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Lagrange Multiplier, max of sinx*siny*sinz
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- Опубликовано: 7 авг 2024
- Lagrange Multiplier, max of sinx*siny*sinz
my previous vid, area of an inscribed triangle, • Area of an inscribed t...
With this vid, we can find the max. area of an inscribed triangle,
blackpenredpen,
math for fun,
blackpenredpen.com/bprplive, / blackpenredpen ,
blackpenredpen@gmail.com
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!
Is this calc 3? In my course it was calc 2. 3 was psychedelic stuff 😂😂
I would love to see you justify/explain/derive Lagrange Multipliers.
go to khan academy if you want to learn
if you want to see it from bprp then idk just wait
It is not so difficult to demonstrate
"Yes! This is Calculus III as there are three variables."
-Blackpenredpen, 2018
Let me guess... x=y=z=pi/3
But... WHY!
XD good one 👍
take the mean ( measure of central tendency )
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.
Samuel Degueldre i wanted to write the same, maximizing the height and the base
More calc 3 questions!!!!!!!!!!!
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.
Yay!! Thank you!
(stares at the screen in confusion)
Oon Han Have you taken Calc III?
no
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
@@jameroth7661 sure, but what fun is that?
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.
I am glad! Wishing calc 3 the best for you!
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)
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!
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).
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:
Wow!! That's amazing!!!
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.
Thank you so much sir for the complete understanding. It really helpful. It was preety much eays too 🙏🙏
This is brilliant!
You are the best!!!
Neat. I'd been hoping for some calc 3+ stuff :)
Thank you sir, you are excellent.
This video was great!
J.J. Shank thanks!
Very nice video and helpful 👍
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)
Can we have more Lagrange and Laplace just like your fun integration drills? 👏👏👏
Oh man, I would love to be in your math classes 🙌💪👏👏👏👏👏👏👏
So Good!
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
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
oh :P
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
+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.
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.
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.
3√3=√3³
can you do a video on stokes theorem?
love your content!
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
U'r awesome !!!!
Appreciated you from INDIA 🇮🇳🇮🇳♥️
i like how your explanation of calc 3 is that it has 3 variables (:>>
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.
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.
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.
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
Beautiful
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.
thanks!!!!
Thank you sir
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!
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?
Is there a video explaining why/how this method works?
first you can do mg
And he's back!
The old Bprp
Márcio Amaral ?
I never left
+blackpenredpen yeah IDK what he meant. Maybe a bit about the chalkboard, which was just a few videos anyway...
Pls can u get a video to make understand partial derivatives
I will be grateful
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
I have question : if the area and perimeter of a triangle are known is the triangle is known?
that end though
also I guessed the answer correctly
if our teacher would explain half the stuff that you do, everybody in my class could be an ace in maths
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?
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.
good
Plz do why the Lagrange formula is true. Love ur videos
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
This reasoning does not exclude the possibility of 3 equal maximums arranged symmetrically.
Love calc 3 stuff
Wow nice question
This qst I was reading in engineering class sir
do some Stokes theorem, Divergence Theorem, Greens Theorem and Line Integrals!!!!
Cool
I knew it had to be an equilateral triangle, it's always an equilateral triangle
If it is calculus 3 question, why I had it on my calc2 quiz?
Would you please prove the Riemann mappings theorem?
I solve it by using the definition of df, subtitute z=π-(x+y), and set it equal to zero
got same answere too
Great
Gotenks good
It is cool.
Please make more calculus 3 videos
Alternatively one can take log of the expression and use Jensen's inequality. This works because log(sin(x) is concave.
Yay!!
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 .
Congratulations Sir, Because Now,
Your subscriber is 97000.😂😂👆👌👌👍👍
What about r? Is this you’re very first Calc III video?
I just recently thought: "Well, I guess I should learn something about multivariable calculus... Hopefully there are RUclips-videos for that..." Now I don't have to search anymore.
Love from India ❤️❤️
Can we find the minimum?
Please try to solve integral from 0 to 1 of (x^2-1)/(ln(x)) and post a video.
#mathchallenge
Super table tennis Can’t be done without numerical integration.
Can anyone tell me why Sin*π/3 equals to sqrt3/2 ?
Thanks 🇮🇳🇮🇳🇮🇳
is it possible to find the largest triangle in a circle using only geometry?
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
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.
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.
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.
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.
I am a middle schooler... i wonder, do you learn calculus 3 first or linear algebra in college?
San Samman
U r only in middle school??!!!!!
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.
What about the minimum value
min of the function = ???? Zeroo???
can it be done without Lagrange Multiplier our teacher gave us without it.
please can you do that?
Often you see symmetry pop up in minimalizing and maximalizing problems
You actually don’t need the x, y, z > 0 constraint(s). x, y, z ≠ 0 is sufficient.
Negative angles exist.
True. But this is meant to be a continuation from my triangle video. So, I put that down., : )
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. 🤔
The easiest way is to notice that tan(x)=tan(y)=tan(z), hence x=y=z
More geometry
The cube has the largest Volume .
You set the derivatives equal to 0, but you theoretically don't know if this is maximum (or maybe minimum) value.
And all of this is true because... Lagrange.
The sam qust in my exeam gahhahahahahha
Yeah I answer it
noyce
*in calc ab* *sees lambda* Oh helllllll nawwwww
Trivial due to symmetry lol
Can’t you use a symmetry argument?
tsujimasen yes. I just wanted to do some calc 3
Second😂😂😂
Or 60° hahahahaha