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Jsdfio K.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

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Nutan C.

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All right, so now we're going to talk about a slightly different type of optimization and an extremely powerful method to solve problems. So the method is called LaGrange multipliers. And so here's the set up. So we're going to continue the theme of looking at just it functions of two variables for now. But the ideas will extend to functions of three or more variables. So we have a function of two variables that we want to maximize or minimize do. In this function, we're going to call the objective function. So the function that we want to maximize and minimize, we call the objective function and F is going to be subject to an equation of this form. So we have a function of X and Y equaling to some constant K, and this is called a constrict. And now we could have more than one constraint. But for now, we'll just consider again functions of two variables for our objective function, and then a single constrict. And so the idea if we sketch a picture so we have the X axis. Why access now? What I'm gonna do is I'm going to first of all, sketch my constraint curve. So this is just a curve. So maybe it looks something like that, I don't know. So here's G equals K, And then the next thing I'm gonna do is I'm going to plot some level curves of my function. Okay, so maybe one is like this. So maybe this is f equals C one. This is f equals C two like that. And so maybe you have another level curve in this level. Curve maybe just barely touches. They constraint like that. So maybe this is f equals C three. So again in green, these air level curves and then here we have our constraint. And so the idea is that, let's say the C one c two c through your increasing so the largest that f could be and still be subject to this constraint so noticed that we have points here and here. Where are function is going to be subject to the constraint. We have points here and here, but we can keep increasing all the way until we just barely touch the constraint. Or, in other words, the level curve is tangent to the constraint. And we actually have another way to write that. So if we look at the Grady in vector of first of all the level curve and then second of all of the constraint, they're going to be pointing in the same direction. So where does this maximum or minimum occur for this objective function? Subject to this constraint? Well, it's gonna occur exactly when the Grady in vectors point in the same direction. Or, in other words, the Grady in of F is equal to some multiple of the gravy int of G. And so it's not that the Grady INTs are exactly the same. They don't necessarily have the same length, but they are pointing in the same direction. And so I'm signifying that by noting that I could re scale the Grady in of G. I can even flip the sign and then have it be equal to the grading of F in this number. Lambda here is called the Look orange multiplier. And so what is the advantage of doing this? Are viewing this problem this way? Well, when we find the Grady in of F and we set it equal to the Grady int of g times this LaGrange multiplier, we get a system of algebraic equations that we can then solved. And when we solve those outbreak equations, uh, the points that we get the X y coordinate points. Uh, that's going to be where either the maximum or the minimum of the functions will occur. We just have to test those individual values and see which is the maximum in, which is the minimum. Okay, so things were gonna work exactly the same if we have a function of three variables subject to a single constraint. So we have our objective function f and then subject to this function g equal to some constant K than to solve. To find the maximum and minimum of this function subject to this constraint, we just set the Grady in of f equal to Lambda some number times the Grady int OMG. And then let's say we have another constraint. So let's say we had h of X y Z is equal to l. Well, then, what I'm gonna dio is I'm just going toe add on a second LaGrange multiplier. Call it mu times the ingredient of H. And so we're now we're gonna have another variable that we're gonna have to solve for but we're going to be able to do that. We're gonna have in total five unknowns. We're gonna have three equations given by the greedy int. And then we're gonna have the two constraints for in five. And so you can view this geometrically. It's a little bit more difficult, but you can at least say that the great you know, F lies in the plane spanned by the grading of G and the grading of H. So you sort of have to view that three dimensions. Thio fully understand it, but the idea works exactly the same way. If there's just one constraint and theoretically there could be a many constraints is you want. And you would just keep adding on the Grady int of the constraint times another LaGrange multiplier, and you would just keep doing that. And then you would just solve the corresponding system of equations. So that's the idea behind the garage multipliers. It's really simple and really powerful and will illustrate how the details go with a few examples

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