Expert Help with High School Geometry

It's been a while since I've had to do derivations so I'm just a little rusty. I need to understand the steps for deriving (Py – 2)(44 – 2Py + Px) to reach Py = 12 + 0.25Px. Py here is price of product y and Px is the price of product x.

I undertsant that I need to take product rule of the first equation and then set it = 0 and solve for Py, but it's finding that derived equation that's troubling me.

How I tried to do it was to first simplify the equation into 44Py - 2Py^2 + PxPy - 88 + 4Py + 2Px. Then taking the derivative and simplifying a little I end up with 46 - 3Py + Px which I would then set = 0 and solve for Py, giving me Py = (46 + Px)/3. Which is obviously not correct.

If you could help me through the steps to reach Py = 12 + 0.25Px I would be very greatful.

I am trying to find a function prescribed in polar coordinates $r=f(\mathrm{\theta <!--\; \theta \; -->})$ that maximizes the following quantity
$\frac{{\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{r}^3\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}d\mathrm{\theta <!--\; \theta \; -->}}{{\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{r}^{4}d\mathrm{\theta <!--\; \theta \; -->}}$
subject to the constraint $r\le <!--\; \le \; -->R\mathrm{\forall <!--\; \forall \; -->}\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->[0,2\mathrm{\pi <!--\; \pi \; -->})$.

I don't even know where to begin to be honest. I went through a couple of books on calculus of variations. I haven't seen anything that comes close to the problem I have at hand.

Extended Pythagorean Theorem
Extended Pythagorean Theorem We all well familiar with the basic Pythagorean statement – I will use a different description that will serve the discussion that follows – Given two points in 2D space, (a,0) and (0,b) with distance d between them, than the relation between a,b, and d follows
$a^2+b^2=d^2$
In 3D this is extended to three points (a,0,0), (0,b,0) and (0,0,c) with an area of s for the triangle created by the three points - than the following equation holds:
${}^1{/}_4(a^2+b^2+c^2)=s^2$
I wonder if this is true for higher dimensions, as example 4D - given four points (a,0,0,0), (0,b,0,0), (0,0,c,0), (0,0,0,d) and the volume for the pyramid created by the four point to be v than
${}^1{/}_9(a^2+b^2+c^2+d^2)=v^2$

Deduce the Pythagorean Theorem [duplicate]
Let V be an inner product space,and suppose that x and y are orthogonal vectors in V.We also know ${\Vert x+y\Vert }^2={\Vert x\Vert }^2+{\Vert y\Vert }^2$.My question is how can we deduce the pythagorean theorem in ${\mathbb{R}}^2$ from it.If possible give geometrical views.
Any help will be greatly appreciated.
thanks!! in advance.

Pythagorean theorem in functional analysis
Prove the Pythagorean theorem and its converse in R : f is orthogonal to g if and only if
$\Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2=\Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2$
LHS -> RHS
$\begin{array}{rl}& \Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2\\ =& (f,f)-<!--\; -\; -->2(f,g)+(g,g)\\ =& \Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2-<!--\; -\; -->2(f,g)+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2\end{array}$
In order for LHS=RHS −2(f,g) has to be 0 which means that f and g are othogonal. If −2(f,g)=0 Then
$=\Vert <!--\; \Vert \; -->f\Vert <!--\; \Vert \; -->+\Vert <!--\; \Vert \; -->g\Vert <!--\; \Vert \; -->$
RHS-> LHS
$\begin{array}{rl}& \Vert <!--\; \Vert \; -->f{\Vert <!--\; \Vert \; -->}^2+\Vert <!--\; \Vert \; -->g{\Vert <!--\; \Vert \; -->}^2\\ =& (f,f)+(g,g)\\ =& \Vert <!--\; \Vert \; -->f-<!--\; -\; -->g{\Vert <!--\; \Vert \; -->}^2\text{ iff }-<!--\; -\; -->2(f,g)=0,\text{ iff }f\text{ and }g\text{ are orthogonal.}\end{array}$

I'm a first year Finance grad student and we're learning utility maximization problem now. We have been assuming that a solution to these problems exists. Are there any general theorem about the existence of maximization solution on ${\mathbb{R}}^n$? I have encountered once that "A utility maximization problem with a continuous utility function on a compact set has solution". Thanks in advance!

Parametrize the contours of integration and write the integrals in terms of the parametrizations. Do not calculate them.
$\int <!--\; \int \; -->\frac{\stackrel{¯<!--\; ¯\; -->}{(}z)}{z^3}dz$
where
$\mathrm{\Gamma <!--\; \Gamma \; -->}$
is the arc of the circle of radius
$\sqrt{2}$
centered at the origin with initial point at 1+i and terminal point at 1-i that lies in the right half-plane and is transversed once.

I asked a question similar to this last night and felt that everyone was very helpful. I'm hoping to see the solution to this problem as I feel completely lost.

Suppose we have a function of the form: $(x_1-<!--\; -\; -->x_2)+(x_3-<!--\; -\; -->x_4)+(x_5-<!--\; -\; -->x_6)$ and we have maximized this summation using linprog (using some constraints which are not important for this matter). This provides us with a value for the different x variables.

The problem I now want to solve is the maximization of the minimum $(x_i-<!--\; -\; -->x_j)$ and with the constraint that the solution $(x_1,x_2,...)$ filled in in the original summation should have a higher or the same value (constraint).

This would distribute the difference between the variables $x_1,x_2;x_3,x_4,...$
How can this problem be solved in Matlab (maximization of the minima)?

Find the condition so that the line $px+qy=r$ intersects the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in points whose eccentric angles differ by $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$
Though I know how to solve it using parametric coordinates, I was wondering if there's an another approach which is less time consuming.

Is it true that for all symmetric $n\times <!--\; \times \; -->n$ matrix, $(a_{ij})$, such that for $x\in <!--\; \in \; -->\{\pm <!--\; \pm \; -->1{\}}^n$
$\underset{ij}{\sum <!--\; \sum \; -->}{a}_{ij}{x}_i{x}_j\le <!--\; \le \; -->1$
there exists a universal constant such that
$\underset{ij}{\sum <!--\; \sum \; -->}{a}_{ij}{x}_i{y}_j\le <!--\; \le \; -->C$
for all $x,y\in <!--\; \in \; -->\{\pm <!--\; \pm \; -->1{\}}^n$ . I tried to use the polarization identity
$⟨<!--\; ⟨\; -->Ax,y⟩<!--\; ⟩\; -->=⟨<!--\; ⟨\; -->Au,u⟩<!--\; ⟩\; -->-<!--\; -\; -->⟨<!--\; ⟨\; -->Av,v⟩<!--\; ⟩\; -->$
where $u=(x+y)/2$ and $v=(x-<!--\; -\; -->y)/2$. However, as $x$ and $y$ vary over $\pm <!--\; \pm \; -->1$ vectors, $u$ and $v$ can be vectors in $\{\pm <!--\; \pm \; -->1,0\}$.

I am given three co-ordinate points of a circle O(Ox,Oy) as a center. Then two Points other points as A(Ax,Ay) & B(Bx,By). Now I have to find the arc length of that circle. Can you please help. Thanks in advance.

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 $A$.

Similar Triangles and the Pythagorean Theorem

Derive the Pythagorean Theorem by eliminating the x.
I have already shown that $\mathrm{△<!--\; △\; -->}BAE$ and $\mathrm{△<!--\; △\; -->}BDE$ are congruent, and that $\mathrm{△<!--\; △\; -->}EDC$ and $\mathrm{△<!--\; △\; -->}BAC$ are similar triangles. However, I am having trouble setting up the resulting proportions and using that to derive the Pythagorean Theorem. I believe the resulting proportions are
$\frac{BA}{DE}=\frac{AC}{DC}=\frac{BC}{EC}.$.
Any help will be appreciated!

Consider the circle of radius r pictured below with central angle , measured in radians, and subtended arc of length s. Prove that the area of the shaded sector is $A=\frac{1}{2}{r}^2\mathrm{\theta <!--\; \theta \; -->}$.

Let's say the entire circle has area $B=\mathrm{\pi <!--\; \pi \; -->}{r}^2$. Using proportionality arguments, the area of a portion of a circle is then $\frac{\mathrm{\theta <!--\; \theta \; -->}}{2\mathrm{\pi <!--\; \pi \; -->}}B=\frac{\mathrm{\theta <!--\; \theta \; -->}}{2\mathrm{\pi <!--\; \pi \; -->}}\mathrm{\pi <!--\; \pi \; -->}{r}^2$ which reduces to $A=\frac{1}{2}{r}^2\mathrm{\theta <!--\; \theta \; -->}$.
Is that right? It seems too easy for a prove. I got this question from a online book which doesn't contain the answer, so I can only resort to stack exchange.

Pythagorean theorem in higher dimensions?
When using vector notation in coordinate systems (Cartesian coordinates) we see that the magnitude of a vector in two dimensions is equal to the square root of its Y component squared added to its X component squared (Pythagorean theorem).But the same calculation is done for a three dimensional vector that has X, Y, and Z components.Is there a triangle that has four sides? (of course not, but how does this right triangle formula work for a calculation that involves more than two dimensions?).

If I know the length of a chord of a circle and the length of the corresponding arc, but do not know the circle's radius, is there a formula by which I can calculate the length of the sagitta?

Find the arc-length of a circle with radius a.
From the equation of a circle, I found out the equation for the one quadrant, which is:
$y=\sqrt{a^2-<!--\; -\; -->x^2}$
I tried solving the problem, and here's what I did:
$A=4{\int <!--\; \int \; -->}_0^a\sqrt{1+\frac{x^2}{a^2-<!--\; -\; -->x^2}}dx$
$A=4{\int <!--\; \int \; -->}_0^a\frac{a}{\sqrt{a^2-<!--\; -\; -->x^2}}dx$
And this is where I need help. How can solve this integral now?

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.