Question

Given that f(x,y)=x^2-y^2+xy-10x+3.
Find the critical point for the given curve and determine whether the critical point maximum, minimum or saddle point.

Answer 1

The graph of this function is not a curve but a surface. So if you want to find the critical point of \[f(x,y)=x^2-y^2+xy-10x+3\] then you should solve \[\frac{\partial f}{\partial x}(x,y)=0\textrm{ and } \frac{\partial f}{\partial y}(x,y)=0.\]
You'll get a system of linear equations with solution x=4, y=2. So the critical point might be (x,y)=(4,2).
To decide whether this point is a maximum, minimum or a saddle point you have to compute the determinant of the second order partial derivatives: \[J=\det(|\begin{array}{cc}\frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x \partial y}\\\frac{\partial^2 f}{\partial y \partial x}&\frac{\partial^2 f}{\partial y^2}\end{array})\] at point x=4, y=2. You'll find J=-5 (in fact J=-5 at every point), so it must be a saddle point. The attached picture is quite convincing.

Answer 2

ive always wondered about 3d equations like this. I was wondering how ignoring one variable would be viable. but it appears that when we hold one variable constant, then if doesnt matter what that "constant" value would be becasue we getwhat, a family of curves right? that all have the same profile. i think i understand it better now, or maybe i just confused myself all over again :)

That J det has me at a loss tho.....