Discover Maximization Techniques in Geometry Triangles

I want to understand how Schnass $†<!--\; †\; -->$ arrived at the maximising objective as described below
$\begin{array}{rl}{\displaystyle \underset{D,A}{min}\Vert <!--\; \Vert \; -->X-<!--\; -\; -->DA{\Vert <!--\; \Vert \; -->}_F^2}& =\underset{D}{min}\underset{n}{\sum <!--\; \sum \; -->}\underset{|I|\le <!--\; \le \; -->S}{min}\Vert <!--\; \Vert \; -->x_n-<!--\; -\; -->D_I{D}_I^{†<!--\; †\; -->}{x}_n{\Vert <!--\; \Vert \; -->}_2^2\\ & =\Vert <!--\; \Vert \; -->X{\Vert <!--\; \Vert \; -->}_F-<!--\; -\; -->\underset{D}{max}\underset{n}{\sum <!--\; \sum \; -->}\underset{|I|\le <!--\; \le \; -->S}{max}\Vert <!--\; \Vert \; -->D_I{D}_I^{†<!--\; †\; -->}{x}_n{\Vert <!--\; \Vert \; -->}_2^2\end{array}$
where $†<!--\; †\; -->$ is the pseudo-inverse, $S$ denotes the cardinality of column vectors an of matrix $A$.

A few weeks ago I took part in a math competition and had this question through some trial and error I got the correct answer 80 as I didn't have time to use maximization using calculus, is there a way to do this question quickly in a rigorous way:
Find the maximum possible value of
$9\sqrt{x}+8\sqrt{y}+5\sqrt{z}$
where x, y, and z are positive real numbers satisfying
$9x+4y+z=128$
Any help would be greatly appreciated, thanks.

Suppose we have a primal problem $P$ which is stated as a maximization problem $max{c}^{T}x$.

The dual problem is defined only for a primal minimization problem.

Then what is the dual problem of $P$ ?

Is it implicit, that the dual problem of $P$ is the dual problem of $P$ stated as the minimization problem $min-<!--\; -\; -->c^{T}x$ ?

Surely these two primal problems are equivalent in the sense that their optimal solution $\stackrel{¯<!--\; ¯\; -->}{x}$ are equal (if it exist). However, the objective values are the same only if we ignore the sign !

Minimize the expression:
$F(a,b)=ab+a{b}^{\prime }+a^{\prime }b$.
(A) $a^{\prime }+b^{\prime }$
(B) $a^{\prime }+b$
(C) $a+b$
(D) $a+b^{\prime }$
I have no idea how to solve this problem. Also how to maximize the expression and what is the value of maximization.

Having the equation
$a^{T}x=b$
where $a$,$a,x\in <!--\; \in \; -->{\mathbb{R}}^n$, $b$ is a number. Can I solve $x$ in terms of $a$,$b$?

Ultimate goal is to find the maximum of
$\frac{1}{2}{x}^{T}Px$
given $a^{T}x=b$, where $P$ is a symmetric positive definite matrix.

I came across a lecture on Support Vector Machines and in the lecture they converted a maximization problem into a minimization problem. I am wondering how it was done...
$Max\frac{1}{||x||}$
is converted into
$Min\frac{1}{2}{x}^{T}x$
How was this step achieved..? Many thanks in advance !

I have no idea how to solve the following question and will be grateful for any help. Suppose we are given $n$ real numbers $y_1,...,y_n$. Is there any simple way to minimize function $f(y)=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}|y_k-<!--\; -\; -->y|$?

The primal assignment problem that I have is:
$\begin{array}{rl}& max\underset{i=i}{\overset{N}{\sum <!--\; \sum \; -->}}\underset{j=1}{\overset{N}{\sum <!--\; \sum \; -->}}{c}_{ij}{x}_{ij}\\ & \text{subject to}\\ & \underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{x}_{ij}=1,j=1,\dots <!--\; \dots \; -->,N\\ & \underset{j=1}{\overset{N}{\sum <!--\; \sum \; -->}}{x}_{ij}=1,i=1,\dots <!--\; \dots \; -->,N\\ & x_{i,j}\in <!--\; \in \; -->\{0,1\}\end{array}$
and I would like to obtain its dual, which should be
$\underset{p_j}{min}\{\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{p}_j+\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}max\{a_{ij}-<!--\; -\; -->p_j\}\}$
Can you give me any hints on how to approach this derivation? I have tried with Lagrange multipliers but I have not been able to reach this expression.

Given $\mathrm{\Omega <!--\; \Omega \; -->}:=[x_a,x_b]\times <!--\; \times \; -->[y_a,y_b]\subset <!--\; \subset \; -->{\mathbb{R}}^2$, consider the problem
$\underset{f\in <!--\; \in \; -->{\mathcal{C}}^2(\mathrm{\Omega <!--\; \Omega \; -->},\mathbb{R})}{max}{\iint <!--\; \iint \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}[f_y(x,y){\int <!--\; \int \; -->}_{x_a}^{x}f(z,y)dz]dxdy$
subject to a constraint of the form
${\iint <!--\; \iint \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}{[{\int <!--\; \int \; -->}_{x_a}^x{f}_y(z,y)dz-<!--\; -\; -->{\int <!--\; \int \; -->}_{x_a}^{x_b}z{f}_y(z,y)dz]}^{2}dxdy=c\in <!--\; \in \; -->\mathbb{R}$
where the subscript $y$ denotes the partial derivative w.r.t. the second variable, i.e., $y$.

I thought the first step would have been to write down the Euler-Lagrange equations, i.e., given the Langragian $\mathcal{L}$,
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}\left(\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}{f}_y}\right)=\frac{\mathrm{\partial <!--\; \partial \; -->}\mathcal{L}}{\mathrm{\partial <!--\; \partial \; -->}f}$
but the Lagrangian is pretty involved and I got stuck. Do you have any hint?

I would like to maximize the Confluent Hypergeometric Distribution in order to apply a Rejection sampling. The formula of the distribution is
$f(x;a,b,c)=K{x}^{a-<!--\; -\; -->1}(1-<!--\; -\; -->x{)}^{b-<!--\; -\; -->1}{e}^{-<!--\; -\; -->cx}$
where $0\le <!--\; \le \; -->x\le <!--\; \le \; -->1$ and $a,b>0,c\in <!--\; \in \; -->\mathbb{R}$
and let us assume that K is constant in terms of x. I would like to maximize this distribution in order to find an upper bound C and apply the rejection sampling.

$K\in <!--\; \in \; -->{\mathbb{R}}_{+}$ and $z\in <!--\; \in \; -->[0,x]\cap <!--\; \cap \; -->[0,K]$ $\mathrm{\gamma <!--\; \gamma \; -->}>0$ one defines
$\begin{array}{r}\mathrm{\varphi <!--\; \varphi \; -->}(z)=(K-<!--\; -\; -->z){\left(\frac{z}{x}\right)}^{\mathrm{\gamma <!--\; \gamma \; -->}}\end{array}$
The maximum is attained at $z=\frac{\mathrm{\gamma <!--\; \gamma \; -->}K}{\mathrm{\gamma <!--\; \gamma \; -->}+1}$. Question: Why is for $x\le <!--\; \le \; -->\frac{\mathrm{\gamma <!--\; \gamma \; -->}K}{\mathrm{\gamma <!--\; \gamma \; -->}+1}$:
$\begin{array}{r}\underset{z}{max}\mathrm{\varphi <!--\; \varphi \; -->}(z)=K-<!--\; -\; -->x\end{array}$
And for $x>\frac{\mathrm{\gamma <!--\; \gamma \; -->}K}{\mathrm{\gamma <!--\; \gamma \; -->}+1}$
$\begin{array}{r}\underset{z}{max}\mathrm{\varphi <!--\; \varphi \; -->}(z)=\mathrm{\varphi <!--\; \varphi \; -->}\left(\frac{\mathrm{\gamma <!--\; \gamma \; -->}K}{\mathrm{\gamma <!--\; \gamma \; -->}+1}\right)\text{ ?}\end{array}$
Best regards.

I need to prove the maxima of the following summation, using Lagrange.
$\underset{x_m}{max}(\underset{m}{\sum <!--\; \sum \; -->}{a}_{m}log(x_m))$
s.t.
$0\le <!--\; \le \; -->x_m\le <!--\; \le \; -->1$
$\underset{m}{\sum <!--\; \sum \; -->}{x}_m=1$
The solution is a closed form $x_m=\frac{a_m}{\underset{m}{\sum <!--\; \sum \; -->}{a}_m}$ .

I formulated the Lagrange equation but I am confused about the signs and the multipliers.

$L(x,\mathrm{\lambda <!--\; \lambda \; -->},\mathrm{\mu <!--\; \mu \; -->})=\underset{m}{\sum <!--\; \sum \; -->}{a}_{m}log(x_m)+\underset{m}{\sum <!--\; \sum \; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}_m(1-<!--\; -\; -->x_m)+\mathrm{\mu <!--\; \mu \; -->}(\underset{m}{\sum <!--\; \sum \; -->}{x}_m-<!--\; -\; -->1)$, Is this formulation correct ? what is wrong ?

note: only one $\mathrm{\mu <!--\; \mu \; -->}$ for one constraint.

I have a non-linear maximization problem and I want to convert it to be a minimization problem, can I do so by multiplying it by a negative sign, or is that wrong; and if that is wrong what should I do?

We have a function:
$f(x,y,z)=x\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+y+{\mathrm{\varphi <!--\; \varphi \; -->}}_1)+z\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t+y+{\mathrm{\varphi <!--\; \varphi \; -->}}_2).$
I want to solve the following maximization problems:
$\underset{x,y,z}{max}\underset{t}{max}f(x,y,z)\text{or}\underset{x,y,z}{max}{\int <!--\; \int \; -->}_{<T>}f(x,y,z)dt.$
Surely, $x$ and $z$ are nonnegative, and $y$ is in $[0,2\mathrm{\pi <!--\; \pi \; -->})$.

Can someone give me hints for solving the problem, or let me know some references to solve this types of problem?

Let $f(x,y):\mathbb{R}\times <!--\; \times \; -->\mathbb{R}\mapsto <!--\; \mapsto \; -->\mathbb{R}$ be a continuous and differentiable function. When can we claim that the following holds true:
$\frac{d}{dx}\underset{y\in <!--\; \in \; -->\mathbb{R}}{max}f(x,y)=\underset{y\in <!--\; \in \; -->\mathbb{R}}{max}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}f(x,y)$
assuming that both $\underset{y\in <!--\; \in \; -->\mathbb{R}}{max}f(x,y)$ and $\underset{y\in <!--\; \in \; -->\mathbb{R}}{max}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}f(x,y)$ exists.

Each cow weighs 300kg. The cost of maintaining a cow is 20€ per day. The rate of change of the cows weight is 4kg a day. The market price is currently 50€ per kg and it is falling 50 cents a day. How much time should the cattleman wait to sell the cows? How much extra money does he make by waiting?

I have read a paper where, at some point, the following optimization problem appears:
$\begin{array}{rl}& \underset{\mathit{x}}{max}{\mathit{w}}^T\mathit{x}\\ & \text{subject to }{\Vert \mathit{x}\Vert }_{\mathrm{\infty <!--\; \infty \; -->}}\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}\end{array}$
where $\mathit{x}$ and $\mathit{w}$ are real vectors and $\mathrm{ϵ<!--\; ϵ\; -->}$ is some given positive constant. The authors claim that the solution for this problem is $\mathit{x}=\text{sign}(\mathit{w})$, which seems clearly wrong to me, since then we would have ${\Vert \mathit{x}\Vert }_{\mathrm{\infty <!--\; \infty \; -->}}=1$.

I believe the correct answer is $\mathit{x}=\mathrm{ϵ<!--\; ϵ\; -->}\text{sign}(\mathit{w})$. However, I am not being able to prove it. I have tried KKT multipliers, with no success. Any ideas?

Consider that I have two items. Let $P_i(x_i)$ denote the profit I get from item i by taking $x_i$ units of it (assume that $x_i$ can be a real number). Let $W_i(x_i)$ denote the weight consumed by item i when I take $x_i$ units of this. Assume that both profit and weight functions are monotonically increasing in the quantity of items picked. The problem I am interested in is
$\begin{array}{rl}\underset{x_1,x_2}{max}& P_1(x_1)+P_2(x_2)\\ s.t.& W_1(x_1)+W_2(x_2)\le <!--\; \le \; -->W\end{array}$
where $W$ is total weight allowed. Thus I need to find $x_i$ (quantity of item $i$) such that the total profit is maximized while keeping the total weight under a given quantity. Let $x_1^{\ast <!--\; \ast \; -->}$ and $x_2^{\ast <!--\; \ast \; -->}$ be the solution to this problem. Is it true that
$\frac{P_1(x_1^{\ast <!--\; \ast \; -->})}{W_1(x_1^{\ast <!--\; \ast \; -->})}=\frac{P_2(x_2^{\ast <!--\; \ast \; -->})}{W_2(x_2^{\ast <!--\; \ast \; -->})}$
Note that ratio is essentially "Profit Per unit weight of an item". Assume that this were different at the optimum and the first item had the larger ratio. Thus I can get more profit per unit weight out of the first item. Then increasing quantity of that should increase profit. In order to balance the weights, I decrease the second items quantity decreasing its profit as well. But since the increase in profits from item 1 is higher, this should compensate for loss from item 2 and make the profits even more higher. What is flawed with this logic?

The maximization problem is:
Maximize $u(x_1,x_2)=min[a_1{x}_1,a_2{x}_2]\text{s.t.}{p}_1{x}_1+p_2{x}_2\le <!--\; \le \; -->$$w$, in which $x_i,p_i$ is the amount and price of good $i$, $w$ is the total budget available.
What I have been told to deal with this min[.,.] function is to solve it graphically. It's very easy to see on the graph that the maximization happens when $a_1{x}_1=a_2{x}_2$. But I wonder if there is a way to solve this algebraically? I'm stumped at the first step, which is to derive ${\displaystyle \frac{\mathrm{\partial <!--\; \partial \; -->}u(x_i)}{\mathrm{\partial <!--\; \partial \; -->}{x}_i}}$.

For some strictly convex functions $f_1(x_1),f_2(x_2),...,f_n(x_n)$, it seems intuitive that for a given sum of $x_1+x_2+...+x_n=X$, where $x_1$, $x_2$,..., $x_n\ge <!--\; \ge \; -->0$
$\mathrm{\exists <!--\; \exists \; -->}k\in <!--\; \in \; -->\{1,2,\dots <!--\; \dots \; -->,n\},f_k(X)=max\{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{f}_i(x_i):\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i=X\text{ and }\mathrm{\forall <!--\; \forall \; -->}i,x_i\ge <!--\; \ge \; -->0\}$
Since these functions are convex, the sum of these functions is greatest when the sum of arguments is equal to the argument of just one function. This seems intuitive, but I can't think of how to argue this in formal maths. Any suggestions?