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Multivariable calculus

Course: Multivariable calculus>Unit 3

Lesson 5: Lagrange multipliers and constrained optimization

Proof for the meaning of Lagrange multipliers

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. Created by Grant Sanderson.

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• If you people are having any trouble understanding it, here;s another analogy.Remeber, how we defined the labour in terms DOLLARS per hour ? and similarly, steel as, DOLLARS per ton of steel? But the max limit was our budget, that was simply dollar. Therefore, the labour, and steel both actually are the functions of dollar, and we have constraint on how much we can spend, that is our budget. Thats why h(b) and s(b) holds true. b represents the dollar that you put in
• Awesome analogy man! Didn't think of it this way, but it makes a lot of sense when we think like this.
(1 vote)
• at , why would lambda* be written as a constant lambda*, rather than a function lambda*(b), as the other variables h*(b) and s*(b)? Thanks!
• Correct me if I am wrong, but probably Grant just missed it. If you watch later on in the video he actually differentiates lambda with respect to b, so that makes it a function of b.
• In the L* function isn't the (B(h*(b),s*(b))-b) term 0? Though we have considered h* & s* to be functions of b, isn't it still true that for every b, B(h*(b),s*(b)) would indeed be equal to b? But then that reduces L* to just R(h*(b),s*(b)) and thus the derivative wrt to b to be 0 which is certainly not the case. Where am I wrong?
• Isn’t it a little hand-wavy to say that ∂L*/∂λ* is the same as ∂L/∂λ evaluated at λ*?
• Is it possible to find an expression for lambda star as a function of b?
• According to this video, I think that `lambda` is the gradient of budget-maximum revenue `b` - `M*` graph.
So, in order to get `lambda` as a function of `b`, you have first to find the general function `M*` of variable `b`, then
``lambda = d(M*)/d(b)``
is what you are looking for.
• looks like traversing with the value of b(constraint) will take you through all the Maxima's that the original revenue function has to offer i can see how machine learning can select the best of the combinations of h and s variable with constraint b :)