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Locate all critical points and classify them:fky=1ly-y-x...

Question

Locate all critical points and classify them:fky=1ly-y-x

Locate all critical points and classify them: fky=1ly-y-x



Answers

Locate all critical points and classify them using Theorem 7.2. $f(x, y)=e^{-x^{2}-y^{2}}$

Gives us a to variable function f of X come alive. And then it asks us to find all the critical points and then state which type of critical points they are, whether that is a saddle point, a local maximum or local minimum. So to do this, we're going to end up using serum 7.2. To do this, we need to take the partial derivatives. So for the partial derivative with respect to X, well, look at what will focus on. We have a function of why here. And so this function will be kept constant. And then we use the chain rule on E to the negative X squared. So first, all right, what's kept constant outside of it and then, all right, our derivative eats a negative X squared times the derivative of this guy right here so that negative two X and then I distributed is going to be negative two x outfront times the quantity of y squared plus one e to the negative x squared f of why our second partial derivative that will be calculating for this one we treat the X function is constant. And then we will differentiate this wife function. So first, right at what kept constant out front. And then it will be multiplied by the derivative of the inside here, which is just too Why so then we have to. Why e to the negative x squared? We're looking for points where the partial derivatives are both equal to zero. And so we can do this easily, most easily by setting them equal to each other. And then this is going to be equal to zero eventually. So we have the partial derivative of F with respect to X. Going to rearrange that a little bit. I'm going to move the E to the negative X squared out front to keep all the excess together. And then these times, why squared plus one equal to two? Why e to the negative x squared. So we already see some terms that we have in common that we could get rid of. They both have a two that made multiplied and they both haven't eat the negative X squared. So then we're left with X times. Why square two plus one is equal toe. Why, this will then be set equal to zero. And so now we're looking out which points of X and Y are going to make both of these equations zero. Well, if we just have the 0.0 common zero, then that would work. So this is a critical point. If you want to look at anything that would make this zero were a bit out of luck there because we would need a negative number for why? Why squared? Instance? We're only using real members. This won't be possible. So we can't really rely on this to give a zero, and then the other functions air just x. And then why? So this is going to be our only critical point. So now we use the 4.2 to classify it. To do that, we're going to have to find the all of the second order partial derivatives so derivative of effort. Respect to X. Twice our original derivative was negative two x e to the negative x squared times y squared plus one. So I'm going to write that up here real quick again. And so for this one again, we're going to set. We're going to keep this y squared, plus one term constant, at least treated at such and then we're going to differentiate all the X terms that will have to do the product rule. Here we have our first being at negative two X. The derivative of that is just negative. Two we have negative to each of the negative X squared for this second term are derivative is going to be e to the negative X squared times negative two x and then we multiply it by our original term Negative two x And so we'll have to do with some simplification here. And so then we have y squared plus one times negative to e to the negative x squared. Then we have negative two x squared. Essentially, this will be plus four x squared e to the negative x squared. We want to plug in our point and see if it's greater than zero less than zero or equal to zero. But that is going to be after we find the discriminative. So the next time we have to look at is f of why, Why taking the second derivative with respect to y So we'll have to look at this again. We have to Why e to the negative x squared. So we treat the two in the E to the negative. X squared is constant. What's the derivative of why it's one? So we just have the remaining terms to e to the negative. X squared is all we have last. We have to compute f of x y and so we Since we have a theory, um, that allows us to take both of these equal to each other. It doesn't matter which one we start with. It looks like starting with F Y X is going to be easier since we have less terms. So why? With respect to X, we have a sub y is equal to two y e to the negative. X squared in this case will keep the two. Why constant or treated as such? So it's out front and then we have the chain role derivative of E to the negative X squared, which we've seen is each the negative X squared times two x and simplifying a bit further, we have the to choose being multiplied and then the negative four negative four x y e to the negative X squared. And so by 4.2 that score imminent is going to be equal to S x x f. Why? Why minus x y squared two s x x times f Why? Why that's going to be equal to to e to the negative x squared, which is the f y y times y squared plus one times negative to e to the negative x squared waas war X squared e to the negative x squared. Then we subtract the f x Y squared formula So that will be 16 x squared. Why squared? He eats the negative two x squared since we swear that we most by the exploded by two and then this is going to be our discriminative now we need to plug in the 0.0 comma zero. This will make a lot of the terms go away. We would have to e to the zero. So just one. Then we would just have one in these parentheses. The next one's we'd have a negative to e to the zero plus four times zero squared, which is zero he to the zero minus 16. We have zero multiplied a few times and then e to zero. Most of these terms end up going away. This becomes one and then we already have one here. This ends up becoming one. This goes to zero and so does this. What is the only term we're left with? It's going to be negative two, which is less than zero. And so that tells us that we're not going toe have any type of extreme here. We're going to have a saddle point. So our critical point is zero comma zero and it's a saddle point.


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