SKKU-Calculus-Sec-13-8 Extrema of Multivariate Functions-New

13.8 Extrema of Multivariate Functions by SGLee - HSKim- SWSun-JHLee, 오교혁
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(Lecture) http://youtu.be/oDZUkOEszOQ
(Exercises) http://youtu.be/FWmk_MasIjE
1. Let
. Find the critical points of
and classify them.
Solve
and
.
So we have critical points,
or
. If
, then
.
If
, then we have
.
The critical points are
,
,
,
.
Next, we consider second order partial derivatives to get
,
,
.
Then
and
thus we obtain
at points
,
.
This implies that
,
are saddle points.
At points
,
, we observe that
.
Moreover, since
,
has a local maximum
at
.
On the other hand, due to
,
has a local minimum
at
.
위에서 구한 를 saddle point, local maximum, local minimum으로 각각 분류하자.
즉, ,
은 saddle point이고,
,
은 local maximum이다.
2. Find the extreme values of the function
when
.
![]()
![]()
![]()
![]()
: critical points
=>
has no local minimum. or maximum at
.
At
,
![]()
, so
has a local maximum
at
.
3-4. Locate the maxima, minima, and saddle points of the functions.
3. ![]()
http://matrix.skku.ac.kr/cal-lab/cal-12-4-3.html
http://matrix.skku.ac.kr/LA-Lab/ms-1.html
4. ![]()
![]()
Try this on your own.
5. Let
. Answer the following:
(a) Find points of local maximum/minimum and a saddle point when
.
(b) Give a condition on
for the case when
has only one critical point.
![]()
(a)
![]()
![]()
![]()
![]()
![]()
=> (1,1) is a saddle point.

=>
are points of local minimum..
(b)
![]()

=> ![]()
If
has only one critical point
has a solution and
should not have a solution. So
.
6. Find maximum value of
on
.
,
.
So the critical point is
and thus critical value is ![]()
Let
,
,
and
.
On
, we have
and
,
.
On
, we have
and
,
.
On
, we have
and
,
.
On
, we have
and
.
.
So the maximum value is 2.
7. Find the absolute maximum and minimum of
in the domain
which is a closed triangle made of three points (0. 0), (2, 1), (1, 2).
![]()
(1) ![]()
=> critical point :
, ![]()
(2) 1.
moves on ![]()
=> The absolute maximum
, and the absolute minimum
on
.
2.
moves on ![]()
=> The absolute maximum
, and the absolute minimum
on
.
3.
moves on ![]()
=> The absolute maximum
, and the absolute minimum
on
.
Hence the absolute maximum is 2 and the absolute minimum is 0.
8. Find the absolute maximum and minimum values
on the disk D:
.
interior of
:
![]()
Then
implies ![]()
If
implies ![]()
Thus, we get the critical points ![]()
If
then
This implies
.
Critical points are ![]()
Thus ![]()
Consider
, boundary of
:
so ![]()
Moreover,
is smallest when
and largest when
But ![]()
Thus on D the absolute maximum of
is
and the absolute minimum is ![]()
9. Find the Taylor series for the function
at the point
.
![]()
,
,
, ![]()
,
,
,
,
,
,
,
.
Therefore
![]()
![]()
![]()
.
10. Expand the Maclaurin series for the function
.
, ![]()
, ![]()
, ![]()
, ![]()
, ![]()
, ![]()
In general,
.