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. .

 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

(1, 0)          local maximum

(-1, 0)         local maximum

(0, 1)          local minimum

(0, -1)         local minimum

4. .

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,