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88213 Partial DerivativesEXERCISESRAIn Exercises 23 find all critica points Determine whether each critical point yields # relative maXimum value, a relative mhinim...

Question

88213 Partial DerivativesEXERCISESRAIn Exercises 23 find all critica points Determine whether each critical point yields # relative maXimum value, a relative mhinimum value. 01 saddle point 1. fl") =42 Hmz 61 Su 2_ f(x>) =42 Hpii 61 Sv Wb 3. f6J) =l 61V Hizvh 6xhHOv 2

882 13 Partial Derivatives EXERCISESRA In Exercises 23 find all critica points Determine whether each critical point yields # relative maXimum value, a relative mhinimum value. 01 saddle point 1. fl") =42 Hmz 61 Su 2_ f(x>) =42 Hpii 61 Sv Wb 3. f6J) =l 61V Hizvh 6xhHOv 2



Answers

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{3}+y^{3}-3 x^{2}+6 y^{2}+3 x+12 y+7 $$

Alright for this problem we are asked to find all the critical points of the function F of x Y equals x, Y squared minus six, X squared minus three Y squared and then determine whether the function is a minimum maximum or saddle at those points. So we start by taking the first partial derivatives which will have why squared minus 12 X. We want that to equal zero, which means that we have y squared needs to equal 12 X. Or Y needs to equal plus or minus the square root of 12 X. Then the partial derivative with respect to why it's going to give us two xy minus six, Y is equal zero. So we can factor at the Y there. Why times two x minus one needs to equal zero. So this is going to equal zero when either Y equals zero or when two x minus one equals zero. So that tells us that we need to x to equal one or X needs to equal one half. Uh huh. So, what that tells us then is that we have three different possible candidates, candidates. So when X equals zero, Or I'll be a little bit careful here, 1/2. So we'll have a solution when x equals zero because when X equals zero, actually, yeah, when X equals zero, then route 12 actually equal zero, which means Y equals zero. So we'll have both of those. Um Both of those will equal zero at 00 And then we'll have the points 1/2 and positive route six as well as one half And Negative Route six. So, having that, we now need to figure out the nature of the function of those points, which we do by taking the second partial derivatives. So the second partial derivative with respect to X is going to be negative 12. Second partial derivative with respect to Y. It's going to be two X -6. And the mixed partial derivative uh is going to be uh to Y. So D of X Y Is going to be negative 12 Times two x -6 -4 Y Squared. Now, all we need to do is calculate an hour D function there at each one of our points, which I'll calculate offscreen. Alright, so at 00 we should get 72. At one half positive route six, We should get 36 and also at one half negative route six, we should get 36 as well. So we also have that the second partial derivative with respect to X is less than zero. So that means that at each one of these points we are going to have a let me double check here. So I don't say something dumb from theorem. See that means that we are going to have a maximum.


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