• anonymous
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.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
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.
1 Attachment
  • amistre64
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.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.