INSTRUCTOR: Find the absolute extrema of the function f of x comma y equal to 4x squared minus 2xy plus 6y squared minus 8x plus 2y plus 3, on the domain defined by 0 less than or equal to x, less than or equal to 2, and negative 1 less than or equal to y, less than or equal to 3.

Let's begin by noting what the domain is. In the xy-plane, this is a rectangle. That's our domain. So we might want to parameterize this. But for the moment, what I'm really more interested in is the boundary and, specifically, parts of the boundary, the corner points, corners of this region, because it is a rectangular region.

So the boundary or the corner points are given by these points, 0, negative 1, 0 comma 3, 2 comma negative 1, and 2 comma 3. Those are the corner points. And those are possible-- again, because of the shape of this region, those are possible locations for extrema.

Now, the other possibility for extrema would be critical values, critical points. So let's take fx. fx is equal to 8x minus 2y minus 8. And if fy is equal to negative 2x plus 12y plus 2.

Now, I'm going to set fy equal to 0, which leads me to say that x is equal to 6y plus 1. I'm going to use this equation as a substitution into fx so that I have 8, open parentheses, 6y plus 1, close parentheses, minus 2y minus 8 equals 0. Again, we want both fx and fy to be equal to 0 here.

And solving this leads us to say that y equals 0 and that x is equal to 1. So our critical point-- I'm going to add this underneath my corner point list-- my critical point is the point 1, 0. Now, to determine whether that interior point, or critical point, is a maximum, minimum, or a saddle point, we'll use the second derivative test.

So I'm going to add to my list of derivatives over here, fxx, which is equal to 8, fyy, which is equal to 12, and fxy, which is equal to negative 2. Putting this information together, D of 1 comma 0 is greater than 0, and fxx evaluated at 1 comma 0 is greater than 0, which means that point 1, 0 is a local minimum. Local minimum.

Now, we want the absolute extrema. So what we're going to do is make a table with these points. So the first column is ordered pairs, x comma y. Next column is f of x comma y. And I'll spare you the algebra as I evaluate these.

But for the first ordered pair, 0 negative 1, f of 0, negative 1 is 7. Next ordered pair, 0 comma 3, f of 0 comma 3 is 63. Our next ordered pair, we have 2, negative 1-- 2 comma negative 1-- and that function value is 11. For our next point we have two comma 3, which evaluates to 51 in our original function. And finally, the point 1 comma 0, which evaluates to negative 1.

This means, because we already said that 1 comma 0 was a local minimum, that comparing those values-- it's the lowest. It's the lowest value. So we have an absolute minimum of negative 1 at the point 1 comma 0. And we have an absolute maximum of 63 at the corner point 0 comma 3.

Now, where we might want to consider the boundary, in addition to this, is if any of our function values at the corners have equal output values. Because that means that the entire edge between those two boundary points, or those two corner points, that boundary between the two is a maximum itself.