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

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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[chrislam]

can you also cover the inequality constraint optimization, please?

[cyancoyote]

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

[davidsweeney]

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!

[xiaoweilin]

@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?

[richardshandross]

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*.

[MauPP]

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!

[DrKvo]

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

[abdullaalmosalami]

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

[petripaavonpoika]

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?

[Dusadof]

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

[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?

[xiaoweilin]

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

[guilhermegondin]

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?

[dionsilverman]

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.

[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?

[cleverclover]

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.

[foxyelen]

The moment you enter b all the other variables become constants

[farissaadat]

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?

[ojussinghal]

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

[atriagotler]

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