I'm gonna give it a name, I'm gonna call this constant lowercase b. So we think of that, and I'm just gonna emphasize here that that's a constant, right, that this b is being treated as a constant right now. You know, we're thinking of h and s and lambda all as these variables, and this gives us some multivariable function.
And if you'll remember from the last video, the reason for defining this function is it gives us a really nice compact way to solve the constraint optimization problem. You set the gradient of L equal to zero, or really the zero vector, right, it'll be a vector with three components here. And when you do that, you'll find some solution, right, you'll find some solution, which I'll call h star, s star, and lambda, here I'll give it that green lambda color, lambda star.
You'll find some value that, when you input this into the function, the gradient will equal zero. And, of course, you might find multiple of these, right, you might find multiple solutions to this problem, but what you do is, for each one of them you're gonna take a look at h star and s star, then you're gonna plug those into the revenue function, or the thing that you're trying to maximize. And, typically, you only get a handful, you get a number, then you can actually plug each one of them into the revenue function, and you'll just check which one of them makes this function the highest.
And whatever the highest value this function can achieve, that is M star, that is the maximum possible revenue, subject to this budget. But it's interesting that when you solve this, you get some specific value of lambda, right, there's a specific lambda star that will be associated with the solution.
And, like I said, this turns out not to just be some dummy variable. It's gonna carry information about how much we can increase the revenue if we increase that budget.
And, here, let me show you what I mean. So we've got this M star, and I'll just write it again, M star here. And what that equals, I'm saying that's the maximum possible revenue. So that's gonna be the revenue when you evaluate it at h star, h star and s star. And h star and s star, they are whatever the solution to this gradient of the Lagrangian equals zero equation is. You set this multivariable function equal to the zero vector, you solve when each of its partial derivatives equal zero, and you'll get some kind of solution.
So when you plug that solution in the revenue, that gives you the maximum possible revenue. But what we could do is consider this as a function of the budget. Now, this is the kind of thing that looks a little bit wacky if you're just looking at the formulas. You wanna ask the question, what happens as you change this b.
Well, the maximizing value, h star and s star, each one of those guys is gonna depend on b, right. As you change what this constant is, it's gonna change the values at which the gradient of the Lagrangian equals zero.
So, I'm gonna rewrite this function as the revenue evaluated at h star and s star, but now I'm gonna consider that h star and s star each as functions of b, right, because they depend on it in some way. As you change b, it changes the solution to this problem It's very implicit and it's kind of hard to think about. It's hard to think, okay, as I change this b, how much does that change h star.
Well that depends on what the, you know, what the definition of R was and everything there. But, in principle, in this context, I think it's quite intuitive. Similar considerations apply to maxima. Note that if there are multiple minima or maxima this should just fall out of the maths. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
Create a free Team What is Teams? Learn more. Why do Lagrange Multipliers work? Ask Question. Asked 6 years, 5 months ago. Active 3 years, 4 months ago. Viewed 3k times. Can someone please give me an explanation. However, as we saw in the examples finding potential optimal points on the boundary was often a fairly long and messy process.
In this section we are going to take a look at another way of optimizing a function subject to given constraint s. We want to optimize i. Again, the constraint may be the equation that describes the boundary of a region or it may not be. The process is actually fairly simple, although the work can still be a little overwhelming at times.
Notice that the system of equations from the method actually has four equations, we just wrote the system in a simpler form. In order for these two vectors to be equal the individual components must also be equal. So, we actually have three equations here. To see a physical justification for the formulas above.
In fact, the two graphs at that point are tangent. If the two graphs are tangent at that point then their normal vectors must be parallel, i. Mathematically, this means,. This means that the method will not find those intersection points as we solve the system of equations.
This is a good thing as we know the solution does say that it should occur at two points. Also, because the point must occur on the constraint itself.
Note that the physical justification above was done for a two dimensional system but the same justification can be done in higher dimensions. For example, in three dimensions we would be working with surfaces. However, the same ideas will still hold. At the points that give minimum and maximum value s of the surfaces would be parallel and so the normal vectors would also be parallel.
Before we start the process here note that we also saw a way to solve this kind of problem in Calculus I , except in those problems we required a condition that related one of the sides of the box to the other sides so that we could get down to a volume and surface area function that only involved two variables.
We no longer need this condition for these problems. Next, we know that the surface area of the box must be a constant So this is the constraint. The surface area of a box is simply the sum of the areas of each of the sides so the constraint is given by,.
Note that we divided the constraint by 2 to simplify the equation a little. There are many ways to solve this system. This gives,. Doing this gives,. This gave two possibilities. This leaves the second possibility. Therefore, the only solution that makes physical sense here is.
We should be a little careful here. Anytime we get a single solution we really need to verify that it is a maximum or minimum if that is what we are looking for.
This is actually pretty simple to do. If the volume of this new set of dimensions is smaller that the volume above then we know that our solution does give a maximum.
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