Find the values of the local minima Select the correct choice below and, if necessary; fill in the answer box

Find the local maxima, local minima, and saddle points, if any, for each function. $z=\frac{y}{x+y}$

For this problem we are asked to find the local maximum local minima and saddle points, if any, exist for the function Z equals X over X plus Y. So what we want to do is to solve for X and Y when the gradient should equal zero. So taking the partial derivative respect to X. And applying the product rule treating it as X times one over X plus Y. You could have one over X plus y minus X over X plus y squared equals zero from our partial derivative respect to X.

And then with respect to why we would get negative X over X plus Y squared 20 0. And if that's the case then well from that second equation we can see that we must have x equals zero and substituting that into our first equation. We would have then that one over why must equal zero. But there does. There is no why that would make that true.

So we would have that. This is not possible, which indicates that there are no extreme A. For the function.

For this problem we are asked to find the local maximum local minima and saddle points if any exist for the function Z equals X times Y times E. To the power of negative X plus Y. So we want to take the partial with respect to X. Which is going to give us well we need to apply the product rule here. So we want Y times E.

To the power of negative X plus Y. And then we would have minus. It would be just let's see here. It would be minus X. Y.

Eat or of negative X plus Y. You want that to equal zero. And with respect to why we would get X times eat are negative X plus y minus uh Why? Or excuse me minus X. Y times each power of negative explains why it's not X plus one, X plus Y. You want that to equal zero.

So we can recognize that the first equation can be simplified down to just y minus X equals or actually would be one minus X equals zero, dividing through both sides by Y equals negative X plus y. And so we would have one minus X equals zero or X equals one. And then same deal for the second equation we would get one minus Y equals zero. So we'd get y equals one. So our candidate point here is not 00, it's the 00.11.

But if we imagine implementing just slightly around that point, for instance taking the point 22 and 00. Well we can see that zet of 00 would be zero. Can see that said of 1, 1 would be one times one times each hour of negative one plus one. So that would be um one over E squared. And that is going to be greater than zero squared.

That's greater than zero. And if we take zet of 2, 2 actually, um let's see here, I want to take something that would give us something lower. So let's say negative one positive one, then we would get negative. Um It would be negative one times each power of zero. So we would have a value there of negative one which is clearly going to be less than zero, Which suggests that the point that we're looking at 00 here or not, 00.

Excuse me. 11 here that actually the relevant thing is negative one has to be less than one over e squared. What we found at that 10.11. So that 0.11 must be a saddle point..

Hi there. In this problem, we're asked to fund any critical points here and then figure out whether we have local men Max Saddle point or neither. So whenever we're refining critical points, the very first thing we do is take the partial derivatives respect to X and Y because we want to set them equal to zero. So let's do that now. Har shows respect.

X now X squared becomes too x x y never retreating. Why, like it's a constant Expect the variable. So the partial derivative with respect to X becomes why and three wide doesn't even have an accident so that become zero partial derivative respect to why Let's go back to the top x squared zero x y that partial driven, different respect. Allies Just X, the partial derivative of three y With respect, why is three So we have our partials the whole point of that. This that we can set these equal to zero to find our critical point.

So let's solve. Each of these equations is easiest to solve of this bottom equation, because the only way expose three could equal zero as if X equals minus three and now, looking mad into the top equation. If X must equal minus three, then we have two times minus three plus y equals zero none. Other words negative, six less viable zero. So why would have to equal six? All right, so there's only one critical point.

So we'll set a critical point and it's at the point. Negative three, comma six. So all that's left to do now is to figure out whether we have a minimum maximum saddle point or neither. So to do that, let's use the second derivative test, which is right at the end of this section. Remember, for the second derivative test, we're gonna need these second powerful derivatives, so we'll need f sub x x.

Let's look at F sub X. That was too explosive. Y said that partial derivative with respect to X is just too. We'll need F some Why, why? No, here's f y right here. So the partial derivative X plus three with respect to why Southerner wise in it.

So that will be zero. And finally we need the mix. Partial derivative f sub x y. Here's FX appear so the derivative partial derivative of F sub X, with respect to why, we'll see. It's just one night that wide termers gives us a one.

And so now we're ready. Thio calculate d notice that none of our second partials here even having extra y in them so they don't even depend on our point, their constant everywhere So our devalue is going to be. It's defined as f sub x x times f sub y y minus have some X y squared. So we have two times zero minus one squared, which is negative one which of course, is less than zero. Now, according to the second derivative test at the end of the section, when D is less than zero, then we can conclude that F has a saddle point at any credible point.

So therefore, this critical point must be a saddle point and we are done..

In this question were given the function I have X y ico to one off x plus x y because one of the why and the first time we need to find the first specially with u F X, They were getting into one of the X Square. Yeah, that's why he coaches zero. So from here, it means that why ICO to one other X square and similarly, the f why we go to, uh, minus one, otherwise square and with the nicotine. So therefore we have the X. We go to one otherwise square.

So we're going to use this choir here, put into this guy, then we should get thanks we go to Nah thanks. Our fallen now and then we have doesn't even only if the X minus expel for e coaches zero So conflict the X outside on. Then we have one minus Expo three e coach is zero. Therefore, we have XY coaches zero Mexico to one when x which is also why we go to under fight. So we'll be, uh, it will be notified in this case here and then we're not considered this one, so no value.

Yeah, and I went actually 21 while you were equal with you one. And therefore, we have only one value off the critical point here. So we have only one. And the second step, we need to find a second brush on the river to f X x. Then we get Nico June, uh, to off Explore three now and then F y why we go thio over.

Why about three on f x y we go to one Therefore add the on 11 Do we ghenda have x x times f y y on the f x y square they will record you now we will have the f x x would other 11 So I'm going to times that we have times two minus no one square So it doesn't include to the three credit and zero and f x X equal to two is created and Sarah Therefore we can conclude that we will have the loco minimum and ex uncle had upon 11 Yeah, yeah,.