INSTRUCTOR: Use the second derivative to find the local extrema of the function f of x comma y equal to x cubed plus 2xy minus 6x minus 4y squared.

Let's begin by finding fx, the partial of f with respect to x, which is 3x squared plus 2y minus 6, and also fy, the partial of f with respect to y, which is 2x minus 8 y. Critical values can be found by making sure that both these are equal to 0. It's easier to solve fy equal to 0 because that would produce an equation that is x equals 4y.

Now, we can take this value x equals 4, this equation, and substitute this back into fx, setting it equal to 0. That will be 3, open parentheses, 4y, close parentheses, squared, plus 2y minus 6 equals 0. All right, now that will become 48y squared plus 2y minus 6 equals 0.

And this will factor-- I'm going to skip some steps here, but this will factor as 2, open parentheses, 8y plus 3, close parentheses, open parentheses, 3y minus 1, close parentheses, equals 0 so that y is equal to negative 3/8 or positive 1/3.

Now, taking each of these values, substituting these again, these y values, substituting that into x equals 4y, we see that this corresponds to an x value of negative 3 over 2 or positive 4 over 3, so that our critical points-- I'm going to mark them here, critical points-- are the ordered pair negative 3 over 2 comma negative 3/8, close parentheses, and the ordered pair 4/3 comma 1/3.

Now, in order to determine, using the second derivative, if these two points are maximums, minimums, or saddle points, we need to find the value of D. All right, now, by definition, D is equal to fxx evaluated at that point, multiplied by fyy evaluated at that point, minus f of xy evaluated at that point, and then squared.

So let's also find fxx. fxx is equal to 6x. fyy is equal to negative 8. And fxy is equal to 2. So we'll want to evaluate each of these functions we've just found at each of our critical points and see what we get. So D of the point negative 3 over 2 comma negative 3 over 8, using all these values, is greater than 0, and fxx at that same point, negative 3 over 2 comma negative 3 over 8, is less than 0.

The D test says that the-- or the second derivative test tells us that this is a local maximum. So we have a local maximum at the point negative 3 over 2 comma negative 3 over 8.

And following similar steps, we find that D evaluated at the point 4/3 comma 1/3 third is less than 0, and fxx evaluated at the point 4 over 3 comma 1 over 3 is greater than 0, which means there is a saddle point at the point 4 over 3 comma 1 over 3.