Hi Whitney. And this first one you solved here for X, which was great, but then when you substituted it, you put it in for Y. So instead we have to put in for X. So we're going to get 3 instead of X. We're going to put a -3 halves y ^2 and square it minus the two Y and that equals 0. So that's going to be +9 fourths and 9 fourths times 3 is 27 fourths Y to the fourth minus two y = 0. I probably am going to multiply through by 4 just because I don't like the fractions. And then I'm going to realize that I can pull out AY and that gives me 20 seven y ^3 -, 8. Now technically that's the difference of two cubes and lots of you won't do it this way, but that technically factors into this. But this portion here are non real roots. So you'd get Y equaling zero and Y equaling 2/3. Now what a lot of people will do is they'll just take this 20 seven y ^3 -, 8 and equals 0, and that'll give us the right answer for real roots. But reality is you're losing your imaginary roots this way. So the cube root of 820 sevenths is 2/3 also. So once we get our Y, we just plug it back into the fact that X equaled -3 halves y ^2, so X and the first one's going to be 0. So we have one point of 00 and the other point our Y is 2/3. So -3 halves times 2/3 ^2. 1 two is going to cancel, 1/3 is going to cancel, so negative 2/3. Now remember, we're not done with this problem. We still have to find our second partials. So FXX, which would be 6X FYY, which would be -6, YFXY, which would be -2 And then when we plug in for the points, we'd have 6 * 0 * -6 * 0 - -2 ^2, and we can see that that's less than 0, and hence a saddle point. And then the other one we're going to do the same way. So we're going to have six times negative 2/3 * 6 negative 6 times 2/3 - -2 ^2 and that's going to give us -4 * -4 positive 16 - 4, so 12. So that's greater than 0 and the -4 is less than 0. So that -4 being less than means that we're going down. So this is going to be a local Max #15 looks pretty good to me. 6 X squared minus two X ^3 + 3 Y squared minus oh, 15 in my paper. I'm not looking at the book, I must admit, but on my homework, that's not what 15 says. It says it's +6 XY at the end. I'll try to go check that in a minute, but I don't have my book with me right now. So you and I have different problems for 15, I guess is the reality. If you have the right problem and I don't, it actually looks just fine. And what you would do is you'd have two points, you'd have 01 half, and you'd have two 1/2. And then you'd find your F, your second derivatives. So your second partial in terms of X, your second partial in terms of Y, your second partial in terms of XY, and sticking your points. But I kind of think that perhaps that's really the right problem. I don't even see a -3 Y -8 close by. 16 was negative three y ^2 -. 8 I don't know. 17, I'm going to get some new papers because I'm running out of space. So 17, you did well. But what I would recommend is to get the two equations to combine. So three X ^2 + 3 Y squared -15 = 0. And if I multiply the second equation through by a negative, I get -6 XY minus three y ^2 + 15 equaling 0. So when I add those we get three X ^2 - 6 XY equaling zero. If I factor out of three, XI get X -, 2 Y so now I can see that X = 0 and X - 2 Y equals 0, so X equal 2 Y. So when X is 0, putting it back into the original either one of them, I get 3 Y squared equaling 15 Y squared equal 5. So Y equal positive negative sqrt 5. So that gives me two points 0 root 5 and 0 negative root 5. Now doing this second one, I'm going to take it into either of the equations. So I'm going to get three times two y ^2 + 3 Y squared equaling 15. Four y ^2 * 3 is twelve y ^2 + 3 Y squared, so we get 15 Y squared equaling 15 Y is going to equal positive or -1. If Y is one, we can see that X is 2 * 1 or two. If Y is -1 we get 2 * -1 or -2. So it was oh, OK, so here we really do have those four points. And then when we substitute into our equation, so we've got to find our FXX and that's going to be 6X and our FYY is going to be 6X plus 6Y and our FXY is going to be 6 Y. So then we would plug those in for each of those points. So FXX for 0 root 5 is going to give me 6 * 0 or 0 * 6 * 0 + 6 times root 5 -, 6 times root 5 ^2, and without even doing the math, we can see that's less than 0 and hence a saddle point. So the other one actually is going to be a saddle point too, because when I stick in zero for X, that's going to make this first term be 0. So then if you look at 21, we'd get 12 * 18 - 6 ^2, I don't know, 9 * 12108. So 216 - 36, that's obviously greater than 0. And our F double X is greater than 0. So that's telling us it's going up. So that means it's got to be a local min. And if we do the same thing for our -2 negative one, we will find it's going to be a local Max because we're going to have -12 * -18 -6 ^2. So that's still greater than 0, but now that's a -12 which is less than 0. If it's less than, we know it's going down, which makes that a local Max. And I went back and I did check an end. This problem was written wrong, so hopefully if you write it correctly, you won't have problems. Hope this helps, have a good night.