13.8     Extrema of Multivariate Functions


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 .


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. .

 (Sage)  http://matrix.skku.ac.kr/cal-lab/cal-12-4-3.html

             http://matrix.skku.ac.kr/LA-Lab/ms-1.html  

var ('x, y, z')

f(x,y)=2*(x^2-y^2)-x^4+y^4

P=implicit_plot3d (z==f(x, y), (x,-1.5,3/2), (y,-3/2,3/2), (z,-1,1), color= 'goldenrod', opacity=0.6)

P.show()

contour_plot(f, (x,-2,2), (y,-2,2), contours= srange(-2,2,0.25), fill=False, cmap= 'cool', labels=True)

                 


var ('x, y, z')

f(x,y)=2*(x^2-y^2)-x^4+y^4

fx=f.diff(x)

fy=f.diff(y)

fxx=diff(f,x,x)

fyy=diff(f,y,y)

fxy=diff(f,x,y)

cpoints=solve([fx==0,fy==0],[x,y],solution_dict=True)

for sol in cpoints:

    if ((sol[x] in RR) and (sol[y] in RR) ):

        print((sol[x],sol[y]))

  (0, 0), (1, 0), (-1, 0), (0, 1), (1, 1), (-1, 1), (0, -1), (1, -1), (-1, -1)


def extreme(f,a,b):

    f11=diff(f,x,x)(a,b)

    f22=diff(f,y,y)(a,b)

    f12=diff(f,x,y)(a,b)

    D=f11*f22-f12^2

    if(D>0):

        if(f11>0):

            return "local minimum"

        else:

            if(f11<0):

                return "local maximum"

            else:

                return "inconclusive"

    else:

        if(D<0):

            return "saddle point"

        else:

            if(D==0):

                return "inconclusive"

table = [["Critical Point", "Type"]]

f(x,y)=2*(x^2-y^2)-x^4+y^4

fx=f.diff(x)

fy=f.diff(y)

fxx=diff(f,x,x)

fyy=diff(f,y,y)

fxy=diff(f,x,y)

cpoints=solve([fx==0,fy==0],[x,y],solution_dict=True)

for sol in cpoints:

    if ((sol[x] in RR) and (sol[y] in RR)):

        a=sol[x].n()

        b=sol[y].n()

        table.append([(sol[x],sol[y]), extreme(f,a,b)])

html.table(table, header=True)

Critical Point   Type

(0, 0)          saddle point

(1, 0)          local maximum

(-1, 0)         local maximum

(0, 1)          local minimum

(1, 1)          saddle point

(-1, 1)         saddle point

(0, -1)         local minimum

(1, -1)         saddle point

(-1, -1)         saddle point

 

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,

          


                              

 

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