If you're seeing this message, it means we're having trouble loading external resources on our website.

Hvis du sidder bag et internet-filter, skal du sikre, at domænerne *. kastatic.org og *.kasandbox.org ikke er blokeret.

Hovedindhold
Aktuel tid:0:00Samlet varighed:6:39

Video udskrift

- [Instructor] So in the last two videos we were talking about this constrained optimization problem where we want to maximize a certain function on a certain set, the set of all points x, y where x squared plus y squared equals one. And we ended up working out through some nice, geometrical reasoning that we need to solve this system of equations. So there's nothing left to do but to just solve this system of equations. I will start with this first one at the top, see what we can simplify. We notice there's an x term in each one so we'll go ahead and cancel those out, which is basically a way of saying we're assuming that x is not zero, and we can kind of return to that to see if x equals zero could be as you wish. And so maybe we'll kind of write that down. We're assuming x is not equal to zero in order to cancel out. And we kind of can revisit whether that could give us another possible solution later, but that will be two times y is equal to lambda times two, and from here, the twos can cancel out, no worries about two equaling zero, and we know that y equals lambda. So that's a nice, simplified form for this equation. And for this next equation, though we can use what we just found, that y is equal to lambda to replace the lambda that we see, and instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y. So that's two times y squared. And I'll leave it in that form because I see that in the next equation, I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and replace that with y squared, so that's two y squared plus y squared equals one. And then from there, simplifies to three times y squared equals one, which in turn means y squared is equal to one third, and so y is equal to plus or minus the square root of one third. Great. So this gives us y, and I'll go ahead and put a box around that, where we have found what y must be. Now if y squared is equal to one third, then we look up here and say, mhm, two times y squared, that's gonna be the same thing as two times one third, so two times one third, so if x squared is equal to two thirds, what that implies is that x is equal to plus or minus the square root of two thirds. And then there we go, that's another, one of the solutions. And I could write down what lambda is, right, I mean, in this case it's easy, 'cause y equals lambda, but all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want, and we just have that pesky, little possibility that x equals zero to address, and for that we can take a look and say, if x equaled zero, you know let's go through the possibility that maybe that's one of the constrained solutions. Well, in this equation that would make sense, since two times zero would equal zero. In this equation, that would mean that we're setting zero equal to lambda times two times y. Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero, so evidently, you know, if it was the case that x equals zero, that would have to imply, from the second equation that y equals zero, but if x and y both equal zero, this constraint can't be satisfied. So none of this is possible, so we never even had to worry about this to start with, but it's something you do need to check, just every time you're dividing by a variable, you're basically assuming that it's not equal to zero. So, this right here gives us four possible solutions, four possible values for x and y that satisfy this constraint, and which potentially maximize this, and remember, when I say potentially maximize, the whole idea of this Lagrange Multiplier, is that we were looking for where there's a point of tangency between the contour lines. So just to make it explicit, the four points that we're dealing with, here I'll right them all here. So, x could be the square root of two thirds, square root of two thirds, and y could be the positive square root of one third, and then we can basically just toggle, you know maybe x is the negative square root of two thirds and y is still the positive square root of one third, or maybe x is the positive square root of two thirds, and y is the negative square root of one third, kind of monotonous, but just getting all of the different possibilities on the table here. X is negative square root of two thirds, and then y is positive, no, negative, that's the last one, square root of one third. So these are the four points where the contour lines are tangent, and to find which one of these maximizes our function, here let's go ahead and write down our function again, it gets easy to forget. So the whole thing we're doing is maximizing f of x, y, equals x squared times y. So let me just put that down again, we're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them is actually greatest. And the first thing to observe is, x squared is always gonna be positive, so if I plug in a negative value for y, right, if I plug in either this guy here or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say that neither of these can be the maximum because it'll be some positive number, some x squared times a negative. Whereas I know that these guys are gonna produce a positive number, and specifically, now if we plug in, if we plug in f of let's say this top one, square root of two thirds, square root of one third, well, x squared is gonna be two thirds, and then y is square root of one third. And in fact that's gonna be the same as what we get plugging in this other value. So either one of these maximizes the function. It's got two different maximizing points and each one of them has a maximum value of two thirds times the square root of one third. And that's the final answer. But I do want to emphasize, that the takeaway here is not the specific algebra that you work out going towards the end, but it's the whole, the whole idea of this Lagrange Multiplier technique to find the gradient of one function, find the gradient of the constraining function, and then set them proportional to each other. That's the key takeaway, and then the rest of it is just, you know making sure that we check our work and go through the minute details, which is important, it has its place. And coming up, I'll go through a few more examples.