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

13.8     Extrema of Multivariate Functions          by SGLee - HSKim- SWSun-JHLee, 오교혁

(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 $x,y$를 saddle point, local maximum, local minimum으로 각각 분류하자.

즉, $(2,0)$$(-2,0)$은 saddle point이고, $(-\frac{2}{3}\sqrt3,-\frac{2}{3}\sqrt3)$$(\frac{2}{3}\sqrt3,\frac{2}{3}\sqrt3)$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.

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,

.