Stewart Calculus ET 5e 0534393217;14. Partial Derivatives ...y1zhao/2009Math231/Sec1508Solution.pdf · f f g(x,y)=8 f(x,y)=c c g(x,y)=8 c=59 c g(x,y)=8 c=30 f g(x,y)=8 59 30 c= 1

f f g(x,y)=8

f (x,y)=c cg(x,y)=8 c=59 c

g(x,y)=8 c=30 fg(x,y)=8 59 30

c= 1 c=1.25

x2+y x

2+y

2=1

f = 2x,1 g= 2 x,2 y 2x=2 x =1 x=0 =1 y=12

x=3

2x=0 y= 1 f

0, 1( )3

2,12

f3

2,12

=54

f (0,1)=1 f (0, 1)= 1

f (x,y)=x2

y2

g(x,y)=x2+y

2=1 f = 2x, 2y g= 2 x,2 y 2x=2 x x=0

=1 x=0 x2+y

2=1 y= 1 =1 2y=2 y y=0 x= 1

f 1,0( ) 0, 1( ) f 1,0( )=1

f (0, 1)= 1 f x2+y

2=1 f ( 1,0)=1

f (0, 1)= 1

f (x,y)=4x+6y g(x,y)=x2+y

2=13 f = 4,6 g= 2 x,2 y 2 x=4 2 y=6

x=2

y=3

13=x2+y

2=

2 2+

3 213=

132

= 1 f

2,3( ) 2, 3( ) f (2,3)=26 f ( 2, 3)= 26

f x2+y

2=13 f (2,3)=26 f ( 2, 3)= 26

1. At the extreme values of , the level curves of just touch the curve with a commontangent line. (See Figure 1 and the accompanying discussion.) We can observe several suchoccurrences on the contour map, but the level curve with the largest value of which stillintersects the curve is approximately , and the smallest value of corresponding to alevel curve which intersects appears to be . Thus we estimate the maximum value of subject to the constraint to be about and the minimum to be .

2. (a) The values and seem to give curves which are tangent to the circle. These values

represent possible extreme values of the function subject to the constraint .

(b) , . So either or . If , then and so

(from the constraint). If , then . Therefore has possible extreme values at the

points and . We calculate (the maximum value),

, and (the minimum value). These are our answers from (a).

3. , , . Then implies or

. If , then implies and if , then implies and thus .Thus the possible points for the extreme values of are , . But while

so the maximum value of on is and the minimum value is .

4. , , . Then and imply

and . But , so has possible extreme

values at the points , . We compute and , so the maximum

value of on is and the minimum value is . 1

Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers

f (x,y)=x2y g(x,y)=x

2+2y

2=6 f = 2xy,x

2g= 2 x,4 y 2xy=2 x x=0

=y x=0 x2=4 y =0 y=0 y=0 g(x,y)=0,

=0 g(x,y)=6 y= 3 =y x2=4 y x

2=4y

2g(x,y)=6

4y2+2y

2=6 y

2=1 y= 1 f 0, 3( ) 2,1( )

2, 1( ) f f ( 2,1)=4f ( 2, 1)= 4

f (x,y)=x2+y

2g(x,y)=x

4+y

4=1 f = 2x,2y g= 4 x

3,4 y

3x=2 x

3x=0

=1

2x2

x=0 x4+y

4=1 y= 1 y=2 y

3y=0 x= 1 =

1

2y2

x2=y

22x

4=1 x=

14

20, 1( ) 1,0( )

1

[4]2,

14

2f x

4+y

4=1

f1

42

,1

42

=2

2= 2 f (0, 1)= f ( 1,0)=1

f (x,y,z)=2x+6y+10z g(x,y,z)=x2+y

2+z

2=35 f = 2,6,10 g= 2 x,2 y,2 z 2 x=2

2 y=6 2 z=10 x=1

y=3

z=5

35=x2+y

2+z

2=

1 2+

3 2+

5 2

35=35

2= 1 f

1,3,5( ) 1, 3, 5( ) f x2+y

2+z

2=35 f (1,3,5)=70

f ( 1, 3, 5)= 70

f (x,y,z)=8x 4z g(x,y,z)=x2+10y

2+z

2=5 f = 8,0, 4 g= 2 x,20 y,2 z 2 x=8

20 y=0 2 z= 4 x=4

y=0 z=2

5=x2+10y

2+z

2=

4 2+10 0( )

2+

2 2

5=20

2= 2 f 2,0, 1( ) 2,0,1( )

f x2+10y

2+z

2=5 f (2,0, 1)=20 f ( 2,0,1)= 20

f (x,y,z)=xyz g(x,y,z)=x2+2y

2+3z

2=6 f = yz,xz,xy g= 2 x,4 y,6 z f = g

=(yz)/(2x)=(xz)/(4y)=(xy)/(6z) x2=2y

2z2=

23

y2

x2+2y

2+3z

2=6 6y

2=6


. If , then implies or . However, if then a contradiction. So

and then . If , then implies , and so

. Thus has possible extreme values at the points , ,and . After evaluating at these points, we find the maximum value to be andthe minimum to be .


. If , then implies . But implies so or and

and so . Hence the possible points are , ,

, with the maximum value of on being

and the minimum value being .

7. , , . Then

, , imply , , and . But

, so has possible extreme values

at the points , . The maximum value of on is , and theminimum is .

8. , , . Then ,

, imply , , and . But

, so has possible extreme values at the points , . The maximum of

on is , and the minimum is .

9. , , . Then

implies or and . Thus implies or

2


y= 1 2, 1,23

2, 1,23

2, 1,23

2, 1,23

f2

32

31 3

f (x,y,z)=x2y

2z2

g(x,y,z)=x2+y

2+z

2=1 f = 2xy

2z2,2yx

2z2,2zx

2y

2g= 2 x,2 y,2 z

f = g =y2z2=x

2z2=x

2y

20 =0

0 x2=y

2=z

2=

13

f

0127

1

3,

1

3,

1

3

1

3,

1

3,

1

3

1

3,

1

3,

1

3

f (x,y,z)=x2+y

2+z

2g(x,y,z)=x

4+y

4+z

4=1

f = 2x,2y,2z g= 4 x3,4 y

3,4 z

3

x 0 y 0 z 0 f = g =1/ 2x2( )=1/ 2y

2( )=1/ 2z2( ) x

2=y

2=z

2

3x4=1 x=

14

3

14

3,

14

3,

14

3

14

3,

14

3,

14

31

43

,1

43

,1

43

14

3,

14

3,

14

3f 3

1

2f 2

1 f 1

x4+y

4+z

4=1 f 3 1

f (x,y,z)=x4+y

4+z

4g(x,y,z)=x

2+y

2+z

2=1

f = 4x3,4y

3,4z

3g= 2 x,2 y,2 z

x 0 y 0 z 0 f = g =2x2=2y

2=2z

2x

2=y

2=z

2=

13

8

f13

. Then the possible points are , ,

, . The maximum value of on the ellipsoid is , occurring when all

coordinates are positive or exactly two are negative and the minimum is occurring when or

of the coordinates are negative.

10. , , .

Then implies (1) and , or (2) and one or two (but not three) of

the coordinates are . If (1) then . The minimum value of on the sphere occurs in case

(2) with a value of and the maximum value is which arises from all the points from (1), that is,

the points , , .

11. ,

, .

Case 1: If , and , then implies or

and or giving the points , ,

, all with an value of . Case 2: If one of

the variables equals zero and the other two are not zero, then the squares of the two nonzero

coordinates are equal with common value and corresponding value of . Case 3: If exactly

two of the variables are zero, then the third variable has value with the corresponding value of

. Thus on , the maximum value of is and the minimum value is .

12. ,

, .

Case 1: If , and then implies or yielding

points each with an value of .

Case 2: 3


012

f12

0 1

f 1 x2+y

2+z

2=1 f 1

13

f (x,y,z,t)=x+y+z+t g(x,y,z,t)=x2+y

2+z

2+t

2=1 1,1,1,1 = 2 x,2 y,2 z,2 t

=1/(2x)=1/(2y)=1/(2z)=1/(2t) x=y=z=t x2+y

2+z

2+t

2=1

12

,12

,12

,12

f f12

,12

,12

,12

=2

f12

,12

,12

,12

= 2

f x1,x

2, . . . ,x

n( )=x1+x

2+ +x

ng x

1,x

2, . . . ,x

n( )=x2

1+x

2

2+ +x

2

n=1

1,1, . . . ,1 = 2 x1,2 x

2, . . . ,2 x

n=1/ 2x

1( )=1/ 2x2( )= =1/ 2x

n( ) x1=x

2= =x

n

x2

1+x

2

2+ +x

2

n=1 x

i= 1/ n i=1 .. . n f

f 1/ n ,1/ n , . . . ,1/ n( )= n f 1/ n , 1/ n , . . . , 1/ n( )= n

f (x,y,z)=x+2y g(x,y,z)=x+y+z=1 h(x,y,z)=y2+z

2=4 f = 1,2,0 g= , ,

h= 0,2 y,2 z 1= 2= +2 y 0= +2 z y=12

= z y=1/(2 ) z= 1/(2 )

x+y+z=1 x=1 y2+z

2=4 =

1

2 21, 2 , 2( ) f 1, 2 , 2( )=1+2 2

f 1, 2 , 2( )=1 2 2

f (x,y,z)=3x y 3z g(x,y,z)=x+y z=0 h(x,y,z)=x2+2z

2=1 f = 3, 1, 3 g= , ,

h= 2 x,0,4 z( ) 3= +2 x 1= 3= +4 z = 1 z= 1 x=2

h(x,y,z)=142

+212

=1 = 6 z=1

6x=

2

6g(x,y,z)=0

y=3

6f f

63

,6

2,

66

=2 6

f63

,6

2,

66

= 2 6

f (x,y,z)=yz+xy,g(x,y,z)=xy=1 h(x,y,z)=y2+z

2=1 f = y,x+z,y g= y, x,0

If one of the variables is and the other two are not, then the squares of the two nonzero coordinates

are equal with common value and the corresponding value is .

Case 3: If exactly two of the variables are , then the third variable has value with corresponding

value of . Thus on , the maximum value of is and the minimum value is .

13. , , so

and . But , so the possible points are

. Thus the maximum value of is and the minimum

value is .

14. ,

, so and . But

, so for , , . Thus the maximum value of is

and the minimum value is .

15. , , , and

. Then , and so or , .

Thus implies and implies . Then the possible points are

and the maximum value is and the minimum value is .

16. , , , , . Then , and , so , , . Thus

implies or , so ; ; and implies

. Hence the maximum of subject to the constraints is and

the minimum is .

17. , , ,

4


h= 0,2 y,2 z y= y =1 x+z= x+2 y y=2 z =z/(2y)=y/(2y)

y2=z

2y

2+z

2=1 y=

1

2z=

1

2xy=1 x= 2

2,1

2,

1

22,

1

2,

1

2f

f 2,1

2,

1

2=

32

f 2,1

2,

1

2=

12

xy=1 f (y,z)=yz+1

y2+z

2=1

f (x,y)=2x2+3y

24x 5 f = 4x 4,6y = 0,0 x=1 y=0 1,0( )

f x2+y

2<16 g(x,y)=x

2+y

2=16 g= 2 x,2 y

6y=2 y y=0 =3 y=0 x= 4 =3 4x 4=2 x x= 2 y= 2 3f (1,0)= 7 f (4,0)=11 f ( 4,0)=43 f 2, 2 3( )=47 f (x,y)

x2+y

216 f 2, 2 3( )=47 f (1,0)= 7

f (x,y)=exy

fx= ye

xyf

y= xe

xy

0,0( ) f (0,0)=1

g(x,y)=x2+4y

2=1 g= 2 x,8 y f = g ye

xy=2 x xe

xy=8 y

exy

= 2 x/y x( 2 x/y)=8 y x2=4y

2

x2+4y

2=1 x=

1

2y=

1

2 2

f1

2,

1

2 2=e

1/41.284 f

1

2,

1

2 2=e

1/40.779

f (x,y)=3.7 f (x,y)=3503.7 350

f (x,y) (x 3)2+(y 3)

2=9

g(x,y)=(x 3)2+(y 3)

2f

x(x,y)=3x

2+3y f

yx,y( )=3y

2+3x g

x(x,y)=2x 6

gy(x,y)=2y 6

g(x,y)=(x 3)2+(y 3)

2=9 f

x= g

xf

y= g

y

(x,y)= 332

2,332

2 0.879,0.879( )

. Then implies , and . Thus or

, and so implies , . Then implies and the possible

points are , . Hence the maximum of subject to the

constraints is and the minimum is .

Note: Since is one of the constraints we could have solved the problem by solving

subject to .

18. , . Thus is the only critical point

of , and it lies in the region . On the boundary, , so either or . If , then ; if , then and .

Now , , , and . Thus the maximum value of

on the disk is , and the minimum value is .

19. . For the interior of the region, we find the critical points: , ,

so the only critical point is , and . For the boundary, we use Lagrange multipliers.

, so setting we get and . The

first of these gives , and then the second gives . Solving this

last equation with the constraint gives and . Now

and . The former are the

maxima on the region and the latter are the minima.

20. (a) The graphs of and seem to be tangentto the circle, and so and are the approximate minimum and

maximum values of the function subject to the constraint .

(b) Let . We calculate , , , and

,

and use a CAS to search for solutions to the equations , , and

.The solutions are and

5


x,y( )= 3+32

2,3+32

2 5.121,5.121( )

f 332

2,332

2 =3512

2432

2 3.673

f 3+32

2,3+32

2 =3512

+2432

2 347.33

P(L,K)=bL K1

g(L,K)=mL+nK=p P= bL1K

1,(1 )bL K g= m, n

b(K/L)1

= m (1 )b(L/K) = n mL+nK=p b(K/L)1

/m=(1 )b(L/K) /n

n / m(1 ) =(L/K) (L/K)1

L=Kn /[m(1 )] mL+nK=p K=(1 )p/nL= p/m

C(L,K)=mL+nK g(L,K)=bL K1

=Q C= m,n g= bL1K

1, (1 )bL K

mb

LK

1=

n(1 )b

KL

bL K1

=Qn

m(1 )=

LK

1 LK

L=Kn

m(1 )b

Knm(1 )

K1

=Q

K=Q

b n / m(1 )( )=

Qm (1 )

bnL=

Qm1(1 )

1

bn1 1

=Qn

1 1

bm1

(1 )1

x y f (x,y)=xy g(x,y)=2x+2y=p f (x,y)= y,x

g= 2 ,2 =12

y=12

x x=y

14

p

f (x,y,z)=s(s x)(s y)(s z) g(x,y,z)=x+y+z f = s(s y)(s z), s(s x)(s z), s(s x)(s y)g= , , (s y)(s z)=(s x)(s z) (s x)(s z)=(s x)(s y) x=y

y=z x=y=z=p/3

. These give

and

, in accordance with part (a).

21. , , .

Then and and , so or

or . Substituting into gives and for the maximum production.

22. , , .

Then and

and so .

Hence and

minimizes cost.

23. Let the sides of the rectangle be and . Then , ,

. Then implies and the rectangle with maximum area is a square with

side length .

24. Let , . Then , . \ Thus (1) and (2) . (1) implies while(2) implies , so and the triangle with maximum area is equilateral.

6


f (x,y,z)=d2=(x 2)

2+(y 1)

2+(z+1)

2f

g(x,y,z)=x+y z=1 f = g 2(x 2),2(y 1),2(z+1) = 1,1, 1 x=( +4)/2 y=( +2)/2

z= ( +2)/2+42

++22

++22

=1 3 +8=2 = 2

x=1 y=0 z=0

d= (1 2)2+(0 1)

2+(0+1)

2= 3

f (x,y,z)=d2=(x 1)

2+(y 2)

2+(z 3)

2f

g(x,y,z)=x y+z=4 f = g 2(x 1),2(y 2),2(z 3) = 1, 1,1 x=( +2)/2 y=(4 )/2

z=( +6)/2+22

42

++62

=4 =43

x=53

y=43

z=113

1,2,3( )53

,43

,113

f (x,y,z)=x2+y

2+z

2g(x,y,z)=z

2xy 1=0 f = 2x,2y,2z = g= y, x,2 z 2z=2 z

z=0 =1 z=0 g(x,y,z)=1 xy= 1 x= 1/y 2x= y 2y= x

=2/y2=2y

2y= 1 x= 1 =1 2x= y 2y= x x=y=0 z= 1

1, 1,0( ) 0,0, 1( ) f f (0,0, 1)=10,0, 1( )

f (x,y,z)=x2+y

2+z

2g(x,y,z)=x

2y

2z=1 f = 2x,2y,2z = g= 2 xy

2z,2 x

2yz, x

2y

2

y2z=1 x

2z=1 x

2y

2=2z y

2z=x

2z x= y 2z/1= x

2y

2/ x

2z( ) 2z

2=y

2

y= 2 z x2y

2z=1 z>0 4z

5=1 ( 2

1/10, 2

1/10,2

2/5)

f (x,y,z)=xyz g(x,y,z)=x+y+z=100 f = yz,xz,xy = g= , , =yz=xz=xy

x=y=z=100

3

f (x,y,z)=xay

bzc

g(x,y,z)=x+y+z=100

f = axa 1

ybzc,bx

ay

b 1zc,cx

ay

bzc 1

= g= , , =axa 1

ybzc=bx

ay

b 1zc=cx

ay

bzc 1

ayz=bxz=cxy x=ayb

z=cyb

ayb

+y+cyb

=100 y=100b

a+b+cx=

100aa+b+c

25. Let , then we want to minimize subject to the constraint . , so , ,

. Substituting into the constraint equation gives ,

so , , and . This must correspond to a minimum, so the shortest distance is

.

26. Let , then we want to minimize subject to the constraint . , so , ,

. Substituting into the constraint equation gives , so ,

, and . This must correspond to a minimum, so the point on the plane closest to the point

is .

27. , . Then implies or . If then implies or . Thus and

imply or , . If , then and imply , so . Hencethe possible points are , and the minimum value of is , so thepoints closest to the origin are .

28. , . Then

, and so and . Also so and

. But implies and . Thus the points are , and theminimum distance is attained at each of these.

29. , . Then implies

.

30. ,

. Then or

. Thus , , and implies that , and

7


z=100c

a+b+c

2x 2y 2z f (x,y,z)=8xyz g(x,y,z)=9x2+36y

2+4z

2=36

f = 8yz,8xz,8xy = g= 18 x,72 y,8 z 18 x=8yz 72 y=8xz 8 z=8xy x2=4y

2

z2=9y

236y

2+36y

2+36y

2=36 y=

1

3y>0

81

3

2

3

3

3=16 3

f (x,y,z)=8xyz g(x,y,z)=b2c

2x+a

2c

2y

2+a

2b

2z2=a

2b

2c

2

f = 8yz,8xz,8xy = g= 2 b2c

2x,2 a

2c

2y,2 a

2b

2z 4yz= b

2c

2x 4xz= a

2c

2y

4xy= a2b

2z =

4yz

b2c

2x

=4xz

a2c

2y

=4xy

a2b

2z

y

b2x

=x

a2y

z

c2y

=y

b2z

x=ayb

z=cyb

a2c

2y

2+c

2a

2y

2+a

2c

2y

2=a

2b

2c

2y=

b

3x=

a

3z=

c

3

8

3 3abc

f (x,y,z)=xyz g(x,y,z)=x+2y+3z=6 f = yz,xz,xy = g= ,2 ,3 =yz=12

xz=13

xy

x=2y z=23

y 2y+2y+2y=6 y=1 x=2 z=23

V =43

f (x,y,z)=xyz g(x,y,z)=xy+yz+xz=32 f = yz,xz,xy = g= (y+z), (x+z), (x+y)(y+z)=yz (x+z)=xz (x+y)=xy (y x)=z(y x) x=y =z

=z z(y+z)=yz z=0 x=y

(z y)=x(z y) y=z =x x x=y=z 3x2=32 x=y=z=

8

6f (x,y,z)=xyz g(x,y,z)=4(x+y+z)=c f = yz,xz,xy g= 4 ,4 ,4 4 =yz=xz=xy

x=y=z=112

c

C x,y,z( )=5xy+2xz+2yz g x,y,z( )=xyz=VC= 5y+2z,5x+2z,2x+2y = g= yz, xz, xy yz=5y+2z xz=5x+2z

xy=2 x+y( ) xyz=V z(y x)=5(y x) x=y =5/z z 0x=y x=y y(2x 5y)=2(2z 5y)

z=52

y52

y3=V x=y=

3 25

V

gives the maximum.

31. If the dimensions are , and , then and

. Thus , , so ,

and or ( ). Thus the volume of the largest such box is

.

32. ,

. Then , ,

imply or and . Thus , ,

and , or , , and the volume is .

33. , . Then

implies , . But so , , and the volume is .

34. , . Then (1) , (2) and (3) . And (1) minus (2) implies so or

. If , then (1) implies or which is false. Thus . Similarly (2) minus (3) implies

so or . As above, , so and or cm.

35. , , . Thus or

are the dimensions giving the maximum volume.

36. , . Then (1) , (2) , (3)

and (4) . Now (1) (2) implies , so or , but can’t be , so . Then twice (2) minus five times (3) together with implies which

gives . Hence and the dimensions which minimize cost are units,

8


z=V1/3 5

2

2/3

x y z

f (x,y,z)=xyz x,y,z>0 L= x2+y

2+z

2g(x,y,z)=x

2+y

2+z

2=L

2

f = g yz,xz,xy = 2x,2y,2z yz=2 x =yz2x

xz=2 y =xz2y

xy=2 z =xy2z

=yz2x

=xz2y

x2=y

2x=y =

yz2x

=xy2z

x=z

x2+x

2+x

2=L

2x

2=L

2/3 x=L/ 3=y=z

L/ 3( ) 3=L

3/ 3 3( )

x y z f (x,y,z)=xyzg(x,y,z)=xy+yz+xz=750 h(x,y,z)=x+y+z=50

f = yz,xz,xy = g+ h= (y+z), (x+z), (x+y) + , , yz= (y+z)+

xz= (x+z)+ xy= (x+y)+ x=y=z=503

xy+yz+xz=2500

3750 x x y x z

z y x( )= (y x) =z y(z x)= (z x) =y y=z= x+y+z=50

x=50 2 xy+yz+xz=750 x(2 )+2=750

50 2 =750

2

23

2100 +750=0 =

50 5 103

13

50 10 10( ),13

50 5 10( ),13

50 5 10( ) f

f13

50 10 3( ),13

50+5 10( ),13

50+5 10( ) =127

87,500 2500 10( )

f13

50+10 10( ),13

50 5 10( ),13

50 5 10( ) =127

87,500+2500 10( )y z

f (x,y,z)=x2+y

2+z

2

g(x,y,z)=x+y+2z=2 h(x,y,z)=x2+y

2z=0 f = 2x,2y,2z g= , ,2

h= 2 x,2 y, 2x= +2 x 2y= +2 y 2z=2 x+y+2z=2

x2+y

2z=0 2(x y)=2 (x y) x y =1

2z=2 1 =z+12

=1 =0 z+12

=0 z=12

units.

37. If the dimensions of the box are given by , , and , then we need to find the maximum value

of ( ) subject to the constraint or .

, so , , and

. Thus and

. Substituting into the constraint equation gives and the

maximum volume is .

38. Let the dimensions of the box be , , and , so its volume is , its surface area is and its total edge length is . Then

. So (1) , (2)

, and (3) . Notice that the box can’t be a cube or else but then

. Assume is the distinct side, that is, , . Then (1) minus (2) implies

or , and (1) minus (3) implies or . So and

implies ; also implies . Hence

or and , giving the points

. Thus the minimum of is

, and its

maximum is .

Note: If either or is the distinct side, then symmetry gives the same result.

39. We need to find the extreme values of subject to the two constraints

and . , and . Thus we need (1) , (2) , (3) , (4) ,

and (5) . From (1) and (2), , so if , . Putting this in (3) gives

or , but putting into (1) says . Hence or . Then (4) and (5)

9


x+y 3=0 x2+y

2+

12

=0

x=y 2x+2z=2 2x2

z=0 z=1 x z=2x2

2x2=1 x 2x

2+x 1=(2x 1)(x+1)=0 x=

12

x= 1

12

,12

,12

1, 1,2( ) f12

,12

,12

=34

f ( 1, 1,2)=612

,12

,12

1, 1,2( )

z=r 4rcos 3rsin +8z=5 z=r

4rcos 3rsin +8r=5 r=5

4cos 3sin +8z=r

=t r=z=5

4cos t 3sin t+80 t 2

f (x,y,z)=z

g(x,y,z)=4x 3y+8z=5 h(x,y,z)=x2+y

2z2=0 f = g+ h

0,0,1 = 4, 3,8 + 2x,2y, 2z 4 +2 x=0 x=2

3 +2 y=0 y=32

8 2 z=1 z=8 12

4x 3y+8z=5 x2+y

2=z

2

42

332

+88 12

=5 =39 8

102 2

+32

2=

8 12

216

2+9

2=(8 1)

239

216 +1=0

=113

=13

=113

=12

x=4

13y=

313

z=5

13=

13

=12

x=43

y=1 z=53

43

,1,53

become and . The last equation cannot be true, so this case gives no solution.

So we must have . Then (4) and (5) become and which imply and

. Thus or so or . The two points to check are

and : and . Thus is the point

on the ellipse nearest the origin and is the one farthest from the origin.

40. (a) Parametric equations for the ellipse are easiest to determine using cylindrical coordinates. Thecone is given by , and the plane is . Substituting into the plane

equation gives . Since on the ellipse, parametric

equations (in cylindrical coordinates) are , , .

(b) We need to find the extreme values of subject to the two constraints

and .

, so we need (1) , (2) ,

(3) , (4) , and (5) . Substituting (1), (2), and (3) into (4)

gives and into (5) gives

or . If then and , , . If then and ,

, . Thus the highest point on the ellipse is and the lowest point is

10


413

,3

13,

513

f (x,y,z)=yex z

g(x,y,z)=9x2+4y

2+36z

2=36 h(x,y,z)=xy+yz=1

f = g+ h yex z

,ex z

, yex z

= 18x,8y,72z + y,x+z,y yex z

=18 x+ y

ex z

=8 y+ (x+z) yex z

=72 z+ y 9x2+4y

2+36z

2=36 xy+yz=1

x y z

x 0.222444, y 2.157012, z 0.686049, 0.200401, 2.108584x 1.951921, y 0.545867, z 0.119973, 0.003141, 0.076238x 0.155142, y 0.904622, z 0.950293, 0.012447, 0.489938x 1.138731, y 1.768057, z 0.573138, 0.317141, 1.862675

f f (0.222444, 2.157012, 0.686049) 5.3506f ( 1.951921, 0.545867,0.119973) 0.0688 f (0.155142,0.904622,0.950293) 0.4084f (1.138731,1.768057, 0.573138) 9.7938 9.7938

5.3506

f (x,y,z)=x+y+z g(x,y,z)=x2

y2

z=0 h(x,y,z)=x2+z

2=4

f = g+ h 1,1,1 = 2x, 2y, 1 + 2x,0,2z 1=2 x+2 x 1= 2 y 1= +2 z

x2

y2=z x

2+z

2=4 x y z

x 1.652878, y 1.964194, z 1.126052, 0.254557, 0.557060x 1.502800, y 0.968872, z 1.319694, 0.516064, 0.183352x 0.992513, y 1.649677, z 1.736352, 0.303090, 0.200682x 1.895178, y 1.718347, z 0.638984, 0.290977, 0.554805

f f ( 1.652878, 1.964194, 1.126052) 4.7431f ( 1.502800,0.968872,1.319694) 0.7858 f ( 0.992513,1.649677, 1.736352) 1.0792f (1.895178,1.718347,0.638984) 4.2525 4.2525

4.7431

f x1,x

2, . . . ,x

n( )=n x1x

2x

n

g x1,x

2, . . . ,x

n( )=x1+x

2+ +x

n=c x

i>0 f

.

41. , , .

, so ,

, , , . Using a CAS to solve these 5equations simultaneously for , , , , and (in Maple, use the allvalues command), we get 4 realvalued solutions:

Substituting these values into gives , , ,

. Thus the maximum is approximately , and themininum is approximately .

42. , , . , so , , ,

, . Using a CAS to solve these 5 equations simultaneously for , , , , and ,we get 4 real valued solutions:

Substituting these values into gives , , ,

. Thus the maximum is approximately , and themininum is approximately .

43. (a) We wish to maximize subject to

and .

11


=1n

x1x

2x

n( )1

n1

x2

xn( ),

1n

x1x

2x

n( )1

n1

x1x

3x

n( ), . . . ,1n

x1x

2x

n( )1

n1

x1

xn 1( )

g= , , . . . ,

1n

x1x

2x

n( )1

n1

x2

xn( )= x

1/n

1x

1/n

2x

1/n

n=n x

1

1n

x1x

2x

n( )1

n1

x1x

3x

n( )= x1/n

1x

1/n

2x

1/n

n=n x

2

1n

x1x

2x

n( )1

n1

x1

xn 1( )= x

1/n

1x

1/n

2x

1/n

n=n x

n

n x1=n x

2= =n x

n0 x

i>0

x1=x

2= =x

nx

1+x

2+ +x

n=c nx

1=c x

1=

cn

=x2=x

3= =x

n

fcn

,cn

, . . . ,cn

x1,x

2, . . . ,x

n( )f f

fcn

,cn

, . . . ,cn

=n c

ncn

cn

=cn

cn

f f x1,x

2, . . . ,x

n( )=n x1x

2x

n

cn

x1+x

2+ +x

n=c n x

1x

2x

n

x1+x

2+ +x

n

nf

cn

cn

,cn

, . . . ,cn

x1=x

2=x

3= =x

n=

cn

f x1, . . . ,x

n,y

1, . . . ,y

n( )=n

i =1x

iy

ig x

1, . . . ,x

n( )=n

i =1x

2

ih x

1, . . . ,x

n( )=n

i =1y

2

i

f =n

i =1x

iy

i= y

1,y

2, . . . ,y

n,x

1,x

2, . . . ,x

ng=

n

i =1x

2

i= 2x

1,2x

2, . . . ,2x

n,0,0, . . . ,0

and , so we need to solve the system of equations

This implies . Note , otherwise we can’t have all . Thus

. But . Then the only point where

can have an extreme value is . Since we can choose values for that

make as close to zero (but not equal) as we like, has no minimum value. Thus the maximum

value is .

(b) From part (a), is the maximum value of . Thus . But

, so . These two means are equal when attains its

maximum value , but this can occur only at the point we found in part (a). So

the means are equal only when .

44. (a) Let , , and . Then

, and

12


h=n

i =1y

2

i= 0,0, . . . ,0,2y

1,2y

2, . . . ,2y

nf = g+ h y

i=2 x

ix

i=2 y

i1 i n

1=n

i =1y

2

i=

n

i =14

2x

2

i=4

2 n

i =1x

2

i=4

2=

12

=12

yi=2

12

xi=x

i1 i n

n

i =1x

iy

i=

n

i =1x

2

i=1 =

12

yi= x

i

n

i =1x

iy

i= 1 =

12

yi= x

i1 i n

n

i =1x

iy

i= 1

n

i =1x

iy

i1

n

i =1a

2

i0

n

i =1b

2

i0

n

i =1a

2

i=0 a

i=0

xi=

ai

a2

j

x2

i=

a2

i

a2

j

=1 yi=

bi

b2

j

y2

i=

b2

i

b2

j

=1

xiy

i=

aib

i

a2

jb

2

j

1 aib

ia

2

jb

2

j

. So and , .

Then .

If then , . Thus . Similarly if we get

and . Similarly we get giving , , and . Thus

the maximum value of is .

(b) Here we assume and . (If , then each and so the inequality

is trivially true.) , and . Therefore, from

(a), .

13


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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives ...y1zhao/2009Math231/Sec1508Solution.pdf · f f g(x,y)=8 f(x,y)=c c g(x,y)=8 c=59 c g(x,y)=8 c=30 f g(x,y)=8 59 30 c= 1