above problem can be written as:
$\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}\underset{x}{min}L(x,\mathrm{\lambda <!--\; \lambda \; -->}).$
We say that strong duality holds at a point when
$\underset{x}{min}\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}L(x,\mathrm{\lambda <!--\; \lambda \; -->})=\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}\underset{x}{min}L(x,\mathrm{\lambda <!--\; \lambda \; -->}).$
By weak duality, the inequality
$\underset{x}{min}\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}L(x,\mathrm{\lambda <!--\; \lambda \; -->})\ge <!--\; \ge \; -->\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}\underset{x}{min}L(x,\mathrm{\lambda <!--\; \lambda \; -->})$
always holds true. My doubt is: suppose there exists an $x^{\ast <!--\; \ast \; -->}$ that minimizes $L(x,\mathrm{\lambda <!--\; \lambda \; -->})$ for a fixed $\mathrm{\lambda <!--\; \lambda \; -->}$, can i say that the following inequality holds true:
$\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}(\underset{x}{min}L(x,\mathrm{\lambda <!--\; \lambda \; -->}))=\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}L(x^{\ast <!--\; \ast \; -->},\mathrm{\lambda <!--\; \lambda \; -->})\ge <!--\; \ge \; -->\underset{x}{min}\underset{\mathrm{\lambda <!--\; \lambda \; -->}\ge <!--\; \ge \; -->0}{max}L(x,\mathrm{\lambda <!--\; \lambda \; -->}).$
If the above is true, then are we saying that if there is a primal variable that attains its minimum in the dual problem, then strong duality holds true? Somewhere this does not seem to add up.

I have a very simple linear problem:
$\begin{array}{rl}\underset{x}{min}& x^2\\ \text{s.t. }& a_1{x}_1+a_2{x}_2=b\end{array}$
Suppose I want to write this problem equivalently as in Find the equivalent linear program. Unlike the problem in the link, I have equality. Can I write it equivalently as:
$\begin{array}{rl}\underset{x,\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}}{min}& x^2\\ \text{s.t. }& a_1{x}_1=\mathrm{\alpha <!--\; \alpha \; -->}b,a_2{x}_2=\mathrm{\beta <!--\; \beta \; -->}b,\mathrm{\alpha <!--\; \alpha \; -->}+\mathrm{\beta <!--\; \beta \; -->}=1.\end{array}$
The converse is intuitive: Given $\{x,\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}\}$ feasible for the second problem, adding the first and second constraints gives the constraint of the first problem. But the forward part is not clear, especially because I have never seen an equality constraint written like this. Any help would be highly appreciated.

Consider a minimization problem:
$\begin{array}{rl}& min\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|x_i|,\\ & Ax=b,\end{array}$
where $A$ is an $m\times <!--\; \times \; -->n$ matrix of rank $m$. I know that the minimum points contains one where at least $n-<!--\; -\; -->m$ components of $x$ is zero and I have proved this conclusion. My problem is that, I think this should be a developed proposition but I am not familiar with optimization theory, so I don't know where to find a reference. Could anybody help me find a reference like a book or article? Thanks!

I have a doubt regarding a constrained optimisation problem.

Suppose my original constrained minimisation problem is
$\underset{x}{min}f(g(x),x)\text{ s.t. }g(x)=3$
I would like to know if this equivalent to solving the unconstrained minimisation problem
$\underset{x}{min}f(3,x)$
If not, when are these two problems equivalent?

Find global extrema of $f(x,y)=\frac{x^2}{2}+3y^2$ on the set $M(g):=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2\mid <!--\; \mid \; -->g(x,y)=0\}$ where $g(x,y)=x^2+y^4-<!--\; -\; -->25$.

$M(g)$ is compact and $f$ continuous so we know that there must exists a global maximum and global minimum. The conditions of the Lagrange multiplier methods are satisfied and if we solve the equations that result from the Lagrange multiplier method we get the following points:
$(0,\sqrt{5}),(0,-<!--\; -\; -->\sqrt{5}),(5,0),(-<!--\; -\; -->5,0),(4,\sqrt{3}),(-<!--\; -\; -->4,\sqrt{3}),(4,-<!--\; -\; -->\sqrt{3}),(-<!--\; -\; -->4,-<!--\; -\; -->\sqrt{3}).$
and
$$\begin{gathered} f(0,\sqrt{5})=15,f(0,-<!--\; -\; -->\sqrt{5})=15, \\ f(5,0)=\frac{25}{2},f(-<!--\; -\; -->5,0)=\frac{25}{2}, \\ f(4,\sqrt{3})=17,f(-<!--\; -\; -->4,\sqrt{3})=17,f(4,-<!--\; -\; -->\sqrt{3})=17,f(-<!--\; -\; -->4,-<!--\; -\; -->\sqrt{3})=17. \end{gathered}$$
So far so good.

However our sample solution says that "from the above values we see that $\frac{25}{2}$ is the global minimum and 17 the global maximum. "

As far as I have understood the Lagrange multiplier method it only delivers a necessary but not sufficient condition. So we don't know if one of the three points $\frac{25}{2},15,17$ is a saddle point. To make sure that the points are indeed extrema we have to resort to another method (e.g. plug in the condition into $f$).

Am I am right or is there something I don't see or didn't understand correctly?

Let $x_1,x_2,x_3,x_4$ be points on the unit sphere ${\mathbb{S}}^2$, that maximizes the quantity
$\underset{i<j}{\sum <!--\; \sum \; -->}\Vert <!--\; \Vert \; -->x_i-<!--\; -\; -->x_j{\Vert <!--\; \Vert \; -->}^2,$
where $\Vert <!--\; \Vert \; -->x_i-<!--\; -\; -->x_j\Vert <!--\; \Vert \; -->$ denotes the Euclidean distance in ${\mathbb{R}}^3$.

Do the $x_i$ form the shape of a Tetrahedron?

I tried using Lagrange's multipliers but this got a bit messy; I got that
$\underset{i\ne <!--\; \ne \; -->j}{\sum <!--\; \sum \; -->}(⟨<!--\; ⟨\; -->v_i,x_j⟩<!--\; ⟩\; -->+⟨<!--\; ⟨\; -->x_i,v_j⟩<!--\; ⟩\; -->)=\underset{i}{\sum <!--\; \sum \; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}_i⟨<!--\; ⟨\; -->x_i,v_i⟩<!--\; ⟩\; -->,$
for every $(v_1,v_2,v_3,v_4)\in <!--\; \in \; -->({\mathbb{R}}^3{)}^4$, but I am not sure how to proceed from here.

Suppose that $x$, $y$, and $z$ are real numbers such that $x+y+z=3$ and $x^2+y^2+z^2=6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\ge <!--\; \ge \; -->(x+y+z{)}^2$, but this doesn't say anything. I also tried some different ways to apply Cauchy, but they all didn't do much.

Let $x_1,...,x_{25}>0$ be such that $\underset{i=1}{\overset{25}{\sum <!--\; \sum \; -->}}{x}_i=4350$ and $\underset{i=1}{\overset{25}{\sum <!--\; \sum \; -->}}{x}_i^2=757770.25$.
From the first equality alone, we know that at least one of the $x_i$'s must be less than or equal to $\frac{4350}{25}=174$. From the second equality alone, we know that at least one of the $x_i$'s must be less than or equal to $\sqrt{\frac{757770.25}{25}}=174.1$, which is less useful than the first bound. My question is whether we can get a better bound, i.e. to find the least upper bound of $min\{x_1,...,x_{25}\}$, when we use both equalities together. I appreciate any comments or hints.

I have a very complex trace function $F(X)=\mathrm{Tr}<!--\; \; -->(AX(X^{T}X{)}^{-<!--\; -\; -->1}{X}^T)$. I would like to find the derivative ${\mathrm{\nabla <!--\; \nabla \; -->}}_{X}F(X)$. I do not know how to do it. I checked the matrix cookbook there is nothing similar to this. Please help!

The definition of a linear program is following:

Find a vector $x$ such that: $min{c}^{T}x$, subject to $Ax=b$ and $x\ge <!--\; \ge \; -->0$.

Generally, $b$ is assumed to be a fixed constant. However is it possible to construct a program where values of $b$ are part of the optimization? Could I included b in the optimization by changing $Ax=b$ to $Ax-<!--\; -\; -->b=0$. If so, would I also be able to place constraints upon $b$ like $\sum <!--\; \sum \; -->b=1$ and $1>b>0$? Finally, would such a program be possible to solve efficiently?

I am trying to solve the linear program for Wasserstein Distance between two discrete distributions. In the standard case, b represents the marginals for each datapoint. I know the marginals for the target distribution but the marginals from my source distribution are unknown. I am wondering if there is an efficient way to optimize the marginals for my source distribution such that the Wasserstein distance is minimized.

I'm using CPLEX through AMPL to solve a problem using two equivalent formulations. The problem is being solved to global optimality. However, one of the formulations is faster than the other. I have checked CPLEX's stats and I have found that the faster formulation has managed to reduce the number of explored branch-and bound nodes by 50%. My question is, what does such reduction in the number of the explored nodes indicate? Does it indicate a smaller search space for example? I would appreciate your feedback.