Discover Maximization Techniques in Geometry Triangles

I am very new to optimization andI have to solve this equation

max U(k)=tlog(1+ yhpg/(pg+s))+mpge^(-ky)) st k>0

can anyybody give me idea where should I start

I know how to solve maximization problems on numbers, and I know how to solve differential equations which are equations on functions, but how do I solve a 'maximization problem' over functions?
Here is a specific problem:
Find a positive real function $F(x)$, continuous and monotonically increasing on the real interval $[0,1]$, which maximizes:
$\frac{F(x)}{F(1)}$
Subject to:
$F^{\prime }(x)=(1-<!--\; -\; -->x)F^{″}(x)$
what is the function $F$ which attains this maximum?

let $T\ge <!--\; \ge \; -->$1 be some finite integer, solve the following maximization problem.
Maximize $\underset{t=1}{\overset{T}{\sum <!--\; \sum \; -->}}$($\frac{1}{2}$)$\sqrt{x_t}$ subject to $\underset{t=1}{\overset{T}{\sum <!--\; \sum \; -->}}$, $x_t\le <!--\; \le \; -->1$, $x_t\ge <!--\; \ge \; -->0$, t=1,...,T
I have never had to maximize summations before and I do not know how to do so. Can someone show me a step by step break down of the solution?

I am trying to solve functional maximization problems. They are typically of the following form (where support of $\mathrm{\theta <!--\; \theta \; -->}$ is [0,1]):
${\displaystyle \int <!--\; \int \; -->[v(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))+u(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))-<!--\; -\; -->u_1(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))(\frac{1-<!--\; -\; -->F(\mathrm{\theta <!--\; \theta \; -->})}{f(\mathrm{\theta <!--\; \theta \; -->})})]f(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}}$
Now one way that was proposed to me was of point-wise maximization. That is you fix a $\mathrm{\theta <!--\; \theta \; -->}$ and then solve:
${\displaystyle argma{x}_{x(\mathrm{\theta <!--\; \theta \; -->})}v(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))+u(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))-<!--\; -\; -->u_1(\mathrm{\theta <!--\; \theta \; -->},x(\mathrm{\theta <!--\; \theta \; -->}))(\frac{1-<!--\; -\; -->F(\mathrm{\theta <!--\; \theta \; -->})}{f(\mathrm{\theta <!--\; \theta \; -->})})}$
Solving this problem would give me a number $x$ for each $\mathrm{\theta <!--\; \theta \; -->}$ and I will recover a function $x(\mathrm{\theta <!--\; \theta \; -->})$ that will maximize the original objective function.

I have two questions related to this:
1) Does such point-wise maximization always work?
2) What happens if rather than doing point-wise maximization I try and take the derivative of the objective function with respect to x(θ) and equating the first order condition to 0? Is this a legitimate way of solving the problem? Can someone show exactly what such a derivative would look like and how to compute it?

Define the function
$f(a,b,c,\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->},\mathrm{\gamma <!--\; \gamma \; -->},x)=max(0,max((a+x)\mathrm{\alpha <!--\; \alpha \; -->},(b+x)\mathrm{\beta <!--\; \beta \; -->})-<!--\; -\; -->(c+x)\mathrm{\gamma <!--\; \gamma \; -->}),$
where
$a,b,c,\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->},\mathrm{\gamma <!--\; \gamma \; -->},x\in <!--\; \in \; -->[0,M].$
Is it true that for any
$\mathrm{\xi <!--\; \xi \; -->}=(a,b,c,\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->},\mathrm{\gamma <!--\; \gamma \; -->}),$
the maximum of $f(\mathrm{\xi <!--\; \xi \; -->},\cdot <!--\; \cdot \; -->)$ occurs either when $x=0$ or $x=M$?

I think the answer is yes, but I have trouble prooving it.
My argument is as follows:
Given any $\mathrm{\xi <!--\; \xi \; -->}$, $f(\mathrm{\xi <!--\; \xi \; -->},c)$ will be
$f(\mathrm{\xi <!--\; \xi \; -->},x)=\{\begin{array}{ll}0& \text{case A}\\ (a+x)\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->(c+x)\mathrm{\gamma <!--\; \gamma \; -->}& \text{case B}\\ (b+x)\mathrm{\beta <!--\; \beta \; -->}-<!--\; -\; -->(c+x)\mathrm{\gamma <!--\; \gamma \; -->}& \text{case C}\end{array}$
Hence, $f(\mathrm{\xi <!--\; \xi \; -->},x)$ is a linear function of $x$ in all three cases, and the result follows.

I think this argument works only if each case is independent of $x$, but this is not the case.

As, when $\mathrm{\xi <!--\; \xi \; -->}$ is given, judiciously choosing $x$ may put $f$ in another case.

What would be a right way to prove this result?

The following problem
$\underset{x,y}{max}f(x,y)=\log <!--\; \; -->x+\log <!--\; \; -->(x+y)-<!--\; -\; -->2x-<!--\; -\; -->3y$
subject to$x\ge <!--\; \ge \; -->0,y\ge <!--\; \ge \; -->0$

is a concave maximization problem, and thus can be numerically solved by convex optimization methods. The optimal solution may be $(x^{\ast <!--\; \ast \; -->},y^{\ast <!--\; \ast \; -->})=(1,0)$.

However, I want to derive the solution analytically. I tried the following approach, for example, by taking partial derivative,
$\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{1}{x}+\frac{1}{x+y}-<!--\; -\; -->2$
$\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}x}=\frac{1}{x+y}-<!--\; -\; -->3$
And there exists no solution that makes both partial derivative to be 0.

Earlier last week I realized I needed to ship a large volume of things domestically. Of course, I decided that I wanted to do so as cheaply as possible.

I first looked at USPS standard post rates. I noted that if the "combined length and girth" of a box exceeds 108", then the package becomes significantly more expensive, as it uses the "oversized" price bracket, so I resolved not to exceed this size limit. The length of a box is defined as its longest dimension, and its girth is defined as the perimeter around the rectangle to which the length is orthogonal – in other words, the girth is $2w+2h$.

However, because I have much to ship, I also would like to maximize the volume of such a box while not exceeding this limit. This turns into a maximization problem with a constraint:

Maximize $V(x)=lwh$, while satisfying $108>=l+2w+2h$.

This seems pretty simple, but I'm not sure how to handle the three nominally independent variables $l$, $w$, and $h$. Because I did not want to wait or rely on my faulty math skills, I ultimately asked Wolfram|Alpha to solve this for me, yielding a box of size 36"x18"x18". This makes intuitive sense, to a degree, but I would like a hint as to how I could have proceeded to solve this on my own.

I Have read that for a matrix of reals $Y$ and a p.s.d matrix $B$ that the

Maximum of $f(Y)=Tr(Y^{T}BY)$ subject to $Y^{T}Y=I$ is achieved when $span(Y)$ equals the span of the first $d$ eigen-vectors of $B$.

What is the reasoning behind this eigen-solution leading to the maximum?

I'd like to compute
$A^{\ast <!--\; \ast \; -->}=\underset{\begin{array}{c}A\in <!--\; \in \; -->{\mathbb{R}}^{d\times <!--\; \times \; -->k}\\ A^{T}A=I\end{array}}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}<!--\; \; -->det(A^T\mathrm{\Lambda <!--\; \Lambda \; -->}A)$
where $k\le <!--\; \le \; -->d$, $\mathrm{\Lambda <!--\; \Lambda \; -->}=\mathrm{diag}<!--\; \; -->({\mathrm{\lambda <!--\; \lambda \; -->}}_1,\dots <!--\; \dots \; -->,{\mathrm{\lambda <!--\; \lambda \; -->}}_d)$ and ${\mathrm{\lambda <!--\; \lambda \; -->}}_1\ge <!--\; \ge \; -->\cdots <!--\; \cdots \; -->\ge <!--\; \ge \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_d\ge <!--\; \ge \; -->0$.

I strongly suspect that $A^{\ast <!--\; \ast \; -->}=[{\mathrm{ϵ<!--\; ϵ\; -->}}_1\cdots <!--\; \cdots \; -->{\mathrm{ϵ<!--\; ϵ\; -->}}_k]$, but after various attempts to prove it I gave up.

I have the following problem. Given this function
$E[\mathrm{\pi <!--\; \pi \; -->}]=(1-<!--\; -\; -->r)[\mathrm{\alpha <!--\; \alpha \; -->}b-<!--\; -\; -->(1-<!--\; -\; -->p)C-<!--\; -\; -->K]+T$
I would like to find the maximum w.r.t. $r$ given this constraint:
$U=(1-<!--\; -\; -->r)b-<!--\; -\; -->T\ge <!--\; \ge \; -->0$
It is an economic problem that I am formalizing, but this is not relevant to its solution, I am only interested in the mathematical resolution of the problem. I provide some background. Our variable $r$ is a number between 0 and 1, $p$ is some probability, $\mathrm{\alpha <!--\; \alpha \; -->},b,C,K$ and $T$ are all positive constants. If it is useful for the resolution of the problem, we can also assume that $\mathrm{\alpha <!--\; \alpha \; -->}$ is between 0 and 1. An important assumption (namely, assumption $\mathrm{\&<!--\; \&\; -->}$) is that $(1-<!--\; -\; -->p)C+K>\mathrm{\alpha <!--\; \alpha \; -->}b$. Readers used to economics may recognize that the objective function is a sort of expected profit and the constraint is a sort of expected utility. I tried with Kuhn-Tucker, but with miserable results since deriving w.r.t. $r$ does not yield any expression with $r$.

Now I'm following a more intuitive approach. I start by assuming that $U=0$ is the constraint, then I can find an expression for $r$ from the constraint and I substitute it in the target function. After easy steps, I get this
$T\frac{\mathrm{\alpha <!--\; \alpha \; -->}b-<!--\; -\; -->(1-<!--\; -\; -->p)C-<!--\; -\; -->K}{b}+T$
At this point, I can use assumption $\mathrm{\&<!--\; \&\; -->}$ to conclude that the first $T$ above will be negative, but I'm not able to conclude whether it will be lower, equal or higher to/than the second $T$ because it is multiplied by some constant.

At this point I'm stuck. I do not know if my approach can work. Could you please suggest a nicer way to proceed? I am on the right track or it's a dead end?

I'm having trouble understanding how to check the second order conditions for my unconstrained maximization problem.

This is the entire problem: Alicia wants to maximize her grade, which is a function of the time spent studying ($T$) and the number of cups of coffee ($C$) she drinks. Her grade out of 100 is given by the following function.
$G(T,C)=50+10T+16C-<!--\; -\; -->(T^2+2TC+2C^2)$
In the first order conditions, I find the partial derivatives and set them equal to zero. I get the following two equations:

$10-<!--\; -\; -->2T-<!--\; -\; -->2C=0$ and $16-<!--\; -\; -->2T-<!--\; -\; -->4c=0$. The first equation was the partial derivative with respect to $T$ and the second equation was the partial derivative with respect to $C$. Solving these two equations, I find that $C=3$ and $T=2$.

Now, I need to check the second order conditions. I know that the second partial derivative with respect to both $T$ and $C$ should be negative. This checks out. I get -2 from the first equation (with respect to $T$) and I get -4 from the second equation (with respect to $C$). The last thing I need to do with the second order condition is multiply these two together (which yields 8) and then subtract the following:
${\left(\frac{{\mathrm{\delta <!--\; \delta \; -->}}^{2}G}{\mathrm{\delta <!--\; \delta \; -->}T\mathrm{\delta <!--\; \delta \; -->}C}\right)}^2$
Please forgive me if this formula isn't displaying correctly. I tried using the laTex equation editor, but I'm not sure if it worked. Anyway, I need to know how to derive this. What is it asking for? I know that this part should be -2 squared, which is 4. Then, 8-4=4, which is positive and tells me that the second order conditions are met.

But where is the -2 coming from? I know within both of the equations, there are a few -2's. But, I'm not sure exactly where this -2 comes from.

Just come across a question regarding sequential maximization and simultaneous maximization, and I do not recall whether there are any established conditions for the equivalence. Anyone has some idea?
$\underset{x}{max}\underset{y}{max}f(x,y)=\underset{y}{max}\underset{x}{max}f(x,y)?$

My initial investment is $\mathrm{\$<!--\; \$\; -->}100$, and I earn $1\mathrm{\%<!--\; \%\; -->}$ interest per day. I can opt for any number of compoundings per day (if twice per day, then the interest rate per compounding period is $0.5\mathrm{\%<!--\; \%\; -->},$ and so on), but I have to pay $\mathrm{\$<!--\; \$\; -->}0.01$ each time my interest is compounded. After 365 days I will close the account.
What would this equation look like, and how should I include this to maximize my total deposit? How to generalize and figure out a good or optimal maximization?

my intuition says I'm right, but I couldn't find or get a proof for it:

Suppose I have a vector function $\stackrel{\to <!--\; \to \; -->}{x}(t)\in <!--\; \in \; -->{\mathbb{R}}^n$. I am interested in finding ${\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}_t||\stackrel{\to <!--\; \to \; -->}{x}(t)|{|}_2$. For some reasons, this optimization problem is hard to solve, what I can solve easily, however, is ${\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}_t||\stackrel{\to <!--\; \to \; -->}{x}(t)|{|}_1$.

I know that 1-norm and 2-norm are equivalent in ${\mathbb{R}}^n$. Does this also include that the solution of these two problems are equivalent?

I am writing a computer program where I have $x$ real positive varying in the domain $[\sqrt{U},U]$. I want the value of $x$ which maximizes:
$(1+\sqrt{U})-<!--\; -\; -->\frac{\sqrt{U}-<!--\; -\; -->1}{U-<!--\; -\; -->\sqrt{U}}x-<!--\; -\; -->\frac{U}{x}$
("visually", it seems to be a bit above $2\sqrt{U}$) what is the best way to find it exactly?

What is the height of a circular cone with surface line ("Mantellinie") s, which has the maximal volumina? My problem here is, that I am not really sure what surface line is..is this the same thing as just to say surface?

Suppose that $f(\cdot <!--\; \cdot \; -->)$ is a quadratic and concave function such that it has a maximum, $x\in <!--\; \in \; -->\mathbb{R}$, $y\in <!--\; \in \; -->(0,+\mathrm{\infty <!--\; \infty \; -->})$. I have to solve a maximization problem that is
$\underset{{\displaystyle \frac{x}{y}}}{max}f(x/y)$
My question is that, since I want to maximize with respect to $x/y$, I will treat this as if $\frac{\mathrm{\partial <!--\; \partial \; -->}f(x/y)}{\mathrm{\partial <!--\; \partial \; -->}(x/y)}$?

Or do i need to maximize with respect to $x$, then in $y$ and use the Hessian matrix?

Or is it something else? Well it is a composition of functions after all, isn't it? I am a llitle confused...

I try to maximize a quadratic function on box constraints as shown below
$\begin{array}{l}max{x}^{T}Qx\\ s.t.a\le <!--\; \le \; -->x_i\le <!--\; \le \; -->b\end{array}$
The length of vector $x$ is approximately 2000 and $Q$ is symmetric. I am wondering if there exists a solver (like Cplex or Gurobi) which can solve this maximization problem efficiently.

"A 10-meter length of wire is available for making a circle and a square. How should the wire be distributed between the two shapes to maximize the sum of the enclosed areas?"

Here's what I have:
$Are{a}_c=\mathrm{\pi <!--\; \pi \; -->}{r}^2$
$Are{a}_s=4r^2$
So, I'm thinking that I need to find the maximized radius size to figure everything else out.
$Are{a}_c+Are{a}_s=10$
$(\mathrm{\pi <!--\; \pi \; -->}{r}^2)+(4r^2)=10$
$\frac{\mathrm{d}}{\mathrm{d}<!--\; \; -->r}[(\mathrm{\pi <!--\; \pi \; -->}{r}^2)+(4r^2)-<!--\; -\; -->10]=2\mathrm{\pi <!--\; \pi \; -->}r+8r$
But here's my dilemma; if I take the derivative of that and solve for r, it comes out 0. So I'm not sure where I'm going wrong. Any advice?

The things we know, usually minimization of a convex function, unique solution will exist.

My question is, maximization of a strictly convex function, will that give an unique maximum? If so how we can proof that. The objective function looks like below.
max $A{e}^{-<!--\; -\; -->(d_0+d_1)}+B{e}^{-<!--\; -\; -->(d_0+d_1+d_2)}+C$
A,B and C are constants and $d_0,d_1,d_2$ are euclidean distances.