Please join.

I am feeling hopeless with the problems in sec 1.1 itself of the book. Please help.

hi

hi

thanks a lot

there is a big prob in the book that u read... there is no definition ...

but, your answer can be explored further. I mean say: there is need for a set of values of dimensions $x,y$ that I requested in the first comment to provide

, as am unable to get on my own

i don't understand your question

what dimension?

sorry, used wrong word in my comment (& here too repeated that),. Your answer stated that to avoid reverse ordering, need to have imposed on the value set $(x,y,r,s)$ with positive values so that $p$ is negative; & hence cannot take reverse ordering. I could not get the idea, and second cannot try out the values stated.

first, we need to agree on a definition... what are the possible values of i

any non-zero, as negative will also lead to proportional shapes

i see

then my answer is not valid

why?

all my working is based on the assumption that $i>0$

if u want to assume $i \ne 0$, to construct an example, try letting $x=y$ and $r \ne s$.

I could not understand a the determinant =0 then

your goal is to map $(x,y)$ to $(r,s)$

$px+u = r, py+u=s$

if $x=y$, then $r=s$, which is a contradiction.

by contradiction, do you mean - it is not possible? Second, why this was not tried by you for $i>0$

because one possible answer suffices. it is a matter of choice.

one counter example suffices

please elaborate why this counter-example (to show impossibility for general mapping) is not shown in your answer for $i >0$ case.

i have shown using another counter example

also, the other is chosen because under that setting, it helps in part d

but since u have chosen $ i \ne 0$, a more obvious choice is to pick something like $x=y$ and $r \ne s$

at least to me

but am unable to see a counter-example, there is a constraint to be imposed by $x,y,r,s$ values to avoid reverse ordering

how will that work is still not clear

which setting are u working on now

i.e., how $p$ would be constrained

have to pick one and stick to it

i > 0

$p=(r-s)/(x-y)$, so I just have to pick $r-s>0$ and $x-y<0$ to make no such $p$ exists

Please elaborate for $i >0 $ case

the $i$ is the $p$ right

yes, it is the proportionality factor, i.e. $i = p$.

if such transformation exist, then we have $px+u=r, py+u=s$

then i solve for $p$ but i show that expression is negative.

hence it doesn't exist

is that also the answer for (c) second part?

there are two transformation, of which later i switch their order and show that they can be reduced down to the first case

so, do you mean that the (c)'s second part does not impose ordering on models, and just asks if any ordering being used, will 2 steps be enough?

sorry, if sound weird; but could not interpret anyway else your last line of response.

my answer is there are things that can't be done within 2 steps

this is (c)'s second part with ----- ordering imposed or not? is the confusion. Is the second part imposing ordering or not?

u have two types of operation right

$P$ and $U$

so in total $PP$ $UU$, $PU$, $UP$

go through each cases if it is not obvious to u?

we have worked out that $UP$ is not enough

and in my answer, I have shown that $PU$ is equivalent to $UP$

$UU$ and $PP$ are quite obvious

yes, 4 combinations for 2 steps. $UP \equiv PU$. But, $PP, UU$ are banned?

show that $PP$ can be reduced to a single $P$ and show that it can't map arbitrary point.

similarly for $UU$

it is a simple matter of *(multiplication) of two $i$ for $PP$, similarly sum two values for $UU$. Yes, it is now clear. But, the matrix used in the answer is not clear------ in terms of taking inverse

i was just solving for $p$ and $u$

given $x, y, r,s$

solve for $p$ and $u$

but, you stated to find values of $x,y,r,s$ i.e. two points (representing rectangles) such that reverse ordering of $UP$ is not possible.

yes, but i didn't know that before hand... I am assuming that such mapping exists

and if it exists and $x \ne y$, then that relationship must hold

and from that working, i know how to pick my $x,y,r,s$ explicitly to make it not work

Please give an example, as you asked on the values of $x,y,r,s$

$(1,2)$ $(2,1)$

so $(x,y) = (1,2)$ and $(r,s)= (2,1), leading to $p = -1, q =3$

but we want $p$ to be positive

but is there no automated way to restraint based on this matrix, i.e. constrain based on some property of matrix. One can be $x!=y $ to ensure non-singularity (i.e. (if, not wrong) the two points $(x,y) & (r,s)$ are not the same.

but, the bigger issue is to constrain values of two points further by getting only positive values of $p/i$.

the matrices are jsut a tool to solve the problem

what is the problem that you are trying to solve

agreed the problem is smaller. Just wanted to know any property of matrices to solve in automated manner.

the matrix method is always applicable, but not in terms of taking the inverse

you can find its rref

my intuition from the working is points on the line $x=y$ can be mapped to each other. points on $x > y$ too. and the final case is $y < x$.

$x,y >0$.

for part (d) it seems for the first part, the answer is still negative.

hmmm dunno what you are talking about

you have stated for part (c), there are 3 cases based on the relative values of $x, y$. If $x>y$ mapping is possible as get positive $p$. Similarly, for $x = y$. But, the final case of $x<y$ is causing negative $p$, and hence not possible.

So, that is just repeated in part (d)

yup

Second part of (d) asks for the minimum number of times the switch between U, P is needed to allow growth

2

i.e. mapping between $(x,y)$ and $r,s)$

but, how to prove it?

for any finite number of transformation

$PP$ can be simplified to be $P$

$UU$ can be simplifed to be $U$

$PU$ can be simplifed to be $UP$

so in the end of the day

just focus on $UP$

Thanks a lot. But, how will rref form help is not clear / known to me.

rref is just a method to solve linear system

fixing $x,y,r,s$, solving for $p$ and $u$ is a lienar system

it is just a tool

if u r not familiar with it, use soemthing simpler like substitution or elimination

no am too much conversant with it. Just its application here is confusing me

i just use it to solve the linear system

I hope you mean that while deriving the rref form, it should be clear if the determinant is $0$ by checking if any row is all zeros, but then I stop

to make any further inference

also, it is a value based method; so one rref form for each value set

anyway, if $x \ne y$, the solution is clear, and if $x=y$, then we quickly reach the conclusion that $r=s$ for it to be consistent

I request another question, that is on pg#12 of the book's first chapter ( worldscientific.com/doi/suppl/10.1142/7810/suppl_file/… ). can we use the map of (p, A) to be drawn somewhere else & find relation. I feel that the book's s/w (Visumatica) or any else cannot give the equation. It is all based on self-estimation.

Although in MODELS/ CH1 the last model 1.8-1 is representing that. But, for me to create that by using multiple values needs some sort of programming, and feeding that to a plotting function.

i will need some time to read the text

no issues

in fact, highly thankful for accepting that

so u r given a parameter and the area?

and then i dunno what is your question

my question is about finding equation of the graph of points (p, A) using 3 given points, or more points with optimal points (i.e. with optimal defined by maximum area for a given perimeter). I feel 3 points are not 'enough' to find shape. And to have many more points there is 'hopefully' need for a program and then again need an estimation. Is there a formula that can be derived in by the plotting program, or by using matrices.

are u referring to a particular question?

on that book?

that question only, as it asks the shape of the 3 optimal points' curve

hmm

on page 12, i see 5 questions

is it one of them?

4th one

I thought if matrices or any way the 3 points be fed and the quadratic form is found by reduction or whatever

$A=f(p)$, solve for $f(p)$?

have you solved it?

please see page 28 question #4(a)

it refers to the same

ya, have you solved for $f(p)$?

you can solve it either by $AM-GM$ or substiution

no, I thought it involved calculus and finding maxima to get area for a given perimeter

sure

u can use calcuplus too

max xy subject to x+y = p

oops, i mean max xy subject to x+y = p/2

dA/ dP = dA/dx*dx/dP

am I correct?

hmmm... i dunno what to say

i would just substituet $y = p/2 -x$ into $xy$ and solve a single quadratic equation

so, $A = 2x - x^2$ for $p=4$ and then derivative $dA/dx = 2 - 2x$ with $ x = 1$, and check for maxima by second derivative of dA/ dx

which is negative = $-1$, so maxima

graphically, this has only one peak as at desmos.com/calculator/ci5pjkfr4h

what you are interested... is to maximize $A$ given an arbitrary $p$

so solve it generally

rather than letting $p=4$.

Sorry, cannot proceed for general P, stuck. But easy to see otherwise that the curve for $A= 2x - x^2$ has peak only once, and any value of $x$ would give a maxima.

$\max x(p/2 - x)$

solve quadratic equation

dA/dx = p/2 - 2x => p = 4x => 2(x +y) = 4x => y = x

yes, so optimal when $x=y$

checking for maxima / minima by second derivative; $dA^2/ dx^2 = -2$, maxima

but, plotting can be only in terms of a single variable, i.e. need a particular value of $x$ to get $A= 2x - x^2$ for a particular value of $P$. I do not know a general way to plot (i.e. in terms of $A, P , x $ all three)

we know $x+y = p/2$ and we know $x=y$, solve for $x$ and $y$.

could not gather

it is a linear system right

yes

you can also use substitution

so can you solve for $x$ and $y$?

remark: leaving in less than 11 minutes... also after early next month might disappear for a long time... i do not know how long. life is tough.

but pottig needs a fixed/chosen value of $P$

yes, solve for $x$ and $y$ first

then note that $P=xy$

so my derivation for $p=4$ for $dA/dx = 2x - x^2$ is correct? Need a family of curves with the peaks to be taken as a connected curve from each curve?

that curve of peak points would form a curve (needed one)

so, need for $p=6,7,3,2$ etc.

don't do it one by one

solve for $x$ and $y$ in terms of $p$

then compute $A=xy$

cannot gather, as need only peaks to plot the needed 'optimal' curve; and you state to find for all the curves (for different $P$ values) by ----- solving $x,y$ in terms of $P$. This last part is not clear.

i need to go soon

but yup, think of how to solve $x+y = 8$ and $x=y$

but, then you are also taking a particular value of $P=16, P/2 = 8$

and then look at your working, replacing $8$ by $\frac{p}2$

ok, need to go

not able to understand, as to how solve by $\frac{P}2$, stuck forever seemingly...........

please plot in desmos whenever you have time

the curve formed by peaks

please give then link here, or in the post

would be highly thankful