Lagrange Multiplier, max of sinx*siny*sinz

[@TheMauror22]

Please do more calc 3 videos! I think that multivariable calculus is very interesting but almost no one does fun videos with it! You should exploit it!

[@_DD_15]

Is this calc 3? In my course it was calc 2. 3 was psychedelic stuff 😂😂

[@jemcel0397]

"Yes! This is Calculus III as there are three variables." -Blackpenredpen, 2018

[@bilz0r]

I would love to see you justify/explain/derive Lagrange Multipliers.

[@ゾカリクゾ]

go to khan academy if you want to learn if you want to see it from bprp then idk just wait

[@yeahyeah54]

It is not so difficult to demonstrate

[@xamzx9281]

Let me guess... x=y=z=pi/3

[@tangentofaj]

But... WHY!

[@jesust162]

XD good one 👍

[@sansamman4619]

take the mean ( measure of central tendency )

[@sdegueldre]

You can easily demonstrate it without calc 3, pick any 2 points on a circle, the triangle with the largest area that can be formed using these 2 points is the one with its third point opposite the first 2 and in the middle of them, because A = bh/2, having it anywhere else would yield a lower height with the same basis, so we know it has to be isoceles, now you do the opposite, fix one point and then pick two other points on the circle, symmetrically from the first since we know the triangle has to be isoceles, put the fixed point at (1,0) in (r,theta) notation, the other points will be at (1,alpha) and (1,-alpha), the height will be 1-cos(alpha) and the basis 2sin(alpha), the area is then sin(alpha)-sin(alpha)cos(alpha) which we try to maximize, the derivative is cos(alpha)+cos(alpha)sin(alpha), ie cos(alpha)(1+sin(alpha)), with only one root in [0,pi] which is 2pi/3, meaning the points are in (0,0),(0,120°) and (0,-120°), ie the triangle is equilateral, QED.

[@xamzx9281]

Samuel Degueldre i wanted to write the same, maximizing the height and the base

[@KamauMayhem]

These questions make me smile, you could also make a symmetry argument about x,y,z at the beginning stating that they are indistinguishable in the equation. Given the constraint, quickly conclude that x=y=z and 3x=pi. It might be the physicist in me though. I do however enjoy formality, good job and keep up this great content.

[@blackpenredpen]

Yay!! Thank you!

[@christianpol260]

I thought I would never see you again when I got through Calculus 2. Currently taking Calculus 3, so I'm happy to see you back at it again.

[@blackpenredpen]

I am glad! Wishing calc 3 the best for you!

[@OonHan]

(stares at the screen in confusion)

[@NotYourAverageNothing]

Oon Han Have you taken Calc III?

[@OonHan]

no

[@jameroth7661]

Much simpler proof: the biggest some of the numbers x*y*z is when x=y=z this can be proven simply by a^2-b^2. you have the numbers a-n; a; a+n mutilping them you get a(a^2-n^2) but this is strictly smaller then a^3 thus the biggest value you can get is when all three are the same. We conclude all have to be pi/3

[@simenjorissen5357]

@@jameroth7661 sure, but what fun is that?

[@maxc6816]

More calc 3 questions!!!!!!!!!!!

[@gordonfreemanthesemendemon1805]

This is random, but i find it kinda nice how sqrt(3) * sqrt(3) * sqrt(3) is the same as sqrt(3) + sqrt(3) + sqrt(3)

[@VidalDuval]

x*sqrt(x) = sqrt(x)^x x*x^½ = (x^½)^x x^(3/2) = x^(x/2) ln(x^(3/2)) = ln(x^(x/2)) 3/2*ln(x)=x/2*ln(x) Now we see the only solution to that is either x=1 or x=3. Neat!

[@animalfarm7467]

blackpenredpen: Another excellent video showing your superior teaching skills. I was hoping you could go through a few more optimization examples of the Lagrange Multiplier? Just as other branches of Calculus III like Euler-Lagrange (e.g. Brachistochrone), the more examples you have to worth with and understand, the more likely you will be able to apply the theory to more abstract problems.

[@alexyoize]

I show that exercise in my vectorial calculus class and i was the hero in my class :3 thank you so much c: Greetings from colombia c:

[@blackpenredpen]

Wow!! That's amazing!!!

[@SteamPunkLV]

hi bprp, I have an interesting Olympiad question for you I call it the "Fibonacci Sequence on Crack" The first term is 1, and the second one is 2 (just like in the Fibonacci Sequence) The third term is the product of the first two (1*2=2) The fourth term is the product of the last two (2*2=4) The fifth term is the product of the last two (4*2=8) The sixth term is the product of the last two (8*4=32) This is where it gets interesting The seventh term is the product of all digits of last two numbers (3*2*8=48) The eighth term is the product of all digits of last two numbers (4*8*3*2=192) The ninth term is the product of all digits of last two numbers (1*9*2*4*8=...) NOTE: Each time that you would multiply by 0 in this sequence - you do not. I forgot to add this little condition which makes the problem a lot harder Find the 2016th term of this sequence (this question is from 2016 :p) I was able to find the answer in ~40 minutes

[@SteamPunkLV]

when olympiads give you numbers like 2018, they are meant to confuse you and make the problem look astronomical when really it's just a simple concept :p

[@SteamPunkLV]

oh :P

[@EnteiFire4]

If you keep going: f(9) = 1*9*2*4*8 = 576 I'll stop there. That means that f(10) will be 5*7*6*n = 10*21n. The last number is a 0, so f(11) will be d1*d2*d3...*0 = 0. So yeah, f(2016) = 0

[@JivanPal]

+Swagger Lemon This ensures that the question is not 100% identical to a question asked in a previous paper, as well as a sort of double-check that you have that year's paper in front of you, and not a previous year's.

[@hendrikvija310]

POSSIBLE SPOILERS? Isn't the answer 0? Because the ninth term is 1*9*2*4*8=576, thus the tenth term contains both 5 and 2 as factors and must be divisible by 10 --> it must end with a 0, from there on it's rather obvious that every term is 0.

[@rishabhdhiman9422]

I do think that Lagrange Multipliers is an overkill for this question, the way more tame Jensen's inequality does the trick. Define: g(x) = ln(sin(x)) g"(x) = - csc^2(x) ≤ 0 for all x Thus, the function is concave. By Jensen's Inequality: f(x) + f(y) + f(z) ≤ 3f((x+y+z)/3) this is equivalent to ln(sin(x) sin(y) sin(z)) ≤ 3 ln(sin((x+y+z)/3)) = 3 ln(sin(pi/3)) since e > 1, we have sin(x) sin(y) sin(z) ≤ sin^3(pi/3) = 3*sqrt(3)/8. By Jensen's we have a nice and easy generalization: sin(x_1)*sin(x_2)*...*sin(x_n) ≤ (sin(pi/n))^n where x_1 + x_2 +.... + x_n = pi. The Lagrange multipliers proof won't be this short for the generalization (there is an obvious pattern in the lagrange proof but who cares when we such a short and simple proof).

[@julian803]

I just noticed something about the final answer to this question. Shouldn't it be (sqrt(3))^3 divided by eight rather than 3(sqrt(3)) divided by eight? Could you also post some more videos of Number Theory proofs? Thanks.

[@elilogan8630]

3√3=√3³

[@callumvlex7059]

When you have the three equations, you could also divide all three by sin (x)sin (y)sin (z) and get that Cot(x)=Cot (y)=Cot (z) so they must be the same sngle (as they are in te range (0,pi)

[@devanshagnihotri3500]

Appreciated you from INDIA 🇮🇳🇮🇳♥️

[@lemonlimeGOD]

Neat. I'd been hoping for some calc 3+ stuff :)

[@mediter123]

Something that was missing was that if you wished to find the largest area of that triangle, you would need to multiply that constant, (3*sqrt (3)/8), By 2r^2. This came from the previous video, this video was just finding the constant and it took me a sec to realize that!

[@FuhrerShattercore]

Can we have more Lagrange and Laplace just like your fun integration drills? 👏👏👏

[@Tumbolisu]

My thought process: (observation:) the function sin(x) is curving downwards between x=0 and x=pi lets start imagining that x, y and z are equal (pi/3). if we increase one, another has to decrease. lets imagine we decrease x by a small amount called h, and then increase z by that amount. because the sine function is curving downwards, sin(x) will shrink faster than sin(z) is growing, therefore our final result is smaller. even if we try to move y and z together, sin(z) shrinks faster than sin(y) grows. Because of all this, x, y and z indeed have to be equal (to pi/3). after you mentioned the largest area of a triangle connection, my thought process became: lets just imagine any triangle inscribed within a circle. lets call the vertices A, B and C. now rotate the image such that AB is at the bottom. when we move C along the circle, the total area shrinks and grows. if we were to move C parall to AB, the area of the triangle wouldn't change. because of this, only the shortest distance between C and AB actually matters. The larger that distance, the larger the area. obviously its the largest when the triangle becomes an isosceles triangle, with C being on the far end of the circle. Now we cna repeat this game with AC being the base and B being free to move. Every time we do this, we have to adjust the points less and less. And if we were to look at the case of an equilateral triangle, we would notice that we don't have to adjust anything. Therefore, all 3 angles are the same.

[@narendra6477]

Very nice video and helpful 👍

[@NotYourAverageNothing]

What about r? Is this you’re very first Calc III video?

[@christopherwilson9358]

I knew it had to be an equilateral triangle, it's always an equilateral triangle

[@retired5548]

if our teacher would explain half the stuff that you do, everybody in my class could be an ace in maths

[@srpenguinbr]

Hey, I came up with a different method. first, make the substitution that z=pi-x-y, from the condition g When doing single variable calculus the function has the minimum/maximum when the derivative is equal to zero. You can use the same concept: find the partial derivative with respect to x and set that equal to 0. Do the same with y and solve for x and y. It works, however you need numerical approximation sometimes

[@danielstone5810]

Oh man, I would love to be in your math classes 🙌💪👏👏👏👏👏👏👏

[@General12th]

This video was great!

[@blackpenredpen]

J.J. Shank thanks!

[@FascinatingBlogs]

Thank you sir, you are excellent.

[@yourtutor_info]

Am I right that using the Lagrange Multiplier method you can only indicate a suspicious points for local extrema? And to be sure that those points are the points of local maximum you have to make an aditional analysis.

[@yusufcakar2312]

Would you please prove the Riemann mappings theorem?

[@marcioamaral7511]

And he's back! The old Bprp

[@blackpenredpen]

Márcio Amaral ? I never left

[@ゾカリクゾ]

+blackpenredpen yeah IDK what he meant. Maybe a bit about the chalkboard, which was just a few videos anyway...

[@sattarbazouneh8867]

I have question : if the area and perimeter of a triangle are known is the triangle is known?

[@tarat.techhh]

all right it means that if we are asked the max value of some symmetric trignometric ratios it comes out to be at x=pie/3 if angles are adding up to pie.. isn't

[@phscience797]

I just recently thought: "Well, I guess I should learn something about multivariable calculus... Hopefully there are KZbin-videos for that..." Now I don't have to search anymore.

[@mathadventuress]

i like how your explanation of calc 3 is that it has 3 variables (:>>

[@borg972]

Is there a video explaining why/how this method works?

[@marcosjuniorquijadagonzale9778]

You are the best!!!

[@ssdd9911]

that end though also I guessed the answer correctly

[@vanshparihar9380]

Thank you so much sir for the complete understanding. It really helpful. It was preety much eays too 🙏🙏

[@bhaskardas8145]

Can we find the minimum?

[@danielescotece7144]

can you do a video on stokes theorem? love your content!

[@kokainum]

There is one problem with this argument. Langrange multipliers find only candidates for local extremes besides boundaries of manifolds. So you still need to check how function behaves near the boundary. Here we have simple situation where function is 0 at boundary and is positive inside our domain so Lagrange multipliers method is gonna find us. But if we had a function that is minus function from this example then there is no maximum, supremum would be 0 and it's value from boundary. Lagrange multiplier method would find us only minimum then. So we can't just throw out z=0 here because it's from outside of our domain. We can throw it because values function attains at boundary are small and therefore maximum can't be there. Also we know supremum can be attained in compact sets so if we add boundary we can find a point where maximum is attained. So standard procedure is to say that maximum is attained in closure of our set and then we say it's not attained at boundary because there is 0 value and we have greater values inside our origin domain and then we know it can be found using Lagrange multiplier. Formally everytime such procedure should be done to have complete proofs although it's ommited oftenly because authors consider it trivial. It's good to remember about it though and know what Lagrange multiplier method actually finds.

[@707107]

This is the very same problem that I am going through as a teacher of Lagrange multipliers in class. But I think the reason that being omitted is not because authors consider it trivial. I think it is because the complete formality is hard to be explained in detail at a calculus level.

[@Whateverbro24]

Please make more calculus 3 videos

[@VonUndZuCaesar]

You can just argue geometrically that you get the largest area with a fixed sum of 2 factors if both are the same. So x*y is max if x=y. So in this case all sin has to be the same. So x, y, z have to be 60 degree

[@Magic73805]

Congratulations Sir, Because Now, Your subscriber is 97000.😂😂👆👌👌👍👍

[@8mice979]

first you can do mg

[@fgzgeimv8u]

is it possible to find the largest triangle in a circle using only geometry?

[@kevincaotong]

Yes, of course there is! Here are 2 proofs, one with algebra, and the other with geometry. We use the triangle area formula A=abc/4R=2R^2 sin A sin B sin C, thus it suffices to maximize sin A * sin B * sin C since R is set as the radius of the circle. By the AM-GM Inequality, we know that (sin A +sin B + sin C)/3>=(sin A sin B sin C)^1/3 since sin A is positive for all 0

[@sergiokorochinsky49]

proof by symmetry: choose 2 points on the circle. the 3rd point travels from the 1st point to the 2nd. at the beginning and at the end of the trip the area of the triangle is 0. in between it grows to a maximum. by symmetry, the maximum is at the middle of the trip, giving you an isosceles triangle with maximum area. again by symmetry, this is also true for the other 2 points. therefore the maximum maximorum is the equilateral triangle.

[@sergiokorochinsky49]

proof by iterations: start like the proof by symmetry. find the isosceles triangle with maximum area. now choose another point as "traveling point". find the new isosceles triangle. keep iterating. it will converge to the equilateral triangle.

[@sergiokorochinsky49]

proof by physics: instead of 3 Points, think of 3 electrons (or 3 particles with the same charge). minimise the electric potential (equivalent to maximise the average distance between all of them). the minimum of electric potential implies a maximum in area.

[@sergiokorochinsky49]

proof by logic: if a non-isosceles triangle has maximum area, then it's mirror image is also maximum. but the curvature of the circle is constant, therefore there can only be one maximum: isosceles. by rotation, it must be equilateral.

[@anwarhassan7454]

can it be done without Lagrange Multiplier our teacher gave us without it. please can you do that?

[@AkshayKumar-et6ij]

What about the minimum value

[@arctan-k]

If it is calculus 3 question, why I had it on my calc2 quiz?

[@erikmingjunma9403]

Alternatively one can take log of the expression and use Jensen's inequality. This works because log(sin(x) is concave.

[@aritradatta6563]

Pls can u get a video to make understand partial derivatives I will be grateful

[@jeffk8019]

So Good!

[@luckytiwari2699]

How can you be so sure that at x=y=z=π/3 , the function (=sinx.siny.sinz) is going to have it's maxima .

[@hamsterdam1942]

Is it true that any function f (x, y, z) (where x, y, z can be replaced in any order) when g(x, y, z)=C (where also x, y , z can be replaced in any order) have a maximum when x=y=z?

[@wduandy]

Beautiful

[@helloitsme7553]

Often you see symmetry pop up in minimalizing and maximalizing problems

[@duncanw9901]

Love calc 3 stuff

[@1willFALL]

do some Stokes theorem, Divergence Theorem, Greens Theorem and Line Integrals!!!!

[@sansamman4619]

I am a middle schooler... i wonder, do you learn calculus 3 first or linear algebra in college?

[@blackpenredpen]

San Samman U r only in middle school??!!!!!

[@blackpenredpen]

San Samman Oh btw, it doesn't really matter. Some ppl do it at the same time. But I would say usually calc 3 first since most ppl finish calc 2 and don't want to forget about the calc material.

[@pablobribiesca2596]

Since x , y and z all behave the same, by symmetry, one can conclude at max (x=pi/3, y=pi/3, z=pi/3) because x+y+z = pi so pi= 3x =3y =3z

[@paulelliott9487]

This reasoning does not exclude the possibility of 3 equal maximums arranged symmetrically.

[@deepanchakraborty7427]

U'r awesome !!!!

[@khushijaiswal24]

Love from India ❤️❤️

[@NotYourAverageNothing]

You actually don’t need the x, y, z > 0 constraint(s). x, y, z ≠ 0 is sufficient. Negative angles exist.

[@blackpenredpen]

True. But this is meant to be a continuation from my triangle video. So, I put that down., : )

[@dimosthenisvallis3555]

Plz do why the Lagrange formula is true. Love ur videos

[@debrajbanerjee9276]

What is the integral of sqroot(sinx) from 0 to pi? I found it 2√(2/π)(gamma(3/4))^2 in wolfram alpha which seems very interesting.can you provide me the magical steps?

[@divyanshirajpurohit6287]

Thank you sir

[@snejpu2508]

You set the derivatives equal to 0, but you theoretically don't know if this is maximum (or maybe minimum) value.

[@himanshumallick2269]

This was simple, because of the symmetry. Without Lagrange multipliers, you may take any two (say y and z) constant, and then .... (I am not writing the whole solution because you must try what should be done next, on your own 😀) show some kind of invariance. And due to symmetry, it has its Maxima at x=y=z= π/3.

[@blackpenredpen]

good

[@doctorb9264]

The cube has the largest Volume .

[@Zonnymaka]

The easiest way is to notice that tan(x)=tan(y)=tan(z), hence x=y=z

[@tsujimasen]

Can’t you use a symmetry argument?

[@blackpenredpen]

tsujimasen yes. I just wanted to do some calc 3

[@yashraj4272]

min of the function = ???? Zeroo???

[@lalitverma5818]

Wow nice question

[@lalitverma5818]

This qst I was reading in engineering class sir

[@unrulyObnoxious]

Only question I have is: how do we know this is the maximum value and not the minimum? Nothing in the solution seems to indicate that. 🤔

[@Galileo2pi]

Cool

[@passionateaboutmath1690]

Please try to solve integral from 0 to 1 of (x^2-1)/(ln(x)) and post a video. #mathchallenge

[@NotYourAverageNothing]

Super table tennis Can’t be done without numerical integration.

[@ゴテンクス-q8q]

Great

[@blackpenredpen]

Gotenks good

[@sergiokorochinsky49]

And all of this is true because... Lagrange.

[@aditya9353]

Thanks 🇮🇳🇮🇳🇮🇳

[@Sangoreborn]

Can anyone tell me why Sin*π/3 equals to sqrt3/2 ?

[@lordphy1513]

I solve it by using the definition of df, subtitute z=π-(x+y), and set it equal to zero got same answere too

[@user-vm6qx2tu3j]

Yay!!

[@vkilgore11]

It is cool.

[@marcoantoniobarreraolea5573]

More geometry

[@yaboylemon9578]

*in calc ab* *sees lambda* Oh helllllll nawwwww

[@blackloop1861]

The sam qust in my exeam gahhahahahahha Yeah I answer it

[@AndDiracisHisProphet]

noyce

[@leonardromano1491]

Trivial due to symmetry lol

[@Magic73805]

Second😂😂😂

[@john-athancrow4169]

Or 60° hahahahaha