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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.

## Want to join the conversation?

• 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 • 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! • 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 :) • When you say the first threem terms of the chain rule evaluate to 0 because the gradient of the Lagrangian is the 0 vector, since we're considering the Lagrangian as now a 4 variable function of h,s,lambda, and b then why doesn't the partial derivative of the Lagrangian with respect to b also evaluate to 0? • But if we have proved that dL*/db = dL/db (and therefore the derivative with respect to b of the function that depends only on b is equal to the derivative with respect to b of the multivariable function), why at minute it expresses lambda as a function of b ?? In the multivariable function isn`t lambda independent from b ??
(1 vote) • if df/dx=lambda(dc/dx) why not just do lambda=(df/dx)/(dc/dx)=df/dc, which translates to dM*/db b/c M*=f(x*,y*) at a maximum, and b=c(x,y) always? Also works if you do it from df/dy.
(1 vote) • Why is it that when he takes the total derivative of L* he uses normal L in each term, for example, dL/dh, not d(L*)/d(h*)?
(1 vote) 