Given the following function, determine the absolute extrema and find where they occur.
f(x,y) = 2x^4 + y^3 - x^2y on the region that is bounded by y = -1 and y = 3 - x^2
The first thing I think to do is find the gradient.
I don't suppose disciminants could be useful, as I think they only apply to LOCAL extrema.....
That is, D(a,b) = f_(xx)(a,b)*f_(yy)(a,b) - [f_(xy)(a,b)]^2, where f_(x), for example, is the partial with respect to x, and then xx is then the partial with respect to x of the one we just found, etc. Then, if D(a,b) > 0 and f_(xx)(a,b) > 0, then f has a local min. at (a,b)... if D(a,b) > 0 and f_(xx)(a,b) < 0, blah blah blah, if D(a,b) < 0, then f has a saddle point at (a,b), and so on. But again, I think this only applies to local.
So then, I think of trying to look at this as I would in Calc 1 finding extrema. That is, find critical points, test those, and test endpoints (which I will assume is the boundaries we are given), etc. We can find crit. points by just taking the partial with respect to x, and the partial with respect to y, and plugging them to 0, I believe, just as you would a derivative.
f_(x) = 8x^3 - 2xy
f_(y) = 3x^2 - x^2
So, f_(x) = 8x^3 - 2xy = 0 ... happens when x = 0 or y = 0
f_(y) = 3x^2 - x^2 .. happens when x = 0
This is where things become a little fuzzy.