help-36

[worthy wedge]

no

no don't

[north bison]

when u say things like 2*3 u mean 2³

[cyan sandal]

means times

[north bison]

2 3 times

is 6

[cyan sandal]

yep so times the numerator by 3 as well

so x/2 = 3x/6

and the same for 2x/3

to get 4x/6

now you can add them to 7x/6 = 21

times both sides by 6

so 7x = 126

divide by 7

x= 18

question asks for half the value, so 18 / 2 is 9

[north bison]

ngl this one was kinda confusing

[cyan sandal]

yeah thats fair its just some algebra

[north bison]

eh ill understand

but ngl in these math questions im just knocking off big or small numbers

to be left with 50/50 chance

And my exam is tomorrow too

[cyan sandal]

i mean its mult choice

you can alweays do trial and error

[north bison]

yea

[cyan sandal]

if and only if your equation doesnt work out

its faster to solve algebraicly

[north bison]

oh theres some other

questions i couldnt quite understand

[cyan sandal]

feel free

[cyan sandal]

lets start with this

so lets solve each inequality

an inequality is when two terms are compared using >, <, <= or >=

so the first one

[north bison]

What does that weird E mean

[cyan sandal]

so thats epsilon

in set theory it means is an element of

[north bison]

Ypsilon?

[cyan sandal]

so like x is a part of which set

[north bison]

variables

[cyan sandal]

a set as an a set of values

for example

each of these choices are a set

[north bison]

yea

[cyan sandal]

so to solve the inequality

cross multiplication 7(x-2) < 2(x+3)

7x-14 < 2x+6

7x - 2x < 6 + 14

5x < 20

x < 4

so x has to be smaller than 4, not including 4

second one 5(2x-4) <= 4(3x-2)

so 10x - 20 <= 12x-8

[north bison]

where do u find 7

or any of these new numbers

[cyan sandal]

so if you do cross multiplication

so if your given a/b = c/d

then ca = bd

[north bison]

So u add the upper stuff with the ones down or

[cyan sandal]

multiply

not add

[north bison]

yea multiply

×

[cyan sandal]

so 8 - 20 <= 12x - 10x

so -12 <= 2x

-6 <=x

so x can be greater than or equal to -6, butless than 4

so its [-6, 4)

which is B

[north bison]

ok fairs short explanation

what about

21

i was almost about to guess it

[cyan sandal]

so 30 is what he recieved, but there was 25% tax

meaning his original income was 40, since 40 * 0.75 = 30

so since its 8%, 0.08 * x = 40

x = 40 / 0.08

,calc 40/0.08

[soft zealot]

Result:

500

[cyan sandal]

so its D

[north bison]

i wont have the oppurtunity to calculator

[cyan sandal]

thats fine also

[north bison]

but yea i understand this onr

too

[cyan sandal]

so 8 is 40/5

so you know its something related to 5

since 0.08 is 8 * 0.01

[north bison]

yea

[cyan sandal]

5 divided by that is 500

[north bison]

alr do 22 and this is gonna be the last one

[cyan sandal]

so we can set up an equation

where x is his money

[north bison]

2/7 2/5 3/1

[cyan sandal]

not quite

its tricky here because it says 2/5 of the REMAINDER

[north bison]

but adding ×

[north bison]

oh

[cyan sandal]

so if he spends 2/7

he has 5/7 lef

t

[north bison]

yea

[cyan sandal]

if he spends 2/5 he has 3/5

so 3/5 of 5/7 is 3/7

and he spends 1/3 so he has 2/3 left

2/3 of 3/7 is 2/7

since 2x/7 = 26

2x = 182

and x = 91

so D

[north bison]

Dam that wasnt hard

[cyan sandal]

yes its confusing because the remainder

[north bison]

Thanks for spending 30 minutes helping me on math tho

[cyan sandal]

no prob

[north bison]

may god help you with other problems u got

!close

[cyan sandal]

its . close

[sturdy flax]

it's .close.

[north bison]

.close

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[north bison]

thx

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[warm python]

So I wanna find the first lagrange polynomial here

[night raft]

the formula for first degree Lagrange polynomial is $P_1(x) = f(x_1) L_1(x) + f(x_2) L_2(x)$

[soft zealot]

MxRgD

[warm python]

yea, was about to type that

cool , so that verifies what I wanted to verify

thanks

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[mint forum]

hi guys could you help me with this sequences and series question please

[drowsy epoch]

Yes, but tell us what you tried and know about logs

[mint forum]

Oh okay cool so let me quickly draw what i did already

and then a = log3/2

r = 8/9

but idk wht they want from me 😭

[drowsy epoch]

hehe

[mint forum]

yess

[urban raptor]

You used that log(a)+log(b)=log(a+b) which is not always true.

[drowsy epoch]

Try to write some terms out and see what happens using that

[mint forum]

oh wait you're meant to multiply?

i just realised

[polar pollen]

logs of products, not logs of sums

[mint forum]

ok let me see

[timber pilot]

You can use this formula
$$ \sum \log(a_{i}) = \log\left(\prod a_{i}\right) $$

[soft zealot]

William James Moriarty

[mint forum]

i did this

[rare girder]

the bottom expression is not the same as the sum; it's actually [\sum_{n=1}^{48}\log_5\left(\frac{n+2}{n+1}\right)=\log_5\left(\frac32\right)+\log_5\left(\frac43\right)+\ldots+\log_5\left(\frac{50}{49}\right)]

[mint forum]

yeah sorry thats way too complex for me

[soft zealot]

Flip

[rare girder]

but with multiplication in place of addition, all is well

[mint forum]

okkk nicee thank u i did that but where do i go from there

[timber pilot]

The result not true

[urban raptor]

But 3, 4, 5, ..., 49 cancels out so you should get 50/2.

[mint forum]

cancels out what

[polar pollen]

the 3 in the numerator of the first fraction cancels with the 3 in the denominator of the second fraction. the 4 in the numerator of the second cancels with the 4 in the denominator of the third so on so forth

[mint forum]

yeah thats where im struggling

[mint forum]

Okay let me try that thank you so much guys

so do i have to write out more terms

[urban raptor]

You do not. You can cross out 5 and 49.

[mint forum]

why though

[polar pollen]

No

[mint forum]

is that just a rule or something

[south dirge]

Just try observing the sequence!

[mint forum]

okay so eventually all the denominators will cancel then?

[south dirge]

yup!!

except 2 ofc

[mint forum]

Ohhhh u smart people

and i can do that anytime

[south dirge]

Along with all numerators asw!

[mint forum]

with any question like that?

[south dirge]

I mean yeah?

[mint forum]

this is crazy

[south dirge]

-# As long as its following the basic rules of multiplication

[drowsy epoch]

🔭

[mint forum]

ok u lot are smart tysmmm guyss 🥰

[south dirge]

Yup!
Its called a telescopic sequence

[mint forum]

really? wow new terminology

[polar pollen]

Yep they cancel one by one all the way along the chain. only things that have nothing to cancel w are 2 on the bottom and 50 on the top no fraction before and after

[mint forum]

niceeeee

this is y i love this server man

ok tysm guys i get it now

[south dirge]

Welp then lets us know your final ans!

[mint forum]

LOL Oh yeah 🤣

ok holdon

[south dirge]

Oh wait it is a prove that Qsn

Yeah then ur settled ig

[mint forum]

Ok yea i got 2

YAYY

nahhh some prove questions r crazy

ty again guys, i'll be back for more 😭

[south dirge]

Sure anytime!

[south dirge]

Your wlcm!
Have a nice day!

[mint forum]

You too!!

.close

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[lofty spade]

Let f : N → N be a bijective function, and define another function g : N → N
by g(x) = xf(x) . We say that g is golden if g is non-decreasing . Determine the number of functions
f : N → N for which g is golden. Take N = {1, 2, 3, . . . , 2026}.

[onyx peak]

Have you tried anything?

If you don't know where to start, I'd start by trying to construct some examples

for example what can f(1) be?

quite interesting question btw, stealing it

[lofty spade]

f(1) can be anytg from 1 to 2026

[onyx peak]

Really?

keep in mind that f is bijective

there will have to be something that'll eventually map to 1

and you need the g to be non-decreasing

[lofty spade]

yea if f(1) is 2026, f(2) = 2025 is perfectly fine and i assume it works out like that

oh nvm

f(2026) goes to 1

then g(x) is also 2026

but g(2) is 4050

ahhh

ok so the highest valaue in this case wud be g(1013) = 1013*1012

f(x) = 2027 - x

[onyx peak]

hm now i realized that i was maybe a bit too strict

[lofty spade]

that doesnt work

f(x) = x is fine

[onyx peak]

yeah, this doesnt work

[onyx peak]

i actually dont think this can work

[lofty spade]

yea yea that is when f(x) = 2027-x

so it wont work

but the thing i cannot always define f(x) as x or 2027-x or smtg in that form

it can also be jumbled

and i dont know how to work it out then

[onyx peak]

I think that f(1) can only be 1 or 2, but i forgot how / if i even proved proved it

[lofty spade]

oh

interesting

[onyx peak]

hmm, suppose f(1) = n

then we have some m for which f(m) = 1

g(1) = n
g(m) = m
so n <= m

[lofty spade]

g(1) = n and g(m) = m, so m>=n

yea ok

[timber pilot]

We can use the fuct that 2026=2×1013
And 1013 is a prime number

g(2) can be 2026

And f(2)=1013 for example

[onyx peak]

what would the other values be?

[lofty spade]

😭

is there a key idea to this q

@onyx peak

[timber pilot]

I think we can assume that f is a linear function

[onyx peak]

nope, i have non-linear sols

i think I know what the solution class is

but i cant prove it

[lofty spade]

what do u think it is

so that i can also think abt proving it

uhh

this is very tempting

y wud u send it tho 😭

[onyx peak]

for @timber pilot

[onyx peak]

you can open it if you wanna see it

im not even sure whether its correct

[rare girder]

it's virtual bubble wrap

[lofty spade]

fr

[rare girder]

you probably would've been better off dming it

[timber pilot]

g is non decreasing?

[lofty spade]

ok imma show self control and not open it

yep

[onyx peak]

yeah, thats in the question

[timber pilot]

xf(x)≤(x+1)f(x+1)

[onyx peak]

<=

[onyx peak]

if you dont know what to do now, just try constructing some sols

[lofty spade]

i did that for smaller sets

[onyx peak]

ill try to find a way to prove that my solution class is correct in the meantime

[lofty spade]

like N = {1,2,3}

it didnt really give me an idea

[lofty spade]

ok

[onyx peak]

very roughly, my idea is that f can never grow too quickly

most solutions will be very close to f(x) = x

[lofty spade]

hmm

[onyx peak]

f(1) = 2
f(2) = 1
f(x) = x
this would be one sol e.g.

[lofty spade]

f(2025) = 2026
f(2026) = 2025 also is fine

basically u can switch any 2 adjacent numbers

ig

[onyx peak]

yeah

now try doing this

f(1) = 3

try completing it

[lofty spade]

but thats not an exhaustive set

[onyx peak]

see if you can

[lofty spade]

i dont think u can put f(1) = 3

[onyx peak]

yeah

and I also dont think you can put e.g. f(3) = 5

or f(2) = 4

[lofty spade]

but can u put f(2024) = 2026

[onyx peak]

...
f(2024) = 2026
f(2025) = 2025
f(2026) = 2024

doesnt work because 2025 * 2025 != 2024 * 2025

[lofty spade]

yea this doesnt work

so the only case is when u swtich 2 adjacent numbers??

damn

is that true

[onyx peak]

the missing part in this argument is that maybeee 2024,2025,2026 could appear somewhere before 2024

[timber pilot]

I think I get an idea

[onyx peak]

but i dont think that can happen, because intuitively, that just makes the problem bigger (more difference between f(x) = x and the function) and delays it

[onyx peak]

so yeah, this is my solution class

swapping some adjacent numbers

[lofty spade]

<@&268886789983436800>

[lofty spade]

but is there a way to prove it

[onyx peak]

maybe william has an idea

but honestly idk yet

maybee some kind of minimal counterexample thing could work

oh maybe considering the loop

f(1) = n

f(n) = m

f(m) = p

...
f(q) = 1

this creates kind of a loop

maybe there could be a way to prove that the loop cant be big

[lofty spade]

uhh how does that exactly help

oh\

i mean

[onyx peak]

if you can prove that the loop is 1 or 2 in length, then youre done

<@&268886789983436800>

[lofty spade]

wth

[timber pilot]

g is non decreasing function
So g(x)≤ g(x+1)
xf(x)≤(x+1)f(x+1)
If we assume that f is decreasing f(x)>f(x+1)
That mean f(x)≥f(x+1)+1

f(x)/f(x+1)≥1+1/f(x+1)
And f(x)/f(x+1)≤1+1/x
So
f(1+x)≥x

[lofty spade]

wait take the loop to be of size 3

[timber pilot]

We can say that f(i)≥i+1

[onyx peak]

that sounds wrong

[timber pilot]

That's contradiction

[onyx peak]

because f(x) = x wokrs

oh wait youre proving that f is increasing?

[timber pilot]

I assume tht f is decreasing
And i get this result

[timber pilot]

Yes

[onyx peak]

f can be decreasing at some points though

for example f(1) = 2, f(2) = 1, f(x) = x

[lofty spade]

for k<m<l
f(k) = m
f(m) = l
f(l) = k
g(k) = mk
g(m) = ml
g(l) = lk

so lk>=ml>=mk

but the first part of the inequality, lk>=ml gives k>=m, but we know m>k

[timber pilot]

f can't be constant
Because f is bijective

[lofty spade]

so its impossible to have a loop of size 3

is this proof enough @onyx peak

[onyx peak]

this deals with loops of size 3, induction can get it to arbitrary size i believe

but there is still sth missing i think

what if you had l < k < m

or some other arrangement

[onyx peak]

yeah, but
f(1) = 2
f(2) = 1
f(3) = 3
f(4) = 4
...
works but isnt increasing

[lofty spade]

in that case wudnt they be part of a bigger loop or smtg like that?

[onyx peak]

f(k) = m
f(m) = l
f(l) = k
WLOG suppose that k is the lowest number

Now there are 2 cases:
k < m < l
k < l < m

you need to deal with both the cases

you did the first one

[lofty spade]

whats WLOG

[onyx peak]

without loss of generality

[lofty spade]

oh k

[onyx peak]

we dont lose anything by that assumption, because we could just rotate the values and get k to be the lowest one

[lofty spade]

ahh ok

WLOG is such a cool abbreviation lol

[onyx peak]

If k < l < m, then
km <= lk <= ml

so m <= l

and thats enoguh

[lofty spade]

yea so it works for the other case also right?

[onyx peak]

yeah

[lofty spade]

so loop size 3 isnt possible

[onyx peak]

now we just need induction for bigger loops i think

[lofty spade]

and im assuming a bigger loop is not possible either

[onyx peak]

there must be a more elegant argument tho

[lofty spade]

wait so r u sure bigger loops are impossible?

[onyx peak]

not yet

but we can try

suppose that loops of size < n are impossible, we will try to prove that loops of size n are impossible as well

[lofty spade]

uhhh

induction here will get messy very quickly tho

right?

[onyx peak]

my idea is that if we get some loop of size n, then we could maybee extract some element safely without breaking the loop-ness

and then we would have loop of a smaller size

contradiction

but it might not work

[lofty spade]

so ig there is a diff more elegant mehtod to thiis

[onyx peak]

idk, the safest element of the loop to remove is the largest one prolly

[lofty spade]

also i didnt really understand this

can u explain it a bit more

[onyx peak]

Are you familiar with proofs by minimal counterexample?

[lofty spade]

no

[onyx peak]

Its basically just contradiction

suppose that there is a loop with size > 3

take the smallest such loop

if we can find a way to make that loop even smaller by extracting some element and rejoining the loop in a way that keeps it working, then thats a contradiction

[lofty spade]

wdym by smallest such loop?

[onyx peak]

because we have found a smaller loop

[onyx peak]

smallest in length

like the shortest

[lofty spade]

hmm ok

it still doesnt make full sense to me tho

[onyx peak]

hm ill just check if it even works and then explain it in more depth if it does

or maybe ill find something more elegant

[lofty spade]

ok

@onyx peak I saw u typing lol

[onyx peak]

yeah im typing out my proofs in here

[lofty spade]

Ooh

Damn

Thank you!

[onyx peak]

they dont work very well

[lofty spade]

It doesn't matter

As long as it leads me in the right direction

Ig

[lofty spade]

Wud u mind dming the proofs tho

[onyx peak]

well, nothing really worked so far

i tried contradiction on f(i) >= i+2

then f(1), f(2), ..., f(i) have at most i-1 elements from {1,2,3,...i+1}

ohhh

that means that 2 elements of {1,2,3,...i+1} will not be mapped to

[final saddle]

@lofty spade Has your question been resolved?

[loud sundial]

\begin{tcolorbox}
Assume that for some $k \in {1, \dots, n}$, the values
$${f(1), f(2), \dots, f(k-1)}=[k-1].$$
Then, either $f(k)=k$ or $f(k)=k+1$ and $f(k+1)=k$.
\end{tcolorbox}
\vspace{0.3cm}
Did ygs establish this yet

[soft zealot]

Civil Service Pigeon

[loud sundial]

I can't really tell if this was mentioned

because I see some loop ideas but idk if this is what y'all ere thinking of

[loud sundial]

oh nvm I think you got the consequence of this anyway (every valid permutation is a product of disjoint consecutive blocks of size 1 and 2, where the blocks of size 2 are adjacent transpositions)

[final saddle]

Channel closed

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[loud sundial]

.reopen

[final saddle]

✅ Original question: #help-36 message

[loud sundial]

\begin{tcolorbox}
Assume that for some $k \in {1, \dots, n}$, the values
$${f(1), f(2), \dots, f(k-1)}=[k-1].$$
Then, either $f(k)=k$ or $f(k)=k+1$ and $f(k+1)=k$.
\end{tcolorbox}
\vspace{0.3cm}
Since ${f(1), \dots, f(k-1)}=[k-1]$, none of $1$, $\dots$, $k-1$ are available at position $k$. If $f(k)=k$, then we are done. \

From now on, we assume that $f(k) \neq k$. Then, $f(k) \geq k+1$. Let $t$ be such that $f(t)=k$. Then, $t>k$. Consider any $i \in [k, t-1]$. It is easy to see that $f(i) \geq k+1$ for every such $i$, so
$$g(t-1)=(t-1)f(t-1) \geq (t-1)(k+1).$$
And since $g$ is nondecreasing with $g(t)=tf(t)=kt$,
$$(t-1)(k+1) \leq sk \implies s \leq k+1$$
and so $f(k+1)=k$. \

Snce $g$ is nondecreasing,
$$g(k) \leq g(k+1) \implies kf(k) \leq (k+1)k \implies f(k) \leq k+1.$$

But we also have $f(k) \geq k+1$, so $f(k)=k+1$ (and $f(k+1)=k$.) \ \

So, either $f(1)=1$ or $f(1)=2$ and $f(2)=1$. Proceeding inductively, the following is trivial:
\begin{tcolorbox}
Every valid permutation is a product of disjoint consecutive blocks of size $1$ and $2$. All blocks of size $1$ are fixed points and all blocks of size $2$ are adjacent transpositions.
\end{tcolorbox}

[soft zealot]

Civil Service Pigeon

[loud sundial]

Ok enjoy my yap

I'd want to guess that all such block decompositions produce a valid f

is there anything in here about that

idt there's anything

Well a block of size $1$ at position $i$ would have $g(i)=i^2$

[soft zealot]

Civil Service Pigeon

[loud sundial]

For a block of size $2$ at positions $i$ and $i+1$, we have $f(i)=i+1$ and $f(i+1)=i$

[soft zealot]

Civil Service Pigeon

[loud sundial]

oh wait $g(i)=g(i+1)$ from this then

[soft zealot]

Civil Service Pigeon

[loud sundial]

ok so g is constant on any block of size 2

[loud sundial]

g is non decreasing

so we just need to show that g does not decrease between blocks

wait that's obvious

oh shoot this is really nice

[loud sundial]

ok I see wym

yeah it's just recursion now

damn unexpected connection

or maybe it is and I'm just dense lol

@lofty spade @onyx peak enjoy my stream of consciousness ig

[karmic glen]

<@&268886789983436800>

[final saddle]

@lofty spade Has your question been resolved?

[final saddle]

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[left trail]

For 31 I think our lower bound step function is 1/floor(x+1) and our upper bound step function is 1/floor(x). I don't see how they can equal a_k and a_k-1 though

[left trail]

We know that for a step function s(x) we have $\int_a^b s(x)dx=\sum_k=1^n s(x_k) (x_k-x_{k-1})$

[soft zealot]

BigBen

[left trail]

I see that we have the $a_k-a_{k-1}$ in both sums for 31 and they represent each sub interval which leaves us with $\frac{1}{a_k}=\frac{1}{\floor{x+1}}$ and likewise for the other sum

[soft zealot]

BigBen

[left trail]

But how could $a_k=\floor{x+1}$

[soft zealot]

BigBen

[final saddle]

@left trail Has your question been resolved?

[left trail]

<@&286206848099549185>

[final saddle]

@left trail Has your question been resolved?

[leaden moon]

well

both fractions have the same numerator

so it has the same denominator

[left trail]

It isn't the same a though

[leaden moon]

elaborate

[final saddle]

@left trail Has your question been resolved?

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[ivory lynx]

whats the difference conceptually?

[worldly spruce]

What do you mean conceptually?

One holds for all, one holds just for sone

[ivory lynx]

like in concept

[sturdy cypress]

it's the real life concept

no one knows how to explain these things

[ivory lynx]

i see

[sturdy cypress]

∀x P(x) = ~(∃x ~P(x))

∃x P(x) = ~(∀x ~P(x))

[ivory lynx]

right right

[barren hound]

∃x means "there exists an x" - implicitly "there exists at least one x"
∀x means "for all x"

[ivory lynx]

can we say that

Universal is also existential, but existential isn't universal

does that make sense 😭

if its universal it must be existential, but it cannot be vice versa

there

[barren hound]

hmmmm well technically no

there's an idea of "vacuous truth" which is where you say like

∀x: P(x) is true if there are no x's to begin with

but.... i see what you mean. Assuming you have at least one thing in your set,
∀x: P(x) ==> ∃x: P(x)
because if it's true for all of them then it's true for at least one of them

[ivory lynx]

i see

thanks guys

.close

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[lofty spade]

if for a 2x2 matrix, A^3 - 3A^2 + 2A = 0, then can i say that A^2 - 3A + 2 is its characteristic eqn??

[desert mantle]

no

[pliant elk]

why not

[desert mantle]

take the zero matrix

from your information you can only conclude that your char poly is one of six different options

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[wary juniper]

Question?

[twin pivot]

How do you get to the formula on the right with that constructed cube?
And how was it constructed?

[pliant elk]

watch the video ig

[wary juniper]

Take a cube of length x, add 3 cuboids of lengths (x,x,1), add 3 cuboids of lengths (x,1,1), and finall add a cube of length 1

[twin pivot]

it isnt a video

[twin pivot]

why

[pliant elk]

maybe
start with a cube of side 1
add x to every side
the equation is just shwoing a relation between the volume of 1st cube , volume of 2nd cube and x

do you understand till here ?

[twin pivot]

no

[pliant elk]

what course are you studying in uni

(so that i can explain with some reference to it)

[twin pivot]

mathematical methods for engineering

?

[barren pebble]

(x+1)^3 is the volume of the whole thing, a cube with side length x+1

The pic shows that it's equal to x³+3x²+3x+1

The blue cube is of side length x, so its volume is x³

The 3 green cuboids are (x,x,1) so their total volume is 3x²

The 3 orange ones are (x,1,1) so their total volume is 3x

And lastly the pink cube is of side length 1, so its volume is 1

[twin pivot]

I was actually interested in knowing how to factor a polynomial in general using that cube, not this particular example.

[barren pebble]

Then you'd need higher dimensions

(for higher degree polynomials)

But uhm... This is only nice here because it's a cube

[pliant elk]

there is no general "method" for factoring a polynomial instead of finding its roots

[twin pivot]

OKAY

Thank you all!

.close

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[pliant elk]

well

you could find a root

and then long divide

and then keep doing it

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[lofty spade]

do u remember the question i asked that day

[lofty spade]

its part of that only

[desert mantle]

the one with AB and BA?

have you thought in the mean time about tr(AB) and tr(BA) ?

[lofty spade]

see the msg in help 23

.close

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[dim flume]

Yooo

[dim flume]

Need help understanding Lagrange multipliers

[blissful meadow]

What do you know so far?

[dim flume]

Yooooo

What’s up man

Nothing 😂😂

[dim flume]

One thing I know though is that Lagrange multipliers are just another method of doing the same exact critical point optimization problems right

[blissful meadow]

They give a more systematic way of solving for the critical points on a boundary, yes

Or on a contraint like the ones you're given.

[dim flume]

Could you use Lagrange multipliers for the problems we did yesterday as well?

Wait can we do it on the box problem

[blissful meadow]

Yes, you would find the critical points of the function in the inside of the region first like you did in the beginning, and then instead of having to parameterize the boundary and all that stuff you can just use Lagrange multipliers to find critical points on the boundary.

And yes that would work on the box

[dim flume]

Can we do the box problem with it then?

So it was L + 2(W+H) = 108

[blissful meadow]

Yes, that's our constraint now.

[dim flume]

Yeah so our volume needs to be in such a way that doesn’t pass that girth

[blissful meadow]

Yeah. The volume is V(W,H,L) = W*H*L and the constraint is L + 2W + 2H - 108 = 0.

[dim flume]

Yes

[blissful meadow]

So first for Lagrange multipliers you would compute the gradients of the functions V(W,H,L) and g(W,H,L) = L + 2W + 2H - 108.

[dim flume]

Why’s that?

[shadow marlin]

https://www.youtube.com/watch?v=5A39Ht9Wcu0 maybe this helps, maybe not

[blissful meadow]

If you have an objective function $f(x,y,z)$ along with a constraint $g(x,y,z) = 0$, then the Lagrange multipliers method says that the extrema on the constraint will occur when $\nabla f (x,y,z) = \lambda \nabla g(x,y,z).$

[soft zealot]

Azyrashacorki

[blissful meadow]

So here your objective function is $V(W,H,L)$ and the constraint is $g(W,H,L) = L + 2W + 2H - 108 = 0$

[soft zealot]

Azyrashacorki

[dim flume]

May I ask for the intuition behind that? Like I’m seeing one gradient vector which is the one of the constraint function being scaled by some value to then become the gradient of the objective function

[blissful meadow]

I think it's better imaged in functions of 2-variables, but it still works for more.

[dim flume]

Were you gonna say something after this

I didn’t wanna interrupt

[blissful meadow]

Yes, but it's not easy to understand what's going on in writing.

I'm trying to think of a picture.

I think we've discussed the fact that the gradient vector is perpendicular to a level set right?

[dim flume]

No not really

Thing is I’m also trying to intuitively understand the gradient vector ngl to you

[blissful meadow]

Like if f(x,y) = k is a level set (the set of (x,y) such that f(x,y) = k, so on this set f(x,y) is a constant k), then grad(f) points orthogonally to the level set since the steepest ascent is orthogonal to direction of no ascent at all, aka orthogonal to the level set.

[dim flume]

Could you please explain to me how the gradient vector behaves in general

I know it’s the components in 2D of the x partial derivative and the y partial derivative

But idk what’s that exactly supposed to mean or look like

[blissful meadow]

The general idea is that the gradient points in the direction where the function increases the fastest.

[bronze grove]

wouldn't it be better to just tell him how to solve lagrange multiples

[blissful meadow]

They asked for an explanation.

[bronze grove]

ohhh are they doing the theory part

makes sense

[blissful meadow]

If you remember the directional derivative of $f(x,y)$ in the direction of a unit vector $\vec{u}$ being $D_{\vec{u}} = (\nabla f) \cdot \vec{u} = ||\nabla f|| \cos(\theta)$ where $\theta$ is the angle made between $\vec{u}$ and $\nabla f$.

[soft zealot]

Azyrashacorki

[dim flume]

Yeah I remember we did directional derivatives but not necessarily the cos interpretation but yeah I did the dot product before

[blissful meadow]

And in particular, since $\cos(\theta)$ attains its maximal value when $\cos(\theta) = 1$, aka when $\theta = 0$, this suggests that $D_{\vec{u}}$ is greatest when $\vec{u}$ is in the same direction as $\nabla f$.

[soft zealot]

Azyrashacorki

[blissful meadow]

aka. $\nabla f$ points in the direction where the directional derivative is the largest : in the direction of fastest increase ("steepest ascent").

[soft zealot]

Azyrashacorki

[dim flume]

Wait so where did |(gradient)f| · cos(θ) come from

[blissful meadow]

That's how you can interpet the dot product.

$\vec{a} \cdot \vec{b} = ||\vec{a}|| \cdot ||\vec{b}|| \cos(\theta)$

[soft zealot]

Azyrashacorki

[blissful meadow]

Here u is a unit vector, so ||u|| = 1

[dim flume]

I think I got lost here

Where you’re talking about cos being at max value and how that correlates to the gradient

[blissful meadow]

Well this dot product |grad(f)| cos(theta) is the directional derivative.

|grad(f)| is a positive number, and -1 <= cos(theta) <= 1.

So the number you get from this dot product is greatest when cos(theta) = 1, which means the angle between the direction you're computing the derivative in and the direction in which grad(f) points is 0.

Which means you compute the biggest directional derivative when the direction you're using is the same direction as grad(f).

Which means that grad(f) always gives the direction in which f is increasing the fastest.

[dim flume]

Oh okay yes I agree

[dim flume]

Yes

[dim flume]

Lost

[blissful meadow]

At a point you can compute the directional derivative in any direction.

[dim flume]

Yes

[blissful meadow]

You agreed that you compute the biggest directional derivative when the direction you're computing in is the same direction as grad(f).

[dim flume]

Yeah

[blissful meadow]

So grad(f) points in the direction in which the directional derivative is the largest

So grad(f) points in the direction in which the function has the greatest rate of change

[dim flume]

I mean I guess I don’t get it because of the following:

you said “points” which suggests a direction

All I know is that the dot product of the gradient vector and the unit directional vector can decrease and increase depending on the angle, but that’s that the scalar value you’d get from the dot product increases/decreases and I don’t really know how that indicates some sort of direction

And lastly I don’t really know what partial of f = <f_y,f_x> are really supposed to represent or look like

[blissful meadow]

grad(f) and u are both vectors. They point in some direction.

So there's an angle between them, theta.

[dim flume]

yes

I’m having trouble understanding grad(f)

[blissful meadow]

If you compute the directional derivative in the direction of u, you end up finding that it depends on this angle theta, and that it's maximal when theta = 0.

So this angle between grad(f) and u needs to be 0.

This means that as vectors, they point in the same direction.

And in particular, amongst all the directions you could've chosen to compute your derivative in, the one which is in the same direction as the vector grad(f) yields the largest rate of change.

[dim flume]

I think my problem may be in not understanding the gradient vector in and of itself

[blissful meadow]

Well you'll agree it's a vector I suppose

[dim flume]

Yes it’s a vector I agree

But I don’t get the components,
Like this first component is the derivative wrt x and the second component the derivative wrt y

But those are just tangent lines

[blissful meadow]

They aren't tangent lines. They're slopes of tangent lines in the x and y direction, respectively.

[dim flume]

Oh

Wait

Is the gradient vector the component of all possible instantaneous velocities of both partials

[blissful meadow]

If you want, the gradient vector contains the information about the tangent plane. It tells you the slopes of the tangent plane at a point in the x and y direction.

This is why you can use the gradient to find an equation to the tangent plane.

So if the gradient vector is (1,2) at a point, you know the tangent plane will have the form 2x + 1y - z = d for some constant d.

[dim flume]

I mean tbh with you I never use the gradient to find the tangent plane I just do it the vector interpretation way 😂

Which is very complicated

But that’s how I learned it

Btw you taught me that the other day Lolol

But yeah

[blissful meadow]

The point is the gradient vector is linked to the best linear approximation (a plane) of f(x,y) the same way the usual derivative is linked to the best linear approximation (a line) of f(x) in the 1-variable sense.

[dim flume]

I guess for now I can just accept the fact that the gradient points to the where the function increases the most

Right

[blissful meadow]

Okay!

So where were we

Right so the gradient points in the direction of fastest increase.

Now say you consider the points on the level set f(x,y) = 1 for a constant 1.

Those are specifically those points (x,y) such that f(x,y) = 1.

If you've ever seen a topographic map of a mountain or something, those are the lines where the height is a certain constant in this case 1, i.e. when f(x,y) = 1

[dim flume]

yeah

[dim flume]

yes

so

are you trying to scale the gradient of the level curve to be the same of that of the objective curve

[blissful meadow]

Well first I wanted to explain why the gradient is always perpendicular to a level set.

[dim flume]

oh my bad

yeah go ahead

[blissful meadow]

So this level set comprises of points (x,y) for which f(x,y) = 1.
Therefore on this level set, the function f(x,y) is constant.

[dim flume]

yes

[blissful meadow]

Since f(x,y) is constant on the level set, it means that f(x,y) doesn't increase at all on it, yes?

[dim flume]

i mean if it was horizontal

the level curve

[blissful meadow]

Yes the level curve is "horizontal" since it lives in the plane z=1 in this case.

[dim flume]

is that always the case

[blissful meadow]

Yes, since you're requiring that z = f(x,y) = constant

So all the points satisfying f(x,y) = constant will be mapped to points on the plane z=constant.

[dim flume]

oh yes youre right

yes

yeah it doesnt increase

[blissful meadow]

Right

So now the thing is that grad(f) points in the direction of greatest increase

[dim flume]

yes

it does

[blissful meadow]

This direction has to be perpendicular to the direction in which f doesn't change.

And specifically, on the level set, f doesn't change.

So grad(f) must point perpendicularly to the level set.

[dim flume]

wdym by this

[blissful meadow]

Intuitively, if you're on a mountain and you make sure to always face the steepest ascent wall to the peak, if you go right or left you shouldn't be going neither up or down.

[dim flume]

true

[blissful meadow]

But in terms of directional derivatives it makes sense as well, since if $D_{\vec{u}}) = ||\nabla f|| \cos(\theta) = 0$ and we assume that $\nabla f \ne (0,0)$, then we need $\cos(\theta) = 0$, which means $\theta =90^{\circ}$.

[soft zealot]

Azyrashacorki

[dim flume]

wait but where did directional derivatives come from

we were discussing gradients on the level curve right

[blissful meadow]

Yes, and the above shows that if you're at a point on a level set and you keep going around this level set, since f doesn't change, you must be going exactly perpendicularly to the direction of greatest increase.

[dim flume]

oh i see yeah

i think

[blissful meadow]

Good. So we've established that the gradient always points perpendicular to a given level curve.

[dim flume]

ill accept that

for the sake of time since i have a test tmrw 💀

[blissful meadow]

Right, so before we go back to the exercise, the mechanism behind Lagrange multipliers is that you have f(x,y) and a constraint g(x,y) = 0 (which you can think of as a level curve of g(x,y)).
What we want is to find the maximum value f(x,y) can attain on this curve, so in particular we want to find a level curve f(x,y) = k which intersects the level curve g(x,y) = 0 (this means that the points we find are actually on the constraint we set) with the biggest value for k.
You can imagine taking a level curve f(x,y) = k which intersects the level curve g(x,y) = 0, probably at multiple points and slowly increasing k up until the level curve you get just kissed the level curve g(x,y) = 0. After this point, the level curve f(x,y) = k won't intersect g(x,y) = 0, so this suggests that this specific k where the two level curves just kiss corresponds to our maximum!
Now this means that if f(x,y) = k is this maximal level curve, it must be tangent to the level curve g(x,y) = 0 at a point! Now remember that the gradient of a function is perpendicular to its level curves. This means in particular that at this point where the level curves are tangent, the gradient vectors of f and g will have to be pointing in the same direction!

But what does this mean? It means specifically that $\nabla f(x,y)$ is a multiple of $\nabla g(x,y)$, i.e. $\nabla f(x,y) = \lambda \nabla g(x,y)$ for some $\lambda$.

[soft zealot]

Azyrashacorki

[blissful meadow]

So the takeaway is that along a constraint, you only need to check those points where this happens

[final saddle]

@dim flume Has your question been resolved?

[dim flume]

can we work on the excercise maybe itll make a bit more sense?

[blissful meadow]

Yes.

So we have a function $V(W,H,L) = WHL$ and a constraint given by the function $g(W,H,L) = L + 2W + 2H - 108 = 0$.

[soft zealot]

Azyrashacorki

[blissful meadow]

Now you have to compute the gradients of those functions

[dim flume]

how should i express the constraint function

in terms of 3 variables?

[blissful meadow]

Like I said your constraint function is g(H,W,L) = L + 2W + 2H - 108

And your objective function is V(H,W,L) = WHL

Its gradient is not 0.

[dim flume]

idk what i was doing 💀

@blissful meadow

[blissful meadow]

Yes. We're not super interested in the actual value of lambda. You can use those equations to figure out stuff about W, H, and L.

[dim flume]

wdym? like solve for each of the three instead?

[blissful meadow]

Well say from the first two equations you get that WL = 2lambda = HL.

Since L>0, what can you say about W and H?

[dim flume]

greater than 0?

yeah idk

<@&286206848099549185>

[blissful meadow]

Sorry I'm back.

You have WL = HL

If L is nonzero.

What can you do

Specifically what can you do on both sides of the equation

[dim flume]

divide by L

so W = H?

[blissful meadow]

Yes

So now we know that any point of extremums on the constraint satisfy W=H

Now from say the first and third equations you know WL = 2lambda = 2WH

Considering W > 0 what do you get ?

[dim flume]

what are doing here though with these equations

i dont get where and why were coming up with them

[blissful meadow]

You need to find W,H and L that render this system of equations possible.

From the first two equations you've already found that you need W=H.

[blissful meadow]

From this one you should get something else

And this will give you essentially all the information you need about the side lengths that respect the constraint which give extremums.

[dim flume]

howd you get 2wh

[blissful meadow]

The first equation gives WL = 2lambda.
The third equation says that WH = lambda, so 2WH = 2lambda, so 2WH = WL.

[dim flume]

oh sure

2H = L?

[blissful meadow]

Yes.

So now you know that you need W=H, and L = 2H.

This is a lot of information, because remember we have our constraint which says that
L + 2H + 2W = 108

[blissful meadow]

Using the two equations then can you solve for H in here?

[dim flume]

H = 108/6?

=18

[blissful meadow]

Okay that's good

What about W and L now?

[dim flume]

L = 36, W = 18?

[silent mango]

i was watching you people type this out and this is correct...👍

[dim flume]

thanks

wait so thats it

[silent mango]

yep

[dim flume]

wait so i just wanna know something

[silent mango]

yes?

[dim flume]

for the sake of time cuz i have a test tmrw i just wanna know if theres like one way into doing this problem that i can just "memorize" the steps of, such as how we did it here for the box problem, and if that method also applies for all optimization problems in 3D

[silent mango]

i believe you just do this:

• Compute ∇f
• Compute ∇g
• Set: ∇f = λ∇g
• Add the constraint equation
• Solve the system
• Check boundaries if needed

[blissful meadow]

Mind you this one was a bit more complicated because you need to translate the whole word problem thing, and it's an objective function in 3-variables, so not exactly easy to deal with at first.

It's still the same process, but you may find it's a bit easier to deal with the exercises you had at the start with just 2 variables first.

[final saddle]

@dim flume Has your question been resolved?

[dim flume]

do you think you could help me with this

[blissful meadow]

Of course.

[blissful meadow]

What do you think your objective function is here?

Hint : ||closest to the origin||

[dim flume]

the plane?

[blissful meadow]

Well what are you trying to minimize?

I give you a point in R^3 and you need to tell me how close it is to the origin

How do you quantify that?

[dim flume]

i dont know tbh theyre just asking for a point closes to the origin idk how that relates to optimization like past problems

[dim flume]

vector?

then measure its length

[blissful meadow]

Okay. Say I give you a point (x,y,z), what's the length of the vector from the origin to (x,y,z)?

[dim flume]

sqrt(x^2+y^2+z^2)

[blissful meadow]

Good. Now you can work with this, but it's much simpler to consider the square of the distance to the point (x,y,z), since if you minimize the square of the distance, you minimize the distance itself.

So you should really be considering x^2 + y^2 + z^2.

Now this is the quantity you want to minimize, so this is your objective function.

The equation of the plane is a constraint.

[dim flume]

wait wdym

[blissful meadow]

We can work with sqrt(x^2 + y^2 + z^2) it doesn't matter.

So let's take the objective function f(x,y,z) = sqrt(x^2 + y^2 + z^2).

What is your constraint function in the form g(x,y,z) = 0?

[dim flume]

why is that our objective function

the objective function is that which is bounded by the constraint function right

[blissful meadow]

The objective function is the quantity you want to find the extremes of

[dim flume]

but we want to find some point on the plane right

[blissful meadow]

In this case "closest to the origin" suggests we want the point with the minimum distance to the origin.

The fact that this point has to be on a given plane is a constraint.

[dim flume]

Ohhh

[severe canyon]

Because you need to minimize it.
The objective function (you probably have seen it also in calc 1) is the function you need to optimise, in other words either minimise or maximise

[dim flume]

So like is our objective function literally the entire space

[blissful meadow]

Yes, you can compute the distance to the origin of any point in R^3.

[dim flume]

Ohh yeah that’s makes sense

[blissful meadow]

We're restricting to points on the plane

[dim flume]

I see now

So we need the gradients right

[blissful meadow]

First answer this

[dim flume]

Wait so shouldn’t our constraint function be = C

Or am I conflicting two things

[blissful meadow]

It's just standard to write it = 0. Notice that in any case we'll be taking its derivatives, so it doesn't matter.

[dim flume]

I thought were taking it’s gradient

[blissful meadow]

For Lagrange multipliers you take the gradient of two functions

An objective function

A constraint function which represents your constraint as a level set.

[severe canyon]

We are indeed
Gradient = vector containing partial derivatives with respect to x and y

[dim flume]

Okay our constraint function is the plane correct

[blissful meadow]

Yes can you write it down? g(x,y,z) = ?

[dim flume]

2x -y +2z = 16 = g(x,y,z)

[blissful meadow]

Okay so let's work with that. g(x,y,z) = 2x - y + 2z.

[dim flume]

Wait why is it in terms of 3 variables

This should be in space right

Why did represent it like that

R4

[blissful meadow]

Your objective function is a function from R^3 to R.
Your constraint will also be a function from R^3 to R, its level sets are surfaces in R^2.

3 variables to optimize => 3 variables in the constraint.

If you had an objective function from R^2 to R, you would haave a constraint which is a function from R^2 to R as well, whose level sets are curves in R^2.

[dim flume]

Oh

[dim flume]

Okay

[blissful meadow]

So now compute $\nabla f (x,y,z)$ and $\nabla g (x,y,z).$

[soft zealot]

Azyrashacorki

[dim flume]

the square root here is kinda ugly 😂 i saw you remove the sqrt earlier why were you able to do that

[blissful meadow]

It's not that ugly to be honest

And it's symmetric you can do it once and then it'll be the same for y and z

[dim flume]

ur right im trippin

[blissful meadow]

Those partials aren't correct

[dim flume]

isnt it just the power rule

[blissful meadow]

Power rule and chain rule

You have $(x^2 + y^2 + z^2)^{\frac{1}{2}}$

[soft zealot]

Azyrashacorki

[dim flume]

yes

isnt that just like (x^2 + a)^1/2 if were doing wrt x

why chain rule

[blissful meadow]

It is just like that.

[blissful meadow]

Because you have a function inside. It's x^2 + a

[dim flume]

Okay so do the same as earlier

[blissful meadow]

Also your partial w.r.t. y for g should be -1

Not 1

[dim flume]

Ooos

You’re right

[blissful meadow]

But yes then it's the same as the other.

You set grad(f) = lambda grad(g) and get three equations.

Along with your constraint equation

And you try to deduce stuff about x,y and z from those.

[blissful meadow]

Okay say equation 1 and 3 what do they tell you?

[dim flume]

Uh

[blissful meadow]

Both of those things should be 2 right?

So they should be equal

[dim flume]

How come

Where’s the 2 coming from

[blissful meadow]

You wrote it in your equations..

2 = this
2 = that

this = that

[dim flume]

Ohhh

I was looking at 1 and 2 💀

Yes they’re equal

But why is that these sort of relationships are really random like it’s not a pattern or anything right

[blissful meadow]

You need to find x,y,z such that the equation
grad(f) = lambda grad(g) holds for some lambda.

Lambda isn't important, the point is that x,y,z should satisfy those equations

[blissful meadow]

Yep.

Now try and see what you can get from equations 1 and 2

[dim flume]

substitution?

idek

[blissful meadow]

Not really no. You have something = 2 and something = -1

What if you multiply the second one by -2?

[cold atlas]

it would be equal to equation 1 and 3

[dim flume]

@blissful meadow my bad i accedentally replied with my other account 🤣

[blissful meadow]

I was like wth

[dim flume]

LMAO

😂

am i cooking

[blissful meadow]

Very much so!

So now you know that z=x=-2y

Use your constraint equation now

[blissful meadow]

Good you're done now. So what's the point on the plane closest to the origin?

[blissful meadow]

-2y = 32/9, so y = -16/9.

[dim flume]

wait how

-2y?

[blissful meadow]

You had -2y = x as one of your equations.

x=32/9

[dim flume]

why can i not just solve for it in the plane equation

[blissful meadow]

You can do, but it's much simpler if you use the fact that -2y = x as you've derived earlier.

In any case you said that $y = -(16 - \frac{128}{9})$ and then wrote $y$ as $16 - \frac{128}{9}$.

[soft zealot]

Azyrashacorki

[dim flume]

oops yeah

blunder

[blissful meadow]

And in general you would want them on the same denominator

[dim flume]

true

[blissful meadow]

So it's just easier using that fact that you know that -2y = x

[dim flume]

is it okay to not understand this intuitively for now 💀

[dim flume]

not this specifically

i mean

the lagrange multiplyers

[blissful meadow]

Yes. It's very ok to accept that it works.

When I first learned about them I don't think I was grasping much of what was going on, but I knew it worked and that it was much easier than substituting on the boundary.

[dim flume]

it really sucks that there are barely any available resources out there that can explain intuition to you

i gave up on reading the book because it doesnt explain intuition almost always 😂

[blissful meadow]

Tbh there are lots of ressources that talk about it, along with loads of MathSE posts discussing it

I wouldn't say they're hard to find

[dim flume]

oh really

for example when it comes to calc 3 is there any good book you would recommend

[blissful meadow]

I think Stewart does Calc 3? It's pretty good for Calc 1 and 2 so I would assume it's also good for Calc 3

But in general textbooks may explain the intuition in a way that it more formal and so may not talk to you very much.

Especially for those types of things videos with nice visuals are golden

[dim flume]

i agree

[blissful meadow]

However, there is a balance in learning about something really deep and actually progressing. It happens to me a lot that I learn something I'm not totally sure I understand but can still use. I feel like using it so it's not such a strange thing anymore is the best way to balance progress and deep understanding.

Often I come back to those things without having done much heavy-lifting in terms of trying to understand what's going on and am surprised they don't seem so strange after all.

[dim flume]

that was worded nicely

i agree with you on that, its just that when you start getting the hang of math you get addicted to wanting to understand everything 😂

but yeah i agree that progress is important so sometimes you just have to accept things

[blissful meadow]

And the thing is as you learn more things and become more knowledgeable and do more exercises working with weird things, you get a new pair of eyes to see what's going on, and often that's just enough to make you understand.

That's not to say that it's not good to seek for deeper intuition but in some cases the best way to understand is to take things as a fact and see where you get.

[dim flume]

i see

i appreciate that

do you think we can go over these? I hope im not taking too much of your time btw

this is a practice test btw not an actual one

[blissful meadow]

No it's fine. I'm in between terms rn so got time

4. first

What do you want to optimize?

[dim flume]

so the surface area is what we wanna optimize

so i assume its equation is the objective function

[blissful meadow]

Yes

What's your constraint?

[cold atlas]

the thing that needs to bound it

[dim flume]

LOL i had the tab open 🤣

[blissful meadow]

Hahah

[dim flume]

its the thing that bounds it

so

i mean i wanna say its the volume equation because thats the only other function but also i dont see how its constraining it

[blissful meadow]

You're given a specific quantity about the cylinder

In the question

[dim flume]

Oh yeah right the volume = 16π

[blissful meadow]

Yep, so now you have a constraint on the volume and an objective function

[blissful meadow]

Looks good!

Now you can take the gradients.

[dim flume]

its the thing where you change their dimensions that get me

like here were just copying down the equations themsleves

but you said earlier that the dimensions must change if im not mistaken

[blissful meadow]

No I said that if you have a 2-variable objective function you'll have a 2-variable constraint.

Which is the case here

[dim flume]

and what about the constraint function

[blissful meadow]

It has 2 variables, h and r.

[dim flume]

is there like a guideline to know if you need to change dimensions or not

kinda like a flowchart

[blissful meadow]

You don't need to change the dimension

Are you talking about the thing I was talking about yesterday about how you start in 3D then get a 2D boundary, then work in 2D and get a 1D boundary?

[dim flume]

this here

like in the plane equation

we made it in R4

even though the plane is in R3

[blissful meadow]

This all means the same as this.

n-variable objective function => n-variable constraint.

[dim flume]

by constraint you mean the constraint function? so basically youre saying the constraint function and the objection function must be the same dimension?

[blissful meadow]

Yes

[dim flume]

why do i always get stuck on this part

[blissful meadow]

Why do we have 42pir

[blissful meadow]

You could always isolate lambda in your second equation and plug that back into your first equation

[dim flume]

well theres one thing tho i always get stuck here like idk what we need to do

i see that were solving for variables but i dont understand under what basis

like what is it that were tryna do

[blissful meadow]

You need to solve the system of equation $$\begin{cases}4\pi r + 2\pi h = \lambda 2\pi h r \ 2\pi r = \lambda \pi r^2 \ \pi r^2 h = 16\pi\end{cases}$$

[soft zealot]

Azyrashacorki

[blissful meadow]

Three equations, 3 unknowns.

[dim flume]

i only have two equations tho

is it always the case that you must solve 3 systems?

[blissful meadow]

You have your two equations from the gradients and one from the constraint

[dim flume]

i mean 3 equations

not systems

[dim flume]

but in the previous problems i had 3 equations already from the gradients

and i dont remember using the constraint

[blissful meadow]

You had 3 equations from the gradients, and one equation from the constraint, solving for 4 variables x,y,z and lambda.

[dim flume]

<@&268886789983436800>

[blissful meadow]

The gradient of an n-variable function will give you n equations, and you'll have one extra equation from your constraint, which will give n+1 equations in n+1 variables x_1, ..., x_n and lambda.

[dim flume]

i dont get it 😭

after you get the system

what do you wanna do

like whats the goal

[blissful meadow]

You have to rearrange stuff in order to solve the system

[blissful meadow]

You can go at it equation by equation. Look at equation 2 here for example.\
This is the same as $2\pi r - \lambda \pi r^2 = 0$, or $\pi r(2 - \lambda r) = 0$.

[soft zealot]

Azyrashacorki

[blissful meadow]

In what cases does this hold?

[dim flume]

wait so i need to solve the system right

[blissful meadow]

Yes. But in particular you can get nice conditions if you consider one equation, see what it takes for it to hold, and working in cases.

So like here, if you have $\pi r(2-\lambda r) = 0$ (this is the second equation rewritten), you'll see that the only way this holds is if $r=0$ or if $\lambda r = 2$, right?

[soft zealot]

Azyrashacorki

[dim flume]

yes i agree

but idk im stuck on the systems part llike right after we solve the system'

[blissful meadow]

Once you get all solutions to the system, you have your extrema.