Point) Find all local maxima , Iocal minima_ and saddle points of each function. Enter each point as an ordered triple, enter comma-separated Iist of ordered triple...

Find the local maxima, local minima, and saddle points, if any, for each function. $z=3 y^{3}-x^{2} y+x$

In this problem we will cover extremely of two variable functions. So to find the critical points of this function, we first have to find the partial derivatives. So we begin with the partial derivative with respect to X. And that's going to be three x squared minus three Y. And or partial derivative with perspective, why is going to be negative three x minus three Y squared there. You have it. And so we want to find values of X and Y that make both of these partial derivatives zero. So we're going to set both of these equal to zero. And for our partial derivative with respect to X. We see that we can rewrite this as Y equals X squared. And so using that we're just going to plug into this equation over here. So we have negative three X -3 -3 times X squared squared zero. This gives us -3 x -3 X to the 4th equals zero. We can factor out a negative three X from both terms and we get one. Sorry one plus x. cute equals zero. And this means that X can equal zero Or X Can Equal -1. And so over here, when we plug in X equals zero, we get y equals zero. And when we plug in negative one we get y was one. So that means that we have two critical points. We have one at 00 and we plug those into our function we get a z coordinate of zero. So we have a critical point there and we have a critical point at Let's see you negative one one. When we plug in those into our function we get a Z value of one. So we have a critical point here to classify what kinds of critical points these are. We need to find that the discriminate it. And we do that by using the second order partial derivatives. So we begin with the second order Partial derivative of respect to X. And this is going to be six X. And at the zero this will be zero. And at the 0.-1 11 it's going to be negative six. Our second order partial derivative with respect to why Is going to be -6 Y. And at the point 000 This will be zero And at the 0.-111, This will be negative six And last are partial derivative with respect why of the derivative respective X is just going to be -3. And now we can compute our discriminate. So Our first discriminate at the zero. So we're going to have let's see zero times zero minus negative free squared which is just going to be negative nine. Since the discriminate here is negative, that means at 000 we have a saddle point and last are discriminative at the point negative 111 is going to be -6 times negative six minus negative three squared. So that's going to be 36 -9 which is 27. The discrimination is positive, but the second order partial derivatives with respect to X, is negative. So that means at negative 111, we have a local maximum.

In this problem we will cover extreme of two variable functions. So to solve this problem we want to find the critical points first and to do so we have to find the partial derivatives. So we begin with the partial derivative with respect to X. And this is going to be three X squared plus three Y. And our partial derivative with respect to why? Which is going to be three x plus three y. I swear here we have it. And so to figure out what are critical points are we have to find values of X and Y. That make each of these partial derivatives equal to zero. So we set them equal to zero. And for our partial derivative with respect to X. We see that one possible combination is when X equals zero, Y equals zero. And we also see here when X equals negative one, Why has the equal negative 1? And the same thing goes for our partial derivative with respect to why we see that X zero, Y S b zero. And we also see that when X is -1 Why has to be negative one as well. So that means we have two critical points. Our first critical point is at 00 and zero. Because when we plug in X and Y equals zero into our function, we get zero in our second critical point. Is that negative one? Negative 11. Because when we plug in those values for X and Y into our function we get one and now to classify what kinds of critical points these are, we need to find the discriminate it. And that means using the second order partial derivatives. So our second order partial derivative with respect to X is going to be six X. When we plug in the zero, we just get zero. And when we plug in the point negative one negative 11 We get -6. Our second order partial derivative with respect to why is going to be six white. When we plug in our zero we just get zero and when we flag in our point negative one negative 11 we get negative six. Again, Unless our partial derivative with respect to why of the derivative with respect to X is going to be three. So now with all of this information we can now compute the disk ruminant. So we begin with the discriminative at the point 000. And this is going to be zero times negative six minus three squared. And that's just going to be -9. And because this is negative, that means we have a saddle point At the zero. And now we compute the discriminate it of the point -1 -11. So that's going to be negative six times negative six minus three squared. Which leaves us with the value of 27. So we see here that the discriminative is positive and we also see that the second order partial derivative with respect to X is negative. So that means that at the point negative one negative 11, we have a local maximum.

Mhm. For this problem we are asked to find the local maxima. Local minima and saddle points if any, for the functions equals Y over X plus Y. So when we take the gradient of this we would have that. Let's see here, we would want to have that. Why? Over or rather negative Y over X plus Y squared Needs to equal zero. or alternatively we need to have that negative. Okay so that was from partial derivative with respect to X. Then with respect to why we need to apply chain ruler quotient rule will apply chain rule, it will be one or not. Chain rule. Excuse me? Product ruler quotient rule. So first taking the derivative of Y and then we'll add on the derivative of one over X plus Y times Y. So that would be negative Y over X plus y squared. We want that to equal zero alternatively. So we can also we can see immediately that we need to have if negative Y over X plus y squared equals zero. Then we need why to equal zero. If that's the case then we need one over X two equals 0 which is not possible. So that tells us then that there are no extreme er

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