Proof for the meaning of Lagrange multipliers | Multivariable Calculus | Khan Academy

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Here, you can see a proof of the fact shown in the last video, that the Lagrange multiplier gives information about how altering a constraint can alter the solution to a constrained maximization problem. Note, this is somewhat technical
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Multivariable Calculus on Khan Academy: Think calculus. Then think algebra II and working with two variables in a single equation. Now generalize and combine these two mathematical concepts, and you begin to see some of what Multivariable calculus entails, only now include multi dimensional thinking. Typical concepts or operations may include: limits and continuity, partial differentiation, multiple integration, scalar functions, and fundamental theorem of calculus in multiple dimensions.
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[@yogeshbhatt7154]

journey with sanderson ends here

[@metuphys5611]

sad moment

[@tentrot4420]

Sad moment

[@scoutgaming737]

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

[@chrislam1341]

can you also cover the inequality constraint optimization, please?

[@poiuwnwang7109]

He explained it in clip 85 briefly. Essential = can be used for

[@cyancoyote7366]

You're back! Can we except more multivariable calculus vids? Love your animations!

[@Mau365PP]

Is this the last video for the Lagrange multipliers??? I wanted to learn the Karush-Kuhn-Tucker conditions for inequality constraints... :( Where can I find a course that is easy to follow?? please HELP!

[@ninatsvetkova]

I was wondering about that, too. It's the last video indeed.

[@richardshandross3453]

Great video and series. One important thing that wasn't covered is whether L* is a differentiable function with respect to h*, s*, and lambda*. I'm not convinced that it is (though I have no reason to believe that it isn't, either). If it's not differentiable, then it doesn't seem valid to consider the derivative of L*.

[@johnnolen8338]

L is differentiable with respect to each of the independent variables in its domain; it's constructed that way from the beginning. Remember, L is a synthetic function put together for the express purpose of solving a constrained optimization problem. There's no requirement that it actually model reality. Therefore as the analyst you are free to choose functions that are smooth and continuous at all points in the domain of consideration. That said, I was not altogether impressed by Grant's proof. It seemed circular and rather hand - wavy to me. Of course the partial derivatives of L are going to vanish at the critical point ( h*, s*, λ*). That is after all what defines a critical point. It is easy enough to show that λ = ∂L/∂b, by appealing to the Implicit Function Theorem; it is not necessary to define another function which explicitly includes b in the list of independent variables upon which L depends in order to do that either. If it bothers you that the dependence on b is not explicit, you can always define a Legendre Transform that is dependent upon {h, s, and b} rather than the Lagrangian itself being dependent upon {h, s, and λ} at which point it becomes clear that λ = ∂(LT)/∂b.

[@davidsweeney111]

Lagrangians are fairly advanced maths, degree level stuff, seem to remember doing this on my physics degree - was really boring, nice to see other applications, this is great!

[@jonhouck4981]

I have a physics degree also and I can't remember using Lagrange multipliers anywhere, except in the calculus where it was taught. However, it might have shown up in the boundary value problems and partial differential equations as well.

[@arunrajbhandari1656]

@@jonhouck4981 It is a very useful tool in econoimics. Especially if you are working in the industry you can use to to figure out a feasible project given the amount of capital the company has.

[@medvedd_]

@@jonhouck4981 CS major here. Lagrangians important for Maximum entropy proof in information Theory to find the probability distribution that maximizes entropy subject to certain constraints. :)

[@xiaoweilin8184]

@15:20, why do we not need to change all the L on the right-hand side to L* here if we change what we are differentiating?

[@DrKvo]

It would be really helpful if you covered the difference between maximizing vs minimizing using Lagrange multipliers / Lagrangians.

[@MightySapphire]

When you maximize F(x,y) you minimize it's negative f(x,y)=-F(x,y). The process is the same.

[@abdullaalmosalami]

Oh I absolutely love this proof! I mean my heart's totally pumping!

[@naru2906]

great video, grant!!

[@guilhermegondin151]

Couldn't we consider the Lagrangian as a 3 variable function, with each being a function of b and use implicit differentiation of the lagrangian on b?

[@zack_120]

Isn't it what is done here?

[@duncanw9901]

WOAH is this 3B1B in the voiceover???

[@cyancoyote7366]

Yep!

[@petripaavonpoika3767]

Suppose I have three interdependent functions: A, B and C. Then, I will write a Lagrangian L1 = A-l1((B-b)-l2(C-c) Extremizing the Lagrangian will give me the multipliers l1 and l2. But then, could I write two more Lagrangians L2=C-l3(A-a)-l4(B-b) L3=B-l5(C-c)-l6(A-a) Extremizing the latter two Lagrangians should give me the multipliers l3, l4, l5 and l6, right?

[@uimasterskill]

At 8:30-8:50, when Grant considers the Lagrangian as a function of functions of b and b itself, why isn't it the case (as it was earlier, when b was fixed) that B(h*(b), s*(b))=b? I mean, aren't h*(b) and s*(b) defined to be precisely such that they maximize R for every b while also making the budget B(h*(b), s*(b))=b? Hence, why doesn't that term cancel?

[@plekton1]

I have the same Question. Although it's been 2 years since you asked the question, could you please help of you figured it out?

[@uimasterskill]

@@plekton1 My god, it's been two years, I forgot having asked this question lol :)) So I've watched it again, and what I'd say now is: yes, it's true that B(h*(b), s*(b)) = b! However, this doesn't mean that when we take the derivative with respect to b, this term can be simply ignored, and this is because we're dealing with a composition of functions, so the L has a lot more freedom to take values in the first three slots than just ( h*(b), s*(b), lambda*(b)) At the very end of the video, he even uses the fact that B(h*(b), s*(b)) = b implicitly to conclude that d M*/d b = lambda*(b), because what he proves until that point is merely that d L*/ db = lambda*(b), but since B(h*(b), s*(b)) = b, it follows that L*(h*(b), s*(b), lambda*(b), b)=M*(h*(b), s*(b))

[@plekton1]

@@uimasterskill Wow thank you so much for replying and even taking out the time to watch the video again! I'll go through your reasoning.

[@desrucca]

Thanks! I just noticed that after ib read this comment

[@Dusadof]

You messed this part up I believe . Appreciate most your effort

[@marcuskissinger3842]

?

[@espectador-]

L

[@cleverclover7]

I think there was a rare, but critical mistake made by Grant in the notation when he didn't close the green parenthesis around another b in L*. around 8:46. Makes the derivative he comes to at the end make sense.

[@BLVGamingY]

he closed it in a different color, critical, moist critical

[@ojussinghal2501]

This took some time, but when it clicked, I suddenly realised this *actually was* mathematically rigorous...

[@mjackstewart]

I know you posted this a long time ago, but; If the gradient of the Lagrangian is 0, And the gradient is the (vectorized) amount of maximum ascent, Does this mean that, by definition, the Lagrangian is maximized since it’s value is 0? In other words, because it’s Lagrangian is 0, it can be maximized no more?

[@james1098778910]

love you all

[@xiaoweilin8184]

@9:30, why here the second term of -lambda*(B(h*, s*)-b) is not evaluated to 0?

[@dionsilverman4195]

Is it not sufficient to say that ∇f(x,y) = λ∇g(x,y) ⇒ λ = ∇f/∇g = df/dg, which is equivalent to saying that λ is the constant of proportionality between the gradients over the (x,y) domain, and therefore the rate of change of f with respect to g? Since the gradients are parallel, they are equivalent to a directional derivative in some direction u⃗, so df/du⃗ = λdg/du⃗ ⇒ (df/du⃗)•(du⃗/dg) = λ ? Also, I'm not sure if I follow why dL/db = dM/db at x⃗* just because L(x⃗*) = M(x⃗*)? I imagine two planes with different slopes passing through the same point somewhere, with different gradients at that point. Love your videos btw.

[@foxyelen1203]

The moment you enter b all the other variables become constants

[@atriagotler]

For me, this series ends here. I'll head to look for the next Grant video.

[@jetfaker6666]

thank you based god

[@niftimalcompression]

First time he mismatched the color of opening and closing brackets!!!

[@itsnotmeitsyou4582]

What’s this? Stats??? 🤔

[@farissaadat4437]

Hasn't this topic already been covered a year ago? Also why is this vid not added to the multivar calc playlist (as far as I cant tell), does it just take time?

[@zes7215]

wrg, no have to worry, no worry nmw, ts not clevx or not,just toolx,do things not worry things

[@tsunningwah3471]

弓弓

[@Hobbit183]

3yellow1brown jk

[@SoumyadeepBanerjee007]

Khub bhalo kintu amar onko korte bhoy lage..

[@fadhel901pdg6]

Gak ngerti aku, bukan jurusan saya

[@COMPTROL]

This is the most important and complicated lecture, yet he doesn’t use the plots, which he is famous for. Big disappointment.

[@espectador-]

?