Discover Maximization Techniques in Geometry Triangles

I have encountered that in some cases maximization of a function had been substituted with a maximization of its monotone transformation.

For example, finding the min or max of $f(x,y)=((x-<!--\; -\; -->1{)}^2+(y-<!--\; -\; -->2{)}^2{)}^{1/2}$ is the same as finding the min / max of $(x^2+y^2)$ and then take the square root of the answer. So, my questions are:
1) Is it true for any monotone transformation $g$?
2) Is the same principle allow us to maximize /minimize $\log <!--\; \; -->g(x)$ instead of $g(x)$ itself?

I have non-negative function $g(y,x)$ that is define using non-negative $f(y,x)$ in the following way:
$\begin{array}{r}g(y,x)={\int <!--\; \int \; -->}_0^{x}f(t,y)dt\end{array}$
I am trying to maximize $\underset{y\in <!--\; \in \; -->S}{max}g(y,x)$. Using Fatou's lemma we have that
$\begin{array}{r}\underset{y\in <!--\; \in \; -->S}{max}g(y,x)={\int <!--\; \int \; -->}_0^{x}f(t,y)dt\le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^x\underset{y\in <!--\; \in \; -->S}{max}f(t,y)dt\end{array}$
I also have that $\underset{y\in <!--\; \in \; -->S}{max}f(t,y)\le <!--\; \le \; -->h(t)$ where ${\int <!--\; \int \; -->}_0^{x}h(t)<\mathrm{\infty <!--\; \infty \; -->}$

My question when does the last inequality hold with equally?

What do I have to assume about $f(y,x)$ and $g(x,y)$ for equality to hold?

Can I apply dominated convergence theorem here?

This is a utility maximzation problem

maximize $x^a+y^b$ subject to $p_{1}x+p_{2}y=w$ (utility maximization problem)

Anyone has any idea, there are no restrictions on $a$ and $b$, as far as i can see it. many thanks!!!

I maximized the function
$f(x,y)=(x+2)(y+1)$
subject to
$4x+6y=130,x,y\ge <!--\; \ge \; -->0$
and I found $(x,y)=(16,11)$. But the question I'm doing asks me to analyse the "sufficient conditions of second order".

So, I thought I should use some theorem that tells me (16,11) is the max value of $f$. I don't know if I could use Weierstrass' Theorem here and I believe the hessian matrix doesn't help me either.

Maximize
$x_2-<!--\; -\; -->x_1+y_1-<!--\; -\; -->y_2$
given that $x_1^2+y_1^2=1$ and $x_2^2+y_2^2=1$.
I was thinking about using Lagrange multipliers, but I only know how that works for a 3-variable function, not 4. Could someone please suggest a way to solve this? Maybe with Lagrange multipliers or some more elementary method?

Suppose I have $n$ integers ${\mathrm{\mu <!--\; \mu \; -->}}_i$, $i=\{1,2,...,n\}$. Define $\stackrel{¯<!--\; ¯\; -->}{\mathrm{\mu <!--\; \mu \; -->}}=\frac{1}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_i$. It is given that all the ${\mathrm{\mu <!--\; \mu \; -->}}_i$'s are either +1 or −1. How can I show that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}({\mathrm{\mu <!--\; \mu \; -->}}_i-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{\mathrm{\mu <!--\; \mu \; -->}}{)}^2$ is maximized only when $\frac{\lfloor <!--\; \lfloor \; -->n\rfloor <!--\; \rfloor \; -->}{2}$ or $\frac{\lceil <!--\; \lceil \; -->n\rceil <!--\; \rceil \; -->}{2}$ are +1 and rest are −1's.

I was thinking about proving this inductively, but I am not sure how.

If $X\sim <!--\; \sim \; -->$ Poisson $(\mathrm{\lambda <!--\; \lambda \; -->})$, prove that $P[X=k]$ is maximized at $k=\mathrm{\lambda <!--\; \lambda \; -->}$

How would I go about proving something like this so that the method would work to prove a maximization point of other random variables like say Binomial is max at $k=(n+1)\star <!--\; \star \; -->p$?

I am trying to solve the following question:
$\text{Maximize }f(x_1,x_2,\dots <!--\; \dots \; -->,x_n)=2\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{x}_i^{t}A{x}_i+\underset{i=1}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}\underset{j>i}{\overset{n}{\sum <!--\; \sum \; -->}}(x_i^{t}A{x}_j+x_j^{t}A{x}_i)$
subject to
$x_i^t{x}_j=\{\begin{array}{cc}1& i=j\\ -<!--\; -\; -->\frac{1}{n}& i\ne <!--\; \ne \; -->j\end{array}$
where xi's are column vectors ($m\times <!--\; \times \; -->1$ matrices) with real entries and A is an $m\times <!--\; \times \; -->m$ ($(n<m)$) real symmetric matrix.

From some source, I know the answer as
$f_{max}=\frac{n+1}{n}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}_i^{\downarrow <!--\; \downarrow \; -->},$
${\mathrm{\lambda <!--\; \lambda \; -->}}_i^{\downarrow <!--\; \downarrow \; -->}$ being the eigenvalues of $A$ sorted in non-increasing order (counting multiplicity). But I am unable to prove it. I will appreciate any help (preferably with established matrix inequality, or Lagrange's multiplier method).

An apartment complex on Ferenginar with 250 units currently has 193 occupants. The current rent for a unit is 965 slips of Gold-Pressed Latinum. The owner of the complex knows from experience that he loses one occupant every time he raises the rent by 2.5 slips of Latinum. What should be our recommendation for the optimal rent?

The equation I got was $(193-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->(965+2.5x)$. I tried to maximize it and got $x=-<!--\; -\; -->96.5$. The optimal rent is $965+2.5(-<!--\; -\; -->96.5)=723.75$.

This answer is not correct? Where did I make a mistake Thanks

Let us consider a linear programming problem shown below:
$\begin{array}{r}\underset{x}{min}f(x)-<!--\; -\; -->c\end{array}$
where, $f(x)$ is a linear function of $x$. Now, can be following maximization problem be considered as an equivalent to the above minimization problem?
$\begin{array}{r}\underset{x}{max}\left(\frac{1}{f(x)-<!--\; -\; -->c}\right)\end{array}$

Consider the example of maximizing $x^{2}yz$ under the constraint that $x^2+y^2+z^2=5$.

One way to do this is to use lagrange multipliers, solving the system of equations
$2xyz=2x\mathrm{\lambda <!--\; \lambda \; -->}$
$x^{2}z=2y\mathrm{\lambda <!--\; \lambda \; -->}$
$x^{2}y=2z\mathrm{\lambda <!--\; \lambda \; -->}$
$x^2+y^2+z^2=5$
However, couldn't you just substitute $x^2=5-<!--\; -\; -->y^2-<!--\; -\; -->z^2$ into the expression you want to maximize to get: $yz(-<!--\; -\; -->y^2-<!--\; -\; -->z^2+5)$and then just maximize that by setting the $y$ and $z$ partial derivatives equal to zero?

Then you just have to solve the arguably simpler system of equations:
$3y^{2}z+z^3=5z$
$y^3+3y{z}^2=5y$
where the $z$s and $y$s cancel out nicely on both sides.

Why is maximize by lagrange multipliers necessary when you can always substitute and maximize the resulting function?

When should you choose one over the other?

Let $w,a\in <!--\; \in \; -->R^n$, and $B\in <!--\; \in \; -->R_{++}^{n\times <!--\; \times \; -->n}$ (the set of $n\times <!--\; \times \; -->n$ positive definite matrices).
We know that the following function (which is a specific form of the Rayleigh quotient) has a unique maximum, and a closed-form solution in $w$:
$f(w)=\frac{w^{T}a{a}^{T}w}{w^{T}Bw}$
It's maximum is achieved at $w^{\ast <!--\; \ast \; -->}=B^{-<!--\; -\; -->1}a$ and its value is $f(w^{\ast <!--\; \ast \; -->})=a^T{B}^{-<!--\; -\; -->1}a$. (by using the generalized eigenvalue decomposition)

Now here is my question:
If I have $w,a_1,a_2\in <!--\; \in \; -->R^n$, and $B_1,B_2\in <!--\; \in \; -->R_{++}^{n\times <!--\; \times \; -->n}$, and the following function:
$g(w)=\frac{w^T{a}_1{a}_1^{T}w}{w^T{B}_{1}w}+\frac{w^T{a}_2{a}_2^{T}w}{w^T{B}_{2}w}$
Then, what can we say about the maximum of $g(w)$, can we still solve for $w$ in closed-form?

Given matrix-valued function $A:{\mathbb{R}}^d\to <!--\; \to \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->n}$ and (tall) matrix $B\in <!--\; \in \; -->{\mathbb{R}}^{n\times <!--\; \times \; -->m}$, where $n><!--\; >\; -->m$, let matrix-valued function $C:{\mathbb{R}}^d\to <!--\; \to \; -->{\mathbb{R}}^{m\times <!--\; \times \; -->m}$ be defined by
$C(x):=B^{T}A(x)B$
I would like to maximize the determinant of $C(x)$ without directly considering $C(x)$. If I maximize the determinant of $A(x)$, does that maximize the determinant of $C(x)$ as well?

If it were $m=n$, then we would have:
$det(C)=det(B{)}^{2}det(A)$
Hence, maximising $det(A)$ would maximize as a consequence $det(C)$. Does it hold something similar in the case with $n><!--\; >\; -->m$? Or is it possible to define some lower bound for $det(C)$? And if so, how to prove it?

If in the previous general case it doesn't hold, maybe it could help that, in my specific case $C$ and $A$ are two positive definite matrices and $B$ is a sparse matrix of zeros and ones. $A$ is also block diagonal. The only way I came up with is using Cauchy-Binet for the determinant of the product of rectangular matrices but I remained stuck with a summation involving the principal minors of the Cholesky factorization of $A$ times minors in $B$.

I need help with the following optimization problem

$max\mathrm{\alpha <!--\; \alpha \; -->}\ln <!--\; \; -->(x(1-<!--\; -\; -->y^2))+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})\ln <!--\; \; -->(z)$
where the maximization is with respect to $x,y,z$, subject to
$\begin{array}{rl}\mathrm{\alpha <!--\; \alpha \; -->}x+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})z& =C_1\\ \mathrm{\alpha <!--\; \alpha \; -->}y\sqrt{x(x+\mathrm{\gamma <!--\; \gamma \; -->})}-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}x& =C_2\end{array}$
where $0\le <!--\; \le \; -->\mathrm{\alpha <!--\; \alpha \; -->}\le <!--\; \le \; -->1$, $\mathrm{\gamma <!--\; \gamma \; -->}>0$, and $x,z\ge <!--\; \ge \; -->0$, and $|y|\le <!--\; \le \; -->1$.

Generally, one can substitute the constraints in the objective function and maximize with respect to one parameter. The problem is that in this way things become algebraically complicated, and I believe that there is a simple solution.