Discover Maximization Techniques in Geometry Triangles

I always saw the algorithm transform its objective function by log. I wonder this is because logarithm function is monotonically increased (logx) to do maximization, or decreased (−logx) to do minimization? So for MM algorithm, any functions should be transformed into log?

Are there also other reasons that log transformation is often used? I wonder log is always work or we have some mathematical operations that can replace log for this purpose?

Let $N=\{1,\dots <!--\; \dots \; -->,n\}$. For a given $(r_1,\dots <!--\; \dots \; -->,r_n)\in <!--\; \in \; -->{\mathbb{R}}_{++}^n$. I need to solve
$\begin{array}{r}\underset{(k_1,\dots <!--\; \dots \; -->,k_n)\in <!--\; \in \; -->{\mathbb{R}}_{++}^n}{max}\underset{i\in <!--\; \in \; -->N}{\prod <!--\; \prod \; -->}[\underset{j\in <!--\; \in \; -->N}{\sum <!--\; \sum \; -->}(r_j-<!--\; -\; -->k_j)+r_i\ln <!--\; \; -->\left(\frac{k_i}{r_i}\right)].\end{array}$
I could not come up with a closed form solution of the maximizers by using first order conditions, because of the log. Is it possible to tell, that there does not exist a closed form solution and we thus have to maximize the product numerically.

How does one maximize the following equation? $(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)(a-<!--\; -\; -->b)$ given that $-<!--\; -\; -->n\le <!--\; \le \; -->c<b<a\le <!--\; \le \; -->n$ for some value of $n$
So far, I have attempted to do this using AM-GM and got $\frac{64}{27}{n}^3$ but I don't think this is correct as some of the values of a, b, and c can be negative and when I tried to find the actual values which satisfied the max value found it gave me a contradiction.

Consider the vectors:
$a_1=\left(\begin{array}{c}0\\ -<!--\; -\; -->1\\ 1\\ 0\end{array}\right),a_2=\left(\begin{array}{c}0\\ 0\\ -<!--\; -\; -->1\\ 1\end{array}\right),a_3=\left(\begin{array}{c}2\\ 0\\ 0\\ 1\end{array}\right)$
Find a single vector $p$ which maximizes $p{a}_i$ for $i=1,2,3$.

To put this in context this is an economics profit max problem where p is a price and each component of the above vectors represents the quantity of the good.

I honestly have no idea how to find this p vector. It doesn't even seem possible to me that a single vector can maximize these three vectors.

Suppose I want to solve the following problem
$\underset{f(\cdot <!--\; \cdot \; -->)}{max}\int <!--\; \int \; -->f(x)g(x)d\mathrm{\mu <!--\; \mu \; -->}(x)$
where the maximization is over measurable functions $f:X\to <!--\; \to \; -->[0,1]$, $\mathrm{\mu <!--\; \mu \; -->}$ is a finite measure and $g$ is measurable.
Can someone come up with an example such that the problem does NOT have a solution? I imagine one can try to maximize point by point and obtain something that is not measurable and then the problem would not have a solution?

I have a function $f=-<!--\; -\; -->20p\cdot <!--\; \cdot \; -->q+9p+9q$. Player 1 chooses $p$ and player 2 chooses $q$. Both $p$ and $q$ are in the inclusive interval [0,1]. Player 1 wants to maximize $f$ while player 2 wants to minimize $f$.

Player 1 goes first, what is the most optimal value of $p$ he should choose knowing that player 2 will choose a $q$ in response to player 1's choice of $p$?

This seems to be some sort of minimization-maximization problem, but I am unsure how to solve it. I was thinking about approaching this from a calculus perspective by taking the partial derivative of $f$ with respect to $p$, but it doesn't seem I get an intuition by doing this, and it seems that $p$ and $q$ are a function of each other. How should I approach solving this problem analytically?

Solve the following NLP:
$\{\begin{array}{cc}min& -<!--\; -\; -->3x+y-<!--\; -\; -->z^2\\ s.t& g(x,y,z)=x+y+z\le <!--\; \le \; -->0\\ & h(x,y,z)=-<!--\; -\; -->x+2y+z^{2}z=0\end{array}$

My attempt
Using kkt conditions, we have 2 possibles situations:
1) If $g<0$: $\mathrm{\nabla <!--\; \nabla \; -->}f+\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\nabla <!--\; \nabla \; -->}h=0,h=0$
$\{\begin{array}{ccc}-<!--\; -\; -->3-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}& =& 0\\ -<!--\; -\; -->3y^2+1+2\mathrm{\lambda <!--\; \lambda \; -->}& =& 0\\ 2\mathrm{\lambda <!--\; \lambda \; -->}z& =& 0\\ -<!--\; -\; -->x+2y+z^2& =& 0\end{array}$
From first and second, we see it is impossible
**2)**$g=0$: $\mathrm{\nabla <!--\; \nabla \; -->}f+\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\nabla <!--\; \nabla \; -->}h+\mathrm{\mu <!--\; \mu \; -->}\mathrm{\nabla <!--\; \nabla \; -->}g=0,\mathrm{\mu <!--\; \mu \; -->}\ge <!--\; \ge \; -->0$
$\{\begin{array}{ccc}-<!--\; -\; -->3-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ -<!--\; -\; -->3y^2+1+2\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ 2\mathrm{\lambda <!--\; \lambda \; -->}z+\mathrm{\mu <!--\; \mu \; -->}& =& 0\\ -<!--\; -\; -->x+2y+z^2& =& 0\\ x+y+z& =& 0\end{array}$
I couldnt solve this last one. Any idea? I've tried to put y,z im function of $\mathrm{\mu <!--\; \mu \; -->}$ and use the last 2 eq., but didnt work (it became complicated).
Thanks in advance!

Find the maximum of
$x_2+y_1-<!--\; -\; -->x_1-<!--\; -\; -->y_2$
if $(x_1-<!--\; -\; -->1/2{)}^2+(y_1-<!--\; -\; -->1/2{)}^2=1/4$ and $(x_2-<!--\; -\; -->1/2{)}^2+(y_2-<!--\; -\; -->1/2{)}^2=1/4$ and $x_1{y}_2=y_1{x}_2$ and all variables are positive.

Is there a way to do this with inequalities? What about Lagrange Multipliers?

I'm having trouble maximizing the norm of $||Ax||$ such that $Bx=b$. I set up a lagrange multiplier so that $L(x,\mathrm{\lambda <!--\; \lambda \; -->})=x^T(A^{T}A)x-<!--\; -\; -->(Bx-<!--\; -\; -->b{)}^T\mathrm{\lambda <!--\; \lambda \; -->}$. (Say that $B$ is not invertible but $b$ is in the colspace so that the problem isn't trivial)
${\mathrm{\nabla <!--\; \nabla \; -->}}_{x}L(x,\mathrm{\lambda <!--\; \lambda \; -->})=0\Rightarrow <!--\; \Rightarrow \; -->2(A^{T}A)x=B^T\mathrm{\lambda <!--\; \lambda \; -->}$
and ${\mathrm{\nabla <!--\; \nabla \; -->}}_{\mathrm{\lambda <!--\; \lambda \; -->}}L(x,\mathrm{\lambda <!--\; \lambda \; -->})=0\Rightarrow <!--\; \Rightarrow \; -->Bx=b$.
If I assume that $A$ has full rank, then $x=\frac{1}{2}(A^{T}A{)}^{-<!--\; -\; -->1}{B}^T\mathrm{\lambda <!--\; \lambda \; -->}$.
If I plug it into the constraint, then $B(A^{T}A{)}^{-<!--\; -\; -->1}{B}^T\mathrm{\lambda <!--\; \lambda \; -->}=2b$. How do I proceed solving for $\mathrm{\lambda <!--\; \lambda \; -->}$ here?

I was just wondering what the steps one would take to maximize the utility of a function of the form U(X,Y) = min{X,Y} + X subject to income I = $p_x$X + $p_y$Y where $p_x$ is the price of X and $p_y$ is the price of Y. I understand that normally this would be solved using the Lagrangian method but obviously you cannot differentiate this particular function.

Consider the following maximization problem.
$\begin{array}{rl}& \underset{x\in <!--\; \in \; -->{\mathbb{R}}^n}{max}{x}^{T}Px\\ & a_i\le <!--\; \le \; -->x_i\le <!--\; \le \; -->b_i,\mathrm{\forall <!--\; \forall \; -->}i\in <!--\; \in \; -->{\mathbb{Z}}_{[1,n]},\end{array}$
where $x_i$ is the $i$-th scalar component of vector $x$ and $P>0$ is positive definite. As the objective is a convex function, the maximizer should be where all constraints are active, i.e. at an extreme point of the domain of the form $x_i^{\star <!--\; \star \; -->}=a_i$ or $x_i^{\star <!--\; \star \; -->}=b_i$. Is the global maximizer the one of such points such that its norm ${x^{\star <!--\; \star \; -->}}^T{x}^{\star <!--\; \star \; -->}$ is maximal?

I am reading a paper and they mention a hump function maximization: I am trying to prove the point of maximization:
$m=(1-<!--\; -\; -->x{)}^{1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}}$ where $x,\mathrm{\sigma <!--\; \sigma \; -->}\in <!--\; \in \; -->[0,1]$
It is said that m is a hump-shaped function of x maximized at $x=\mathrm{\sigma <!--\; \sigma \; -->}$, where $\mathrm{\sigma <!--\; \sigma \; -->}$ is a parameter that is assumed to be fixed here.

My attempt:
First derivative with respect to x: $(1-<!--\; -\; -->x{)}^{-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}}+(1-<!--\; -\; -->x{)}^{1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}=0$
$1+(1-<!--\; -\; -->x{)}^1{x}^{-<!--\; -\; -->1}=0$
Second derivative with respect to x: $-<!--\; -\; -->(1-<!--\; -\; -->x{)}^{-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}}+(1-<!--\; -\; -->x{)}^{-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}-<!--\; -\; -->(1-<!--\; -\; -->x{)}^{-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->1}+(1-<!--\; -\; -->x{)}^{1-<!--\; -\; -->\mathrm{\sigma <!--\; \sigma \; -->}}{x}^{\mathrm{\sigma <!--\; \sigma \; -->}-<!--\; -\; -->2}$
I couldn't get to the given result based on the above. Any clarification would be appreciated!

I am trying to find the maximum of a hermitian positive definite quadratic form $xQ{x}^H$ (where $Q=Q^H$ and all eigenvalues of $Q$ are non-negative) over the complex unit cube $|x_i|\le <!--\; \le \; -->1$, $i=1,\dots <!--\; \dots \; -->,n$ where $x=(x_1,x_2,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->{\mathbb{C}}^n$.

This is the problem of minimizing a concave function over a convex domain. I have read that this problem is NP-hard but there exist some bounds on the optimum. What global optimization technique would your recommend to tackle this problem numerically? Since I am new to the field of optimization, I would appreciate every answer, thanks!

I have an exam in a few hours. I need to understand the solution to the following question

Find the Maximal to the the following $2x_1+3x_2$ is the objective function.

The constraints are
$4x_1+3x_2\le 600;$
$x_1+x_2\le 160;$
$3x_1+7x_2\le 840;$
$x_1,x_2\ge 0.$
Also the answer should be in the "Dual". If its possible please do it in the Algebraic method. If not I would just like the solution using the tableau method and how do you arrive to the solution.(PS: Any help would be great. )

You are a jeweler who sells necklaces and rings. Each necklace takes 4 ounces of gold and 2 diamonds to produce, each ring takes 1 ounce of gold and 3 diamonds to produce. You have 80 ounces of gold and 90 diamonds. You make a profit of 60 dollars per necklace you sell and 30 dollars per ring you sell, and want to figure out how many necklaces and rings to produce to maximize your profits.


Clearly this is a maximization problem, and this is how I formulated it.

Let $n$, $r$, $g$, and $d$ represent units of necklaces, rings, gold, and diamonds, respectively.

Then, $n=4g+2d$, and $r=g+3d$.

Our profit can be defined by $p=60n+30r$.

Our constraints are $g\le <!--\; \le \; -->80$, and $d\le <!--\; \le \; -->90$.

I tried to approach solving this by substituting n and r for their production functions in the profit function. But, on second thought, I'm not sure if plugging them in maintains an equivalent profit function, because the gold/diamond amounts are tied together for each necklace/ring made.

So, I'm not sure if $p=60n+30r$ and $p=60(4g+2d)+30(g+3d)$ are equivalent statements.

Am I approaching this problem correctly? If not, how should I think about the problem? Thank you.

We are given a ratio:
$\frac{g(x)}{f(x)}$
where:
$g(x)\in <!--\; \in \; -->{\mathbb{R}}^{+}$
$f(x)\in <!--\; \in \; -->\mathbb{N}\cap <!--\; \cap \; -->f(x)\ge <!--\; \ge \; -->2$
So $g(x)$ returns values in $[0,+\mathrm{\infty <!--\; \infty \; -->}]$ while $f(x)$ returns values in $\{2,3,4,\dots <!--\; \dots \; -->\}$.

I am looking for a confirmation about a very simple question: if I maximize ${\displaystyle \frac{g(x)}{f(x)}}$, do I also maximize ${\displaystyle \frac{g(x)}{f(x)-<!--\; -\; -->1}}$ in this very particular case?

We need to prove that when we apply the Newton-Raphson method to strictly quadratic concave function. It will converge in one step.

How to apply this method to maximization of
$f(x)=4\cdot <!--\; \cdot \; -->x_1+6\cdot <!--\; \cdot \; -->x_2-<!--\; -\; -->2\cdot <!--\; \cdot \; -->x_1^2-<!--\; -\; -->2\cdot <!--\; \cdot \; -->x_1\cdot <!--\; \cdot \; -->x_2-<!--\; -\; -->2\cdot <!--\; \cdot \; -->x_2^2$
I did not understand how to apply method to this function? what should be the interval?

I have a simple question. This might be a theorem somewhere, but I do not know the appropriate keywords to find it. Please help.

Say there is a function $G(k,x)={\int <!--\; \int \; -->}_a^{x}f(k,t)dt$, and I wish to maximize $G(k,x)$ w.r.t. $k$. Under what conditions is this maximization problem equivalent to maximizing $f(k,t)$ w.r.t. $k$?

In short, when is the following true:
$\underset{k}{max}\{{\int <!--\; \int \; -->}_a^{x}f(k,t)\}dt\iff <!--\; \iff \; -->\underset{k}{max}\{f(k,t)\}$

A coffee company sells two types of breakfast blends. They have on hand 132 kg of dark roast and 84 kg of hazelnut. One breakfast blend will contain half dark roast and half hazelnut and will sell for $\mathrm{\$<!--\; \$\; -->}7$ per kg. The other breakfast blend will contain 3/4 dark roast and 1/4 hazelnut and will sell for $\mathrm{\$<!--\; \$\; -->}9.50$ per kg.

My approach:
I see this is an optimization problem and need to show that $132x_1+84x_2$ needs to be optimized but confused on the other constraints
I used the convex optimization polygon theory of checking on the endpoints and computing the maximal value, can someone check and elaborate further on the reply below?

We’ve been given a triangle $ABC$ with an area = 1. Now Marcus gets to choose a point $X$ on the line $BC$, afterwards Ashley gets to choose a point $Y$ on line $AC$ and finally Marcus gets to choose a point $Z$ on line AB. They can choose every point on their given line (Marcus: $BC$; Ashley: $CA$; Marcus: $AB$) except of $A$, $B$ or $C$. Marcus tries to maximize the area of the new triangle $XYZ$ while Ashley wants to minimize the area of the new triangle. What is the final area of the triangle $XYZ$ if both people choose in the best possible way?