Courses on Khan Academy are always 100% free. Start practicing-and saving your progress-now: www.khanacademy.org/math/mult... Working out the algebra for the final solution the the example from the previous two videos.
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@collinsmith75727 жыл бұрын
this was the BIGGEST help for understand the intuition behind this principle. thank you so much haha. I have a final for this class tomorrow morning and now I feel prepared for the langrange problem I'm betting is going to be on there. thanks again.
@alejrandom65923 жыл бұрын
3:06 "since lambda equals y" Wait! that fact came from the assumption that x is not equal to zero, which means the first equation doesn't give you any information if x=0 and we have to work with only the last 2 equations. If we plug x=0 into the last one, we get y=+-1, now we plug x=0 and y=+-1 into the second equation and we get that lambda equals 0. (0,1) and (0,-1) are still worth checking. I just saw the graph on the first video again and they happen to be local maxima and minima...
@aylora70332 жыл бұрын
nicely explained bro. when i looked back on the graph it basically outlined the diameter of the circle so i couldnt figure out if its local maxima or minima but i looked at it with regard to part of the function that is above z axis that had the projection of the circle bended upward - in this case local minima and below z axis that had the projection of the circle bended downward - which is local maxima
@user-hb2wq1pl7u7 ай бұрын
i said the same thing i think he made a mistake
@fahrenheit21017 ай бұрын
didn't he say this in the video...
@juv70265 ай бұрын
first of all, he talks about it in the video. Second of all, you're wrong
@mintylemon662 ай бұрын
@@juv7026 would you please tell me where he said it? thank you!
@cameronbaird5658 Жыл бұрын
I have used the Lagrange multiplier so much in the last few months and I finally understand why. Thank you Grant for yet another masterclass
@irtazaisawsome6 жыл бұрын
3Blue1Brown! I thought you sounded familiar. :)
@garlic70995 жыл бұрын
Thanks A lot! One thing worth mentioning is that the 2 values you crossed out at 5:27 MINIMIZE the function on the constraint.
@LeopoldoTejada7 ай бұрын
I would say there is a mistake when checking for the possibility of x=0. It uses the fact that y=lambda, but this identity holds only if x is not zero. Actually the values x=0, y=1 and lambda=0 seem to be a valid solution, unless I made a mistake
@1Happy_Singh6 ай бұрын
Even If We Don't Say That Lambda = y , The Fact That Y Has To Be Zero In The Second Equation Can Be Concluded , Lambda Can't Be Equal To Zero As This Would Mean That There Is No Gradient , Please Let Me Know My Mistake , Or My Lack Of Understanding
@momobeb73614 ай бұрын
You'r right, I will just say that if x=0, y can be equal to 0, but the constraint will not bé satisfied. I have also a question, because it's thé first time i saw a langrangian multipliar. Why have Lambda =0 a problème ? Why not having a gradient is false here ?
@gemacabero64823 жыл бұрын
One question. When you analyze the case x = 0, why do you say that lambda has to equal y ? I didn't understand this part. Lambda could equal zero and then y = 1. So you have x = 0 and y = 1, which do satisfy the third equation. Why would this be wrong? Thank you!
@antonzamay44823 жыл бұрын
You are right, there is the mistake on the video. We get lambda = y only when we can divide on x (x 0).
@atifzaheer26713 жыл бұрын
Yeah, he made a mistake there.
@rajmehta46432 жыл бұрын
Yes, but lambda cannot equal zero, otherwise grad(f) = 0 for all (x,y) implying f is a constant function, which is not true. Hence, if x = 0, y must equal 0 as well.
@vladislav_sidorenko2 жыл бұрын
@@rajmehta4643 The original equation which must be met is not ∇f(x, y)=λ∇g(x, y), but ∇f(x_m, y_m)=λ∇g(x_m, y_m). We evaluate it at the specific point (x_m, y_m), so λ being equal to 0 implies not that the entire ∇f(x, y) is a constant (0, 0), but that ∇f(x_m, y_m) is (0, 0), which is very much possible. In fact, the original equation has the critical points (0, 1) and (0, -1), which are missed by dismissing that λ can be 0. They, of course, don't turn out to be the solution in this case, but they are potential solutions. While the gradient value of (0, 0) may seem that it cannot mean that the contour line is perpendicular to the graph, so what? In general, it may also mean that the contour is also a single critical point, or there's not enough info to calculate the line (As with the contour line of 0 of (x+y)^2). In both cases, the gradient is (0, 0).
@GaborGyebnar7 жыл бұрын
At 3:01 you say lambda equals y, so both of them need to be zero. But that's only true for x != 0, so "y=lambda" doesn't stand here. Actually, y=+-1, x=0, lambda=0 satisfy all three equations.
@aeroscience98347 жыл бұрын
Maxwell Einstein what are you talking about? If x=0 you can't divide the first equation by x in the first place, that's Gabor's point. Going back to the original system of equations, x=0, y=1, lambda=0 satisfied all three original equations.
@iwtwb87 жыл бұрын
Lambda = 0 is a "trivial solution" (although of course that doesn't mean it's not important). Those come up all the time. He should have mentioned it.
@Ropbastos6 жыл бұрын
What you say is true, I think, but as I see it, lambda can't be zero because we know none of the gradients is.
@krishanudasbaksi95306 жыл бұрын
Gabor: You are right.. He made a mistake
@alessioulivi67345 жыл бұрын
@@Ropbastos That's not true, in the case of (0,±1) the gradient of f is exactly (0,0), so it is true that grad(f)=0*grad(g).
@kamitube10594 жыл бұрын
Wish I found this video earlier. You just explained what my teacher took.a semester do so. Thank you
@Xilotl5 жыл бұрын
What a big help!! Now, I have a test to take on this in 1.5 hours....
@sxd62593 жыл бұрын
whats the result
@Xilotl3 жыл бұрын
@@sxd6259 I don't remember that was a year ago
@ranitchatterjee55523 жыл бұрын
Khan Academy never disappoints ❤
@bennicholl76435 жыл бұрын
What if their are numerous places that the two functions are perfectly parallel. Would you just iterate through each solution where the partial derivatives of f(x) = partial derivative's of lambda * g(x), than the solution with the max output?
@jasdeepsinghgrover24706 жыл бұрын
Thanks a lot for making great videos. Does this method always work. Like maximize z=-(x-1)^2 given (x-1)^2+y^2=1. The contours seem to be perpendicular here
@ethanlawrence2825 Жыл бұрын
Am I correct that following the Legrange Multiplier method only gives you values of x and y that make the function STATIONARY (so could be minimum or maximum).
@mikeywatts36611 ай бұрын
It might be interesting to compare this solution with another method: equating dy/dx of the two given equations. (For f(x, y), we get d(f(x,y)) = 2xy*dx + x^2*dy. But since we're looking at a contour, d(f(x,y)) = 0; similarly, x^2+y^2 = 1 can be differentiated implicitly to get dy/dx = -x/y).This gives -x/y = -2xy/x^2. Solving, we get the same solution as in the video.
@scariuslvl99873 жыл бұрын
hi, I have a question: at 2 min. 30 you say lambda=0 if x=0, but that's assuming you can cancel out x in the first equation if x=0, but wouldn't that mean that you assumed 0/0=1? Or am I looking at it the wrong way? edit: If x=0, then 0*2y = 0*2*lambda; 0^2 = y*2*lambda; 0^2 + y^2 = 1 => y=1; lambda=0 This is my reasoning, it doesn't change much because it isn't a relevant solution but I want to know if my way of thinking is wrong, please help!
@smahire2994 жыл бұрын
let's say within the area of the circle I want to find the maximum point, then do we have to evaluate each and every contour ranging from 0 to 1 for determining max point?
@mahakjauhri67594 жыл бұрын
Clear enough! But your voice 😍
@vvmcmurdo7 жыл бұрын
@Gábor Gyebnár, True, but lambda can't be zero. That's why that possibility is not useful.
@martingutlbauer90717 жыл бұрын
@chittaranjan: why not? if lambda equals zero means that the "red" side is zero, but there are solutions to the gradient of f where it equals zero, so that the equations holds. I don't see why we should exclude lambda=0? Can you explain?
@adityaprasad4654 жыл бұрын
@@martingutlbauer9071 Remember what we're trying to find: places where the gradient vectors of the two functions are a constant multiple of each other, and hence point in the same direction. If one of those vectors is zero, then it's not meaningful to say that the other is a multiple of it.
@hakeemnaa2 жыл бұрын
the direction is the slop, ratio of change y/ change of x we equal the direction of both functions Y1/X1=Y2/X2 of course Y1≠ Y2 but we can say Y1= λY2, X1=λX2 ( change of x and y ) so
@saurabhsingh-ow7ue4 жыл бұрын
thank you sir.........
@vladimirkolovrat28466 жыл бұрын
Very clear! Thank you.
@piotrzalewski9381 Жыл бұрын
There is solution for x=0 (it is not true that y=lambda if x=0). The (two) solution(s) are [x,y,lambda] = [0,1,0] or [0,-1,0] and both were easy to spot on the graph in the previous part. These are local (constraint) max. and min respectively. The reason (in the context) is that if grad(f)=0 (as it is for x=0) than grad(f)=0*grad(g) for every grad(g). The series is excellent anyway (as usual).
@ankitapaul64263 жыл бұрын
Thank you sir
@osamasahib77414 жыл бұрын
Thanks a lot
@kavinduadhikari5881 Жыл бұрын
This is the greatest
@k.markendahl40635 жыл бұрын
Is the maximum lokal or global?
@user-hb2wq1pl7u7 ай бұрын
3:32 IF X = 0 we said that it should satisfy the constraint why can't y be plus or minus 1
@izabilska72784 жыл бұрын
thank you for being good at math unlike my professor :)
@riddhimasapra53582 ай бұрын
Thank youuu ❤️
@hakeemnaa2 жыл бұрын
this is high school math, it is just a different notation if you take the derivative of a function and equal it to zero, you will have no change nearby, so you are at max or min zone here you don't equal it to zero, you equal it to another derivative, you put limda because the derivatives are not equal in magnitude the problem will be that you will have many same directions of the derivative ( same slop or same ratio if change y over the change of x) between two functions even if x and y of one function are far away from the other function so you have to but the constraint
@CalleTful2 жыл бұрын
Which playlist is this in?
@yveltal70125 ай бұрын
thankyousomuch
@DianaMorales-sk7uf5 жыл бұрын
helped with hw
@amol51463 жыл бұрын
Can't we just substitute x^2 = 1 - y^2 into the function and set the derivative = 0?
@amaarquadri3 жыл бұрын
At 3:00, when you are analyzing the case where x = 0 you state that lambda = y. But since x = 0, lambda = y no longer applies (that would be like saying 0 * lambda = 0 * y => lambda = y, which is false). In fact, the x = 0, y = 1, lambda = 0 seems to be a solution to the system of equations. It wouldn't be the maximum though, since the objective would evaluate to 0.
@gemacabero64823 жыл бұрын
Hi! I had the same question, but why wouldn't y = 1 and x= 0 be a possible solution? It does satisfy the third equation right? Thanks!
@CSBAMBATIPUDIRISHIVINAYAKA2 жыл бұрын
Is it a coincidence that we got minimum value of function for lambda < 0 case, or can we have maximum value even when lambda < 0?
@praneelmadhuvanesh37708 ай бұрын
huh? When checking the x=0 possibility, lambda would no longer equal y. So lambda could equal 0 and y could equal 1. Right?
@Bunk_Moreland7 жыл бұрын
so, where is the use of lagrange multipliers (lambda) in this method?
@thisiskartik2 жыл бұрын
Is it grant Sanderson in this video?
@FineFlu Жыл бұрын
Can someone explain why lambda is seemingly irrelevant in the end?
@5612dag Жыл бұрын
3:06 well, if x=0, we can no longer be sure that lambda equals y! If x is zero, you can't really say anything about the relationship between y and lambda in that equation. (take for example 0*3 = 0 = 0*5, then it is obvious that 3 =/= 5, even though 0*3 equals 0*5. )
@MoguMogu8184 ай бұрын
thank you 3b1b.
@aidaasekenye35405 ай бұрын
Utitarianism w1(x1)=square root of x1
@-mwolf2 жыл бұрын
I love you bro
@funnzone7945 Жыл бұрын
you great
@nguyennguyenkhoi99882 жыл бұрын
3:28 if x is zero y not need to equal lamda so your conclusion if x = 0 there is no y is totally wrong
@amaljeevk395011 ай бұрын
❤
@mrwess19275 жыл бұрын
What do you mean by maximizing the function f(x,y)?
@ConceptualCalculus3 жыл бұрын
Find the pair (x, y) that gets the largest possible value out of the function.
@SakuraxStars4 жыл бұрын
This voice!!
@kumarparshotam6624 жыл бұрын
3 blue 1brown
@sxd62593 жыл бұрын
brandt sanderson >>>>>>>>>>>>>>>> any maths teacher