Optimizing multivariable functions
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Second partial derivative test example, part 1
- [Voiceover] So one common type of problem that you see in a number of multivariable calculus classes will say something to the effect of the following: find and classify all of the critical point of, and then you'll insert some kind of multivariable function. So first of all, this idea of a critical point basically means anywhere where the gradient equals zero. So you're looking for places where the gradient of your function, at some kind of input, some specified input X and Y that you're solving for, is equal to zero. And as I talked about in the last couple of videos, the reason you might want to do this is because you're hoping to maximize the function or to maybe minimize the function. And now the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something, or the gradient equals zero, you want to be able to determine, is it a local maximum, is it a local minimum or is it a settle point? So let's go ahead and work through this example. The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line, we take both partial derivatives. So the partial derivative with respect to X is, well, this first term, when we take the derivative of three X squared times Y with respect to X, the two hops down so we have six times X times Y. Y cubed, well, Y looks like a constant so Y cubed looks like a constant minus three X squared, so that two comes down. So we're subtracting off six times X. Six times X. And then again, this three Y squared term, Y looks like a constant so everything here looks like a constant with zero derivative, as far as the X direction is concerned. And we do partial of f with respect to Y, then this first term looks like some sort of constant, three X squared. X looks like a constant so some kind of constant times Y. So the whole thing looks like three X squared. The second term, minus X, minus Y cubed, excuse me, looks like minus three Y squared when we take the derivative. Minus three Y squared. And then this next term only has an X so it looks like a constant as far as Y is concerned. And then this last term, we take down the two because we're differentiating Y squared, and you'll get negative six Y, negative six times Y. So when we are finding the critical points, the first step is to set both of these guys equal to zero. So this first one, when we do set it equal to zero, we can typify a bit by factoring out six X. So this really looks like six X multiplied by Y minus one. And then that's what we're setting equal to zero. And what this equation tells us is that either it's the six X term that equals zero, in which case that would mean X is equal to zero; or it's the case the Y minus one equals zero, in which case that would mean that Y equals one. So at least one of these things has to be true. That's kind of the first requirement that we found. Let me scroll down a little bit here. And for the second equation, when we set it equal to zero, it's not immediately straightforward how you would solve for this in a nice way in terms of X and Y, but because we've already solved one, we can kind of plug them in and say, for example, if it was the case that X is equal to zero, and we kind of wanna see what that turns our equation into, then we would have, well, three X squared is nothing that would be zero and we'd just be left with negative three Y squared minus six Y is equal to zero. And that, we can factor out a bit, so I'm gonna factor out a negative three Y. So I'll factor out negative three Y, which means that first term just has a Y remaining and then that second term has a two, a positive two, since I factored out negative three. So positive two and that equals zero. So what this whole situation would imply is that either negative three Y equals zero, which would be, which would mean Y equals zero; or it would be the case the Y plus two equals zero, which would mean that Y is equal to negative two. So that's the first situation where we plug in X equals zero. Now alternatively, there's the possibility that Y equals one, so we could say Y equals one; and what that gives us in the entire equation, we still have that three X squared because we're kind of solving for X now, three X squared, and then the rest of it becomes, let's see, minus three times one squared, so minus three, we're plugging in one for Y then we subtract off six, plugging in that one for Y again, and that whole thing is equal to three X squared then minus three minus six. So I'm subtracting off nine. So from here, I can factor out a little bit, and this will be three multiplied by X squared minus three. And what that implies then, since this whole thing has to equal zero, what that implies is that X squared minus three is equal to zero so we have X is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found. One of them was in the circumstance where X equals zero, and then the other was where we found in the circumstance where Y equals one. So this give us a grand total of three different critical points because in the first situation, where X equals zero, the critical point that we have, well, both of them are going to have an X coordinate of zero in them, X coordinate of zero; and the two corresponding Y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another two possibilities here, where if Y is equal to one, when Y is equal to one, we'll have X as positive or negative square root of three. So we have positive square root of three and Y equals one, and then we have negative square root of three and Y equals one. So these, these are the critical points, critical points, which basically means all partial derivatives are equal to zero. In then in the next video, I will classify each of these critical points using the second partial derivative test.