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2. Find all critical points for the function f(x,y) = &x' _ 3x4 + 48xy -12y2 and label them as local maxima, local minima; or saddle points. Show your work...

Question

2. Find all critical points for the function f(x,y) = &x' _ 3x4 + 48xy -12y2 and label them as local maxima, local minima; or saddle points. Show your work: (10 pts:)

2. Find all critical points for the function f(x,y) = &x' _ 3x4 + 48xy -12y2 and label them as local maxima, local minima; or saddle points. Show your work: (10 pts:)



Answers

Find the critical points and classify them as local maxima, local minima, saddle points, or none of these. $$f(x, y)=x^{2}-y^{2}+4 x+2 y$$

In this problem we will cover local extremists. So first, in order to find the critical points of this function, we must find the partial derivatives with respect to X and Y and set them equal to zero. So with Beginning with the partial derivative with respect to X, which is going to be two X -2. Why if we set this equal zero, we see that. We can simplify it. two x equals two y which leads to X equals Y. So we keep that in mind. Are partial derivative of respect. So why is going to be negative two? X Plus 6? Y -8. We set this equal zero and we're going to use our X equals Y from here. So we're going to have zero equals negative two times Y plus six Y minus H. And this leads to zero equals four, y minus eight, which leads to why equals two. And because X equals Y over here, that means that X has to equal to And so we know that are critical points is going to be at 22. Now to classify what kind of critical point this is. We have to use the discriminate. And that means finding the second order partial derivatives. So we find the second order partial votive of respect X. And that is going to be too The second order partial devoted its perspective. Why is going to be 6? And the partial derivative of respect why of the derivative of X is going to be negative too. And we plug in these values into the discriminate. So we have D Equals two times 6 minus negative two squared. That's going to equal 12 -4, which equals eight. And because the discriminate is positive and the second order partial derivative with respect to X is positive as well. That means at our critical .22 we have a local minimum.

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.

Either in this problem were asked to find all the critical, uh, points for this function and then figure out whether each is a local men Max saddle point or none of those things. So to find critical points, we take the partial derivatives with respect to X. And why said them equal to zero and solve. So we always like to start by taking a partial derivatives. So let's take the partial derivative with respect to X and that will give us some two X here and derivative of four X with respect to X is for the derivative of y squared with respect to access just zero. In the partial respect to why going back up to the top here, X squared doesn't have a wife so that 04 x doesn't have a wife. So that zero why Squared has a partial derivative to why Okay, so now we want to set each of these equal to zero and then solved, and we can do this quickly by inspection. This first equation. The only way to expose for equal zero is if X equals negative, too. And the only way to wipe ALS zero is if y equals zero. So we only have one critical point critical point, and it's at the point. Negative two comma zero. Okay, so now all that's left is to determine whether we have local men. Local Max Settle Point are neither of those. So that's used the criterion right at the end of the chapter, the second derivative test it's called, So it involves the second partial derivatives. So they use the to figure out whether we have local max or men or anything like that. We're going to need the second partial derivatives, so we'll need F X X. Let's go ahead and do that. FX was up here to expose for the derivative of that with respect to X is just too okay. Why why? Well, here is F supply just to are the derivative of two y with respect to wise again just to, and now we need the mixed partial so f X y. So here's a partial respect, except here there are no wise in it. So the derivative is zero, and the second derivative test tells us that we need to look at this capital D, which is defined as FX x times F Y y minus affects. Why squared? I didn't write that point X comma y for each of these. But we're evaluating these at our critical point. Negative to zero. Okay, so let's figure that out now. In this case will notice, uh, none of our partial. None of our second partial derivatives even depend on accent. Why so? Well, let's get a constant number here, and that will answer everything for us. So we have to times two minus zero way. Get four. Which is positive, of course, greater than zero. So if we look at the second derivative tests, we have degraded than zero and we also need to look at x X here, which is also greater than zero. So D is greater than zero and no second partial f xx eyes also great in zero in the second derivative test tells us that when this result happens, um, than F has a local minimum at a critical point. Therefore, we can safely say this critical point is a local minimum, and we're done hopefully

In this problem we will cover local extreme A. So first to find the incredible points of dysfunction will have to find the partial derivatives and set equal to zero. So we begin with the partial derivative with respect to X. And that's going to be three X squared minus three. We set this equal to zero. We see that we could factor the right hand side To me at zero equals 3 times X squared minus one. And we will get that X equals Plus or -1. Are partial derivative with respect to why is going to be three, Y squared minus three. Set this equal to zero. And we see that we can factor the right hand side, we get zero equals three times why squared -1. And we get that Y equals plus or minus one. So that means that we have for critical points We have 1 1, one, negative one, -1, And -1 1. And so to classify what kinds of critical points these are, we have to find the discriminate for each point. And to do so we must find these second order partial derivatives. So we know that the second order partial derivative with respect to X Is going to be six x. And when we plug in the point 11 It's going to be six. Plugging in the point 1 -1 is going to be six. Again, The 0.-1 -1. That's going to be -6. And at the point negative one one negative six. Again, our second order partial derivative with respect to y is going to be six. Why? So at the .11 This derivative is going to be six At 1 -1 it's going to be positive six. Or not positive 6 -6. At the 0.-1 -1, it's going to be -6. And at the 0.-1, one is going to be positive six. And our last partial derivative second order partial derivative, it's just going to be zero. So we can ignore that. And now we just want to find the discriminate at each point. So the indiscriminate of the point 11 Is going to be six Times six and that's 36. And since the discriminative is positive and the second order partial derivative respect the X. Is positive. That means we have a local minimum at this point. Now for the discriminate at the point one negative one That's going to be six times negative six. Which gives us -36. And this is going to be a shadow point because the discrimination is negative. Now we want the discriminate at the 0.-1 -1. And that's going to be -6 times negative six. Which gives us a positive 36. And since the discriminate it is positive, yet the second hershel derivative with respect to X, is going to be negative. We have here a local maximum. Our last point is going to be -11. And when we plug in negative six six we get negative 36. And because the discriminate is negative, that means this point there's going to be a saddle point.


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