# The Critical Point Test

Prequisites: Partial Derivatives

So, Say a point *P*_{0}=(*a*_{1},*a*,_{2},...*a*_{n
,})
is a maximum point (or a minimum pont), of the equation
*y*=* f*(*x*_{1},*x*,_{2},...*x*_{n,})
. Then if we hold (*x*_{2},...*x*_{n})
fixed at (*a*_{2},...*a*_{n}) ,
*f* becomes a function of *x*_{1} only, with a maximum at
* x*_{1}=*a*_{1}.
Therefore at this point .

Likewise,
*f*_{x2}=0,
*f*_{x3}=0, ...
*f*_{xn}=0 at point *P*.

A ** Critical Point ** is any point that satisfies these *n*
equations. Alternately, the critical point is any point on our
*n*-dimensional surface whose tangential *n*-1 surface is
parallel to the *x*_{1}-axis, *x*_{2}-axis,
... , and
*x*_{n}-axis. We can determine critical points
(*a*_{1},*a*_{2},... *a*_{n})
by solving these *n* equations for the *n* unknowns .

For example, say
*f*(*x,y*)=*x*^{5}+*y*^{4}-5*x*-32*y*. Then,

## ,
.

Solving for real critical points, we get (1,2) and (-1,2). But
are these points minimas or maxima, or neither? We notice
that the point *x*=1 is a minimum for the first equation and the
point *y*=2 is a minimum for the second equation. Therefore, the
point (1,2) is a minimum for
*f*(*x,y*)=*x*^{5}+*y*^{4}-5*x*-32*y*
. On the other hand, the point *x*=-1 is a maximum for the first
equation. Thus, this surface, viewed along the *x*-axis, comes to
a maximum, while viewed along the *y*-axis, comes to a minimum. The
critical point of (-1,2) is neither a minimum nor a maximum point for
the surface. It is a ** saddle point **. Below are images of a
minimum, a maximum, and a saddle point critical point for a
two-variable function.

##

In
the next section
we will deal with one method
of figuring out whether a point is a minimum, maximum, or neither.

Vector Calculus Index |
World Web Math Main Page

watko@athena.mit.edu
Last modified November 5, 1998