Stewart Calculus ET 5e 0534393217;14. Partial Derivatives...
Transcript of Stewart Calculus ET 5e 0534393217;14. Partial Derivatives...
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
5. , , . Then implies or
. 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 .
6. , , . Then implies or
. 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
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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. , , ,
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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.
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
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Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
=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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers
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
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; 14.8 Lagrange Multipliers