Multi-variable Optimization & the Second Derivative Test

in this video we're gonna study optimization if I have some particular function that depends on an x and y how do I find where this function is a Mac so how do I find where it has a minimum and sometimes how do I find where has this interesting new phenomenon called a saddle in the single variable case there were actually three different major possibilities the first of them was you have a minimum so some point for example a function x squared and it would go up at the value of x equal to 0 you would have a minimum location the derivative was 0 you had a horizontal tangent line but function appear to sort of bought an out and flatten off at that point likewise we had maximums and also a third case where it appears at the derivative was 0 you had a horizontal tangent line at the value of x equal to 0 but in this third case an example of which is someone like Y it would be X cubed but it was neither a maximum or a minimum it went up and I leveled off and then went other day no the first thing to say is all three of these scenarios are peep themselves in the multi variable calculus case you can have a minimum you can have a maximum and that neither situation can also occur you can be going not flattening off and then going up again but there's a new situation that can happen in the multi variable calculus case that couldn't happen before and that's the idea of a saddle when I have a saddle when I look at one of those directions for example I've plugged away if I go along the x axis I get this yellow curve then it appears to be a minimum according to the yellow curve but if you go parallel now to the y axis at the point 0 0 you get this blue curve which is going downwards and then it appears to be a maximum the idealist saddle is that in one direction it appears to be a maximum the other appears to be a minimum when you both those properties you get a saddle and single variable isn't relevant cuz you only ever had one direction that you could go now these two curves the yellow and blue ones be yellow one is what happens when I just plug in the value of y it would is around if I have a multivariable function f of X & Y and I plug in y producer I know these single variable function and you can see its graph likewise for the blue one if I plug in now this time X equal to zero I also get a single variable function f of zero because there are single variable functions I can use all the analysis from first or calculus but where to find that something was a minimum you took the derivative set it equal to 0 and those gave you candidates to be minimums maximums or 9 so in both of these cases I can pick the derivative with respect to X of the yellow curve or the derivative with respect to Y of the blue curve and both of those are derivatives of single variable functions now one of the big features of single variable calculus was just because the derivative was 0 didn't tell you it was a max or a minimum or neither Q had to do more and else beyond that but setting the derivatives equal to 0 gives you candidates to be maximums and candidates three minimums in the same exact story it may be true here is just that when I take the derivatives of these yellow and blue curves the derivatives say of f of X comma 0 that is going to turn into a partial derivative in the multivariable case so this is my statement of the so-called first derivative test it says if you have a local Max or min at a point then the following must be true it must be the true that both of the partials are 0 in that particular point so this is the exact analog of the previous theorem in single variable calculus saying that if they function head of maximum at one point the derivative had to be 0 at that particular point now let's see an example of this so I need to give the parabola x squared plus y squared so I want to go and figure out well where is the parcel throat is equal to 0 where is my candidates to be a maximum or a minimum well you take the first part so privileged is to exit that serum like you do the next is equal to 0 taking the second partial derivative this is going to do to wife take it respect to Y it said that it would is really good wide loser just because I tell you that the partials are both so doesn't tell you it was a maximum or a minimum or neither we have to investigate a little bit further but in this particular case if I just look at what the function is well the function is always positive or equal to zero of this one spot zero zero so that one spot zero zero discovered has to be a minimum because the thumb just always bigger around it but notice how I'm doing a bit of an ad arguing here I had to go and investigate the certain specifics of the function you changed function and this analysis might be harder or smaller it's easy to find candidates it's easy to find candidates to be maxims or minimums that the partial is equal to zero but how do I really know that they're going to be maximums or minimums in a general case where I can't just do this easy little bit of analysis but for that we have the analog of this second derivative test and Europeans it's a lot too much so we have to unpack it basically what it says in the following first of all assume that your functions are nice so what an Imagi first and second partial derivatives all are going to exist in there it would exist in some open disk around your point so you've got some nicely functions then it gives you conditions to tell you whether you've got a max a min a saddle or when the test fails and is inconclusive and you cannot use the test so the test is not completed tells you sometimes you can use this test and sometimes it's inconclusive but a good number of times we can use the test so it's worthwhile knowing okay the conditions here are a little bit weird there's something about the second derivative at X X being negative being positive but then there's also this weird expression in pink the second partial respect to x times the second partial respect to Y minus the square of the mixed partials F XY and whether this is greater than zero or less what's going on so let's try to break it down piece by piece first I'm gonna study that's an example of a minimum here now what I've highlighted in yellow here is the curve when I plug in y equal to zero so I have f of X zero then the second derivative of that curve tells me the concavity of it if the second derivative is positive it tells me that this yellow curve is facing off it's concave if the yellow curve here is concave down the second derivative will be negative indeed we use this analysis of concavity in single variable calculus to tell whether or not our particular point that was a candidate remax or a minimum was a maxima or minima and so indeed it makes a lot of sense that we demand that the second partial is going to be positive at this point down at the bottom is going to be a minimum and likewise if it's going to be a maximum you'd expect the exact opposite that the second partial has to be negative which means it's concave down however studying just the person with the X is not enough you might think okay well why don't you go and also look at the second partial factor Y so if you go the other direction if you fix your x value or Y is changing look at the second derivative that is a concave up concave down so that is actually done a little bit in the middle of the pink expression the pink expression is where FY y first appears but it's a little bit complex so let's switch to the third exam area that could happen to see why we might have this let's look now at the stat well in the case of the saddle what we have is that my our second partial respective X is positive but my second partials but 2y is negative so this means it's sort of concave up in one way concave down the other consistent with my notion of a saddle point then if I go and look at what the product of these two things are the partial of F respective X twice the partial of Y with about two y twice the product of those things is sort of a code for whether the two partial derivatives have the same sign or opposite sign so for instance in the case when they have opposite sign like one is concave up in the one direction and concave down in the other so positive second derivative the one Russian negative second derivative the other then the product FX and at YY will be negative since we're then subtracting something square is always positive in the case of a saddle what you have is that if these mixed partials or opposite directions for sure it's going to be a saddle so we think it's so far we justify why we have the FX X being less than zero and greater than zero with differential between maxims minimums and we've also seen the sort of first part of my pig expression the F xx times FY why why the sign of that might be important for telling us whether or not we have a saddle but what about the other part of this expression consider the following weird example ok so this is a new example it is a weird type of saddle it's actually the function x squared plus y squared minus 3x y know if I go along here and I specify that I am on the x axis I plug in Y for the 0 then what I just get is the curve x squared that's what I've highlighted in yellow and x squared its second partial derivative is just 2 which is positive it's concave up as you go along the x-axis alright but what happens when you go along the y-axis well I lose the exact same thing if you're going along the y axis of this now it's also concave up indeed the function is y squared along the y axis and its second derivative expected Y is 2 so whether you all on the X whether you go on the Y is concave up in both directions which might make you think that it should be a minimum and yet if I choose this somewhat clever third what if I go along the path y equal to X map so I'm here I'm restricting to y equal to X and if I plug it back in then what I get out of this is x squared plus x squared minus 3x squared I get minus x squared minus x squared is concave down so basically when I want to be x axis with concave up going on the y axis is concave up but going along y equal to X it was concave down now watch what happens to the mixed partial the f sub x y well first of all I'm going to say that that's taking the first version but the Y of I pick or soap derivatives for 2x 2x minus 3y I take that parcel and I get the value of minus 3 when you swear that you're gonna get nine which is just much bigger than the two times two you have for FXX times FY y so in this case considering that FX y the so-called mixed partial squared even allowed us to see that this was a saddle even though the analysis along x and y respectively could not see that with a saddle indeed it can be going down in some direction that's not the X and the y axis so the way to think about the mixed partial is that as I change my access I in for example parts of respect to X partial derivative with respect to Y now as I studied that it changes as I move off of the x axis and indeed you might recall that F X Y was equal to my ex so likewise you can study the partial respect to Y and then see how that changes if you move the X which is now moving off of the y axis so the mixed partials are a way of figuring out what happens when we're not just constrained to the axes that's what FXX I have to why why does the mixed partials tell us well how big are these changes if I go off a little on X and then I globe along Y that captures changes for example in y equal to X which is not along either of the axes and so if this you change X and then as you change Y which is picking you off the axes if that change is very large then the FX Y squared is gonna be very large and indeed then you can have a saddle even way along both axes it appears to be a maximum or appear to be a minimum now exactly proving this that the grotesque feel on the intuition that I've given you is beyond the scope of this particular video I can tell you that it comes from the generalization of Taylor's theorem we've seen in single variable calculus to the multivariable calculus case and indeed we're gonna have this expression is related to something called the hestian and that will be our way to formally demonstrate that these conditions are valid ways to decide maximum or minimum or a sample from the computational perspective our method is now relatively straightforward you find all the candidates by the first derivative test that is you go along and you figure out where is the partial present at X equal to 0 partial reified to y 0 and when both of those occur then the second derivative test gives us the conditions to classify those when is it the case is gonna be a maximum a minimum a saddle or unfortunately sometimes the test fails and we have to go further we have to do more analysis that we can get from just this test if you have a question about this video leave it down in the comments below real mathematicians here we appreciate algorithms so let's just hope the YouTube l know about by giving this video a like and finally if you want to watch more multivariable calculus videos this video is part of a larger playlist a multivariable calculus so you can check out those videos here and we'll do some more math in the next video

Channel: Dr. Trefor Bazett

Views: 17,583

Rating: 4.973856 out of 5

Keywords: Math, Solution, Example, Maximum, Minimum, Optimize, Optimization, Saddle Point, Critical Point, Hessian, Second Derivative Test, 2nd Derivative Test, 3D, multi-variable, matrix, partials

Id: _Ffcr98c7EE

Length: 13min 36sec (816 seconds)

Published: Sun Nov 24 2019