# Obtain the extremum of f(x,y,z)

by s3a
Tags: extremum, obtain
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,564 You have $f(x,y,z)= 2x^2+ y^2+ 2z^2+ 2xy+ 2xz+ 2yz+ x- 3z- 5$ and assert that $f_x= 4x+ 2y+ 2z+ 2y+ 1$ Where did that second "2y" come from? The other two derivatives are correct. As for determining what kind of extremum, there are two things you can do. The most basic is to look at values around the (x,y,z) point you get. Since there is only the one extremum, if other values of (x,y,z) give smaller values for f, you have a maximum, if higher, a minimum. Of course, there is the possibility of a saddle point but since the problem asks for an "extremum" that shouldn't happen. More sophisticated is to form the matrix of second derivatives: $$\begin{bmatrix}f_{xx} & f_{xy} & f_{xz} \\ f_{yx} & f_{yy} & f_{yz} \\ f_{zx} & f_{zy} & f_{zz}\end{bmatrix}$$ evaluated at the critical point. Because of the equality of the "mixed" derivatives, $f_{xy}= f_{yx}$, etc., that is a symmetric matrix and so has all real eigenvalues. If all eigenvalues are positive the extremum is a minimum, if all eigenvalues are negative, it is a maximum, and if there are both positive and negative eigenvalues, there is no "exteremum" but a saddle point.