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Felicia S.

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

00:56

00:50

Masoumeh A.

02:07

Fangjun Z.

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All right. So here we have a very simple optimization question. So we want to minimize this function, which is our objective function, and then subject to this constraint, X plus y equals one. But we want to do it in two ways. So the first way we wanna actually reduce this down to a one variable optimization problem and then second, we want to use the method of LaGrange multipliers. So we're just gonna solve it the way we probably already know how, by reducing to one variable optimization case and then use the garage multipliers just to illustrate how the method works. So Method one what we would do we want toe optimize this function or sorry, minimize dysfunction subject to this constraint. Well, I can solve and just say why is equal toe one minus X? But then, if I plug this in, I get a function of one variable. So if I plug in one minus x for why I get X squared plus one minus x squared notice. This is now a function of one variable. And then I take the derivative. Here we go. And then I said it equal to zero to find my critical points. And so I have two X minus two and then plus two x zero Or in other words, for X minus two is zero and then we can solve this to get X is equal to one half. And now notice that if I take the second derivative, what am I going to get? I'm just going to get for And so what that means is that my function is always con cave down, which means there could Onley be one minimum. So I know that a minimum is going to occur at X equals one half. And so what is that minimum? Well, I'm gonna have one half for X. Why is going to be one minus one half? So also one half. And so what is this? One half squared is 1/4 plus one. Half squared is another 1/4. So our minimum value actually is one half. So that was just a quick Calculon review of how you would solve this optimization question. But let's see how it goes with the garage multipliers. So what we want to do is realize we have our objective function here and notice that we have this constraint function that needs to be one. So we have g of X y is equal to my constraint function, which is X plus y, and we require the g of X y be one. Okay, so let's set the Grady in of f equal Thio are LaGrange multiplier Lambda Times the create Ian of G. So the grading of s is two x a partial with respect to X comma Two Why Partial with respect to why and that should equal Lambda Times the Grady in of G, which is one one. Okay, so what we see is we actually get to equations. We get two X equals Lambda for the X coordinate. Let me get to y equals Lambda for the y coordinate. And now if we subtract thes two equations, we get two times x minus y equals zero. But the only way this is true is if X is equal to why. But I know that X plus why has to equal one. So if x equals y and X plus y equals one, X equals y equals one half. But this is the same thing we got here. So we take effort one half one half and we get one half, so we do indeed get the same minimum value. Now. It's debatable which one of these methods is quicker in this particular problem. But what is going to be true is that the second method is going to apply to a lot, uh, broader range of problems than the first method. And the first method can get extremely messy, whereas the second method keeps things relatively simple and relatively outbreak as we'll see.

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