Proof for the meaning of Lagrange multipliers | Multivariable Calculus

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

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Күн бұрын

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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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Пікірлер: 53

@yogeshbhatt7154 4 жыл бұрын

journey with sanderson ends here

@metuphys5611 Жыл бұрын

sad moment

@tentrot4420 5 ай бұрын

Sad moment

@scoutgaming737 2 ай бұрын

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

@chrislam1341 4 жыл бұрын

can you also cover the inequality constraint optimization, please?

@poiuwnwang7109 11 ай бұрын

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

@cyancoyote7366 6 жыл бұрын

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

@davidsweeney111 6 жыл бұрын

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 5 жыл бұрын

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 3 ай бұрын

@@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_ 7 күн бұрын

@@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. :)

@richardshandross3453 6 жыл бұрын

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 2 жыл бұрын

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.

@xiaoweilin8184 Жыл бұрын

@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?

@Mau365PP 5 жыл бұрын

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 3 жыл бұрын

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

@DrKvo 3 жыл бұрын

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

@MightySapphire Жыл бұрын

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

@abdullaalmosalami 3 жыл бұрын

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

@petripaavonpoika3767 Жыл бұрын

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 6 жыл бұрын

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 4 жыл бұрын

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 4 жыл бұрын

@@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 4 жыл бұрын

@@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 2 жыл бұрын

Thanks! I just noticed that after ib read this comment

@guilhermegondin151 5 жыл бұрын

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 2 жыл бұрын

Isn't it what is done here?

@Dusadof 10 ай бұрын

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

@marcuskissinger3842 5 ай бұрын

@espectador- 5 ай бұрын

@mjackstewart 3 жыл бұрын

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?

@naru2906 2 жыл бұрын

great video, grant!!

@xiaoweilin8184 Жыл бұрын

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

@duncanw9901 6 жыл бұрын

WOAH is this 3B1B in the voiceover???

@cyancoyote7366 6 жыл бұрын

Yep!

@dionsilverman4195 4 жыл бұрын

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.

@cleverclover7 2 жыл бұрын

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 Жыл бұрын

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

@ojussinghal2501 4 жыл бұрын

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

@james1098778910 5 жыл бұрын

love you all

@atriagotler 2 жыл бұрын

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

@foxyelen1203 4 жыл бұрын

The moment you enter b all the other variables become constants

@jetfaker6666 2 жыл бұрын

thank you based god

@farissaadat4437 6 жыл бұрын

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?