I
L-ag~tipierscch.lk 8)
.
In Ch 14.7 we learned how to
.
and
maxima of
the domain of
f
a
function f fax g) in
goal
find minima
-
.
or in a c closed)
set in the domain
The main
find minima
of f
bounded
.
is to learn home to
here
and maxima
of
a
function
along specified curves ehthe
domain off This is called
minimizing
or
maximizing f subjectto constraint
of f ex
=
)
y
.
'
.
a
⑦ Extremeing subject to
a
constraint
The main idea
Assume that
ant to find the
wee un
extreme
values of ffax y) subject to
.
a
constraint of the form gcx yl
.
In
other
words ,
we
-
k
.
seek the extreme
2
values
is
of f fix g)
-
when the
restricted tolie on the lead
gCx y) = th
,
fee
,
yl
÷÷i÷sg÷*÷¥"
c
st
-
.
the
curve
fix g)
.
gcxcy)
picture that
curves
i c
" "" "
.
have to find the largest value of
the level curve
the level
ourye
the level curves
T
we
,
.
←
So
point ex y)
this
just touch
=
k
=
.
appears from
when these
It
happens
each
intersects
c
other
,
that is
,
3
hear
we
theyhave
a
common
otherwise the value
tangent line
of e could be
increased
further
point say
(
)
At such
.
a
,
.
exo . so)
line
.
the same normal
so
Ofcx
or
they have
,
o ,
yo )
ll
Ogexo
yo)
,
equivalently
,
,
D
f exo
y o)
e
2
=
Og Ko go)
,
Lagrange
multiplier
for
some
a
ECR called
.
we
note that the
same
argument applies
problem of finding the extreme
values of fcx y z)
subject to the
to the
constraint
k Instead of level
gcx
,
,
y,t ) =
,
curries
argue
,
in
we
.
consider level
the same
Ofcxocyo.to)
=
surfaces
and
wayto reach
d
OgCio fo
i
ft o)
.
4
procedure based
above equation is as follows :
To,
conclude
,
the
on
the
of Lagrange Multipliers
find the extreme values of foxest)
Method
To
,
subject to
(
the constraint gcx y 2-I - k
,
.
assumingthat these extreme values exist
Ogfo along the surface §Cx g.Zt k) :
step find all values of x,y t and a S f
and
,
:
⑦{
-
,
Of ex sit)
,
L gc x y
,
steps
2-I
a
=
Ogex y iz)
,
k
Evaluate f at all the
Cag 2-I
.
,
=
found
in
.
step i
.
points
The
largest
of these values is the maximum value
of f ; the smallest isthe minimum
value
of f
wee
note that if
wee carrife
equation by components
the vector
,
⑨ becomes
5
fx
a
=
gx , fy
a
-
gy , fz =
a
ga
,
g- k
equation with 4 unknowns
Also , for functions of a variables
⑨ is replaced by
so we
get
4
.
,
of
f.
g
Example
:
k
=
Ss
g
-
-
=
a
gx
asy
be
find the extreme values
Lex
on
{
09
-
L
fx
,
?
't ay
y) =
the circle
of
x
't 5--1
x
.
-
gCx i S)
¥
me
first
4.
Lx
①
Usingthe
=
a
I
method
ofLagrange multipliers
solve
gx
⇐
s.
×- o
¥
①
o
⇐ gas
I ⇐
t't
.
,
82=1
③
y, ±,
x=±#
6
f
has
(Oil)
CO
,
f coil)
f
possible
Clio)
-
-
,
-
)
l
,
Cl
O)
,
floc
Z
,
I
extreme values at
e
-
f
C
-
e
,
-
,
C- 1,0)
al
← max
2-
-
ol =
:
fcoite)
wnI off : fcti
L
.
win
x2ty2=1
2
01=1
Ma
,
Xc
,
-
^gz=x2t2y2
#•
y
1. ←
-
maxeoff
2-
.
s
L
.
-
7-
Example
:
Lex y)
,
on
SI
=
{
't y?
is
I 1
X
ex ,y)
't
x
:
'
y
}
E e
continuous
f attains its
in D
.
is
a
R2
closed set in (
bounded
f
of
x2 tag
=
the disk
D
extreme values
find the
abs
on
is
and win
max
.
.
/
.
aiticdpoiu
Is:
Lc
point
o, ol = o
.
←
win
-
Next ,
me
in:i
⇐
from the previous example
know that
Max
.
Max
value of f ord d
'
.
-
fc o I )
e
,
•
2
=
-
win
value
.
-
In
of f
conclusion
ondd :
fc It
,
ol
=
l
-
8
maxe of f or D
f co it c)
D:
wiue of f
f co ol
:
on
Ee5=
,
'
\
←
/
m•a×
the
=
=
2
o
.
geometry
behind the method :
✓
min
the extreme values
min
off correspond
•
•
to the level curves
that touch the
r
mat
I
find the
Eixample
circle
points
on
the
Etf
-
-
e
.
sphere
I get 2-2=4
x
that
are
point
Soli
closest to and
(3 ,i
We
from
a
to the
d
=
.
-
)
farthest from
the
e
.
have to
point
extoeuiize the distance
cxis ,Z)
on
the sphere
point Cs e i ) that is
T
t (Zf c)2
(
332 t Cy )
,
x
-
.
-
,
-
n
Z
g
or
,
equivalently find the
of
lues
Lex get D=
Cx
-
,
subject to
Ski's ,Z)
pliers
,
He
-
3ft
Cy
-
: =
aa
-
RtcZf 112
'
the
←
xIgIz2=
method
Usingthe
{
extreme
,
4
pants are
on
the
of Lagrange
sphere
multi
-
first solve
÷÷÷÷⇐÷÷÷
Lx
-
;
22×2+82+72=4
agx
2 (2- te)
⇐
/
/
x ee
-
S (e
a)
-
2-
Ca
'
¥
=3
a) = I
-
a) =
-
x
2-
get
① ② and
,
-
3-
①
c- a
1×2 t y
't 2-2=4 ④
Plugging none
-
y
I
2
=
=
=
1-
②
e- a
-
1i
-
③
a
③ into ④
me
.
'°
c¥ptc÷t
u ca
ii.
e
-
If
a
If
a
=
=
i
-
at
I
-
a = I*
5=41
a
Ez
i
=
x
-
I
¥
fu is Fair
-
i
i
?÷
=
15
=
3¥
-
4
-
x
-
-
fi
,
5- -2€
-
-
to
Tf
←
,
tf
-
Fe
ti
-
-
=
=
In
i
3£
+3¥
,
-
5t
.
2€
-
,
,
-
t St
conclusion :
-
win
'
←
z
,
tf ¥ ti
⇐ E) EE -354¥
Er
⇐
is't Eat it
-
S l
-
'
Life ⇐ Ea) CE -354¥
,
'
a- as
.
e- a
II
Iz
l
4
=
F
wax
t
#
ti
-
liter HI
.
tf tf t I
,
"
theepoint is Cfa
the
fenpoint is
C
-
.
If
i
-
¥)
.
¥ ⇐ Za)
,
-
.
.
Az
kx
•
I
7
⇐ Er
'
,
i
-
EI
,
I2
⑤ Extremezing subject totwo constraints
The
main
idea
Assume that
values of
findthe extreme
subject to Tono
we want
f Six y z)
to
constraints
and
of the form gcx
-
,
,
,
he x cy,ZI
=
C
.
In other words
values
y ,zk k
,
we
seek the
extreme
of f fcxiy z) carhen the point
-
,
restricted to lie
Cx ,y H
.
intersection
surfaces
of
is
C
the level
on
the
curve
g-
of
k and
h
=
e
.
13
Suppose that f has such
some
point P C
Xo .go ,Zo)
Of
exo
yo
e
,
Zo)
Then
.
extrema at
an
.
L C
but also
Og exo yo Zo) t C
,
so
,
Oh cxocyo
.
of cxocyoezo) is inthe plane determined
by Ogexo yo Zo)
,
,
and Oh
Cxo ,yo ,Zo) Cassa
.
wing that these two vectors
In
are
,
=
a
Ogexo yo fo)t y 0h (Xo Yo Zo)
,
.
in this case the method
multipliers
consists in
,
of Lagrange
looking for
extreme
values by
solving
Of
=
Ogt p oh
a
a
E
:
fz
(
=
a
g- k
h
=
C
-
and X)
to
other beards
Ofcxocyo Zo)
so
Zo) IC
,
gz t yht
14
evaluate f at the solution
spot the largest
and
values
Exempt
function
,
if
any
smallest
and
.
find the maximum
fcx.yi.tk
value of the
dy t ?Z
xt
-
We
so wee
Of
g
=
-
i
-
.
,
first solve
a
,
og f
y oh
he
i
⇐
agxtyhx
¥÷÷¥¥i
f- x
=
method
the
use
of Lagrange multipliers
{
the
.
h
Sod
on
of intersection of the plane x y t z
T
the cylinder x7g2= 1
-
curve
and
points
:
/
a
1
I5
f
e
at
-
2=
-
y
a t
13-
&
X
-
y
=
①
2x
y 2y
②③
ayy
.
x=
z
=
-
=
5
y
-
f⑥
Ey
-
②
③
5h
2ft
⑤
I
-
\
y
-
y
'
i
¥s
FI
=
x
:
=
-
et
Roze
Egg)
x
:
'
-
Eg
=
'
-
is
-
z
=
's
-
-
¥g
i
z= l
Egg
-
i
i t
Fas
s-
,
t 3t
Efg
Eg
3tTg ← wax
-
3
the wax value
3t
-
Ffg
Ing tf
-
-
In conclusion
=
Eg
a
=
Eg) Fes Test 3
t
=
curve
=
⑤
=
fl Fas
fat Fff
⑤
f- C Eg Eg
given
y
y t 2- = I ④
t
'
2x
.
.
Tvs
.
-
← win
off
Ing
.
on
the
16
Exercises :
-
Each
13-147
of these extreme value
problems
has
solution mishboth
a
value and
a
Max
a
min value
-
.
.
Lagrange
Use
multipliers to find the extreme values
of the
function subject to the given
constraint
gcxcg)
.
④ fax
me
legs
f
g
③
,
y) =3Xt y
the
use
pliers
x
→
a
=
-
-
=
method of
so we
,
X2tg2=
,
{
gx
co
¥5②
?
Iy
x
=
f
-
\
a
a
-
=
=
I
-
co
⑨⑤
z④
-
.
①
④
¥ Faz t Jaz
multi
Lagrange
first solve
as.
22
co
x
E①
-
-
⑤
.
I
#
X =3
x-
-
3
=
to
,
y
,
=
yi
-
i
I
17
yes
LC
-
,
)
=3 3 t t
e
3,
-
.
al
-
So
Max
win
3
-
.
.
.
C-3)
yalue
=
-
Max
←
'S
fc3
off :
,
=
e) =
.
to
lo
of f : f C -3 l)
the Ftl) t lucy't l) Huettl)
value
④ fcx.ge-21
win
←
I
=
-
-
.
.
yet 2-2=12
XZ t
-
g Cx is it)
He
f.
fx
If
a
-
on
=
-
a
ga
z t
①② ②
o
L
-→
o #z
Casey if
.
2x
①
→ ②
12
a-
multi
first solve
dgx
-
fz
g
so we
.
of Lagrange
the method
use
pliers
.
*
x't
x
=
=
a
.
2Z
③
It 2-Zaz ④
y= 7=0
impossible
for y fo and 2- to
⑤→
IET
①② ③
-2
¥2
¥2 ¥zz
:
=
=
=
-
18
-
it x
x
x'
=
the
y
y
=
72
E
3x
52=2-2--4
possible points for
sa
Z
a
x
values of f
( I2
BUT
It
=
z2⑤ €5
'
2
=
'
i t
=
.
of the
one
I2
I
,
form
← 8
2)
extreme
points
the same value at these
f has
points namely
.
LC t 2
,
I 2, I
2)
Casezi if
leet's
say
'
y
-
ZZ
=
3In 5
-
exactly
x=
of 4.83
← what
o
of
one
¥÷¥
¥
To
z
a
y7- a 2
=
X
,
y,Z
is Zero
etty ettzz
←
y2=6
To
possible
points for extreme values of f
y- I
z
-
±
the
are
co
,
± To
,
I 56
similarly , other
( ± of o , I 56
Eu points
,
)
←
a
points
possible points are
)
,
C ITG I To
.
E 4
)
co
points
,
19
f has the same value
points namely
atthese
But
e2
,
f
co
,
=
It
f(
.
I
56)
f CITf
s
ITG , ITG) =
say
values of
co
s
similarly
o
,
,
f
o
,
z
)
at all these
6
In
is
It)
soIn 5 ,
the
2-
the
are
Zero
,
extreme
points
← 22
T
c
2
O ,o
are
)
.
points
the same value
has
points namely
,
=
f
f CI R2 ,
conclusion
,
2-e- ± Q2
C ± T2
,
points
,
y
possible points
other
f
=
x.
iz
I TE )
But again
f- C O , o
.
are
I Tiz , o
E
I Tf )
-
possible points for
the
C
-27
x - y- o
,
2 In 7- re 3.89
o c
Case If exactly 2 of
,o
,
C
o,
l
oco
the
I Tiz ,
=
)
o
lU 13
=
win
d
I 2.56
-
maxesof f
wiue off
is 1h13
.
.
20
⑤ the method of
multipliers
exist
extreme
Lagrange
values
that the
assumes
BUT that is NOT
,
always the case Shove
the cuinimuwi
that the problem of
finding
?
value of fex y) x7y subject to the
-
,
constraint
.
-
x
y
-
I
be solved using
can
Lagrange multipliers but f does NOT
,
have
maximum value with that
constraint
a
.
with gex ,y) =
x
y
apply the method
me
of Lagrange multipliers
Is
f
x
=
a
÷
gx
s.
I I
-
From ⑦
me
a
②
5=5
see
%
①
③
x
-
x
o
-
2£ ⑤
=
a
:
⑥
-
.
is to
④
El ⑦
x
s -x
we
a
⇐
\ y=
that
solve
and
x
③
-
2=1
*
impossible
hate the
21
following candidate points for extreme
C
-
fc
I
-
s
C
,
l
=L , f C
)
ne
-
Clearly
there
,
x
large
so
ie e)
-
is
or
y
is
the
117-202
,
.
=L
NO maximum Value
f
)
e
can
fC
=
I ⇐ y
-
to become
In confusion
fce
)
e
=
allows
,
e
constraint x y
the
since
l)
i
-
,
arbitrarily
be made arbitrarily
-
l)
=
1
minimum value
Find the
I
off
.
extreme values of
f
subject to both constraints hey y z ,
⑧ fcxcy H Z ; x't y?Ed ; xtTE= 24
,
,
-
.
*
xZe
2-ZZ
y
o
-
gCx ,y t )
.
Hlith
=
x
t
g
Cx ,y , 2- I
y t 2-
,
we
'
=
XZ t y
-
ZZ , hex.y,Hi
apply the meted of
Lagrange multipliers
and
solve
.
22
first
{ fy
fx
-
gxtyhx
>
gytyhy
>
=
=
I:÷÷÷¥÷:
a 2x t
y
d. 2's t
y
-
"
×2+y2
x
-
l
①
l
②
"
so
-
.
2-2=0
⑨
⑤
test 7--24
①
⇐
✓
y
a-
\
=
⑥
2×2
✓
\
-
x
22cg
say
-
-
-
=
24
=
-
2x
22=2×2
96
x
96 X
48
i
=
o
⑧
→ 2-
&
XZ
=
4=0
⑤
x
576
⇐
za x
0£08
⑦
ya
-
-
x +
t
t
(
impossible
24-2×172×2
(
4×2=2×2
576=0
288
=
2- =
o
=
-
24-2482
48+755-+24+1252 ⑤ JF⑧
2
x
x) so
-
h8-r
=
24
-
12 Fr
⑦
=
z
Z-
-
241-24 Tz
y
d)⑧
23
So
- me
have
2
the extreme values
C24 tI252 , 24 ti2 Fe
C 24
F
-
,
24
C 2 h t i252
=
f
T2
l2
C 24
=
24
-
-
-
12
,
1252
Tz T
T2 , 24
24
,
-
24 Fz)
-24
T2
-
,
I2T2
,
24
-
24
←
.
←
)
.
Tz) =
win
Zh t 2 4
-
9 gy
E
,
t 2 E TE
-57.54
-
24 t 2 4 52
for
of f
-
,
2E t 12
24
-
-
point
candidate
Yafue
52)
i
hee
wax 6 a
.
-
④ f cx
Xt
y itI
=
y Z
=
,
-
-
3x
-
o
y
;
SCx i ] ,Z)
-
?
x
3 Z 's
t
22-2
=
-
h C X e Tu2-I
1
.
applythe method of Lagrange
multipliers
thus
we
and
solve
:
÷÷÷÷÷÷÷÷÷÷÷÷÷÷÷
⑤
"
②
⑤
a=
y
=
⑤
=
56
-
-
x
EES
ro
f- CE
Ll
y
③\
8¥ tf tf
- y
\
-
Y
-
.
=
Xi
=
Z
y
=
I
¥
,
Zz
z-
,
Fei ¥) if
-
Fa Fo TE)
i
,
121-2-37 find the
=
-
-
Z
x
y)
,
?
ty
-
for
{
D is
its
ex, y)
⑤
ily ⑧
-
us
€6
=
←
¥ -2¥
=
-1oz
Max
←
.
win
.
value
.
f
on
inequality
It
.
"
4×-5
X
16
:
.
value
extreme values of
x
-
Let's denote
D=
-
=
-
'
2×2+3y?
I
2-=
I
the region described by the
④ fcx
-214 ⑦
42=6
⑤
tf y Ze
z-
-
-
x
X-2
Rt GZ
closed
and
boundary is
E
h
}
16
bounded
>
4
-
4
and
"
f
OD=
f
ex
,
y)
I 's 16
y
continuous
is
its abs
.
points ofD
abs win at
.
.
4×-4
i
69
y=
,
)i
o
Next
,
OD=
point
4
-
.
-
5
=
-
f
←
min
•
we
values
He
2 to
ED
sis'm:*:i
⇐
f Cl
some
.
ariticelpoiutsfxf
Hs's
=
f attains
on D
and
max
}
x
:
have to
of f
on
{
:
Cxcy
)
find the extreme
the
Z
x
t
boundary
'
y
=
-
}
16
gexcy)
use
HE
the method of Lagrange multipliers :
o "
② l:
2×2
7③
y
t
16
26
①
x
ca
-
a)
=
②
y es a)
④
x
-
x
.
C
2
3)
y= I 253
So
points for
C h
-
-
4
fC 4
f
.
,
) , C 4 , o)
)
O
=
)
=
I
=
4
x =
z
? s2
y
2
-
following candidate
C 2
-
2 I 6 to
.
.
⑤
.
,
2 I6 t
⑨
o
a =3
the extrema
o
o
.
=
have the
me
-
\
x
-
T-
-
= o
716
⑤
fC
2
O
-
-
,
of f
253)
h C h)
-
4
.
4
-
-
5
on
8D
C-2 ,-2Fs)
,
5=433
=
wax
ll
de
-
C- 2
,
I 253)
=
2 4 t 3 12
.
.
-
4 C 2) 5
.
-
-
In conclusion the extreme values
,
f.
on
:
D
Max
win
.
.
of
are :
value
:
value
:
④ Find the
fC
-
2
,
-447J
07=1-72
±2B)
f ee
maximum
,
and minimum
volumes
of rectangular box whose sur
a
-
"
face
area is 1500
total edge Penghu
Let
denote by
us
and 2-
Cwf and whose
is 200
its
}z
.
f
¥1
--
-
-
-
,
nolame :
-
.
X
,
g.2-I
Xyz
=
its surface
area
gcx yet) 2x T t 2x Zt ZYZ
its total edge length
-
,
•
.
.
the dimensions of
Six
•
.
!
x. y
the box and :
•
.cm
=
1500
.
he x y
,
So
,
Z)
4 Xt
=
4Gt 4 2-
=
zoo
.
have to find the extreme
values
off sing Lagrange multipliers
we
*
gxtyhx
Ly ingy t y try
+ Tht
fx
-
>
-
I
1
'
.
:
2
4
tf 4
-
a .C2yt2Z)
=
at 2x t 27)
:
①
ty 420
.
anxiety
XY t X
X t
Z t y 2- = 750 ④
y
tZ
=
50
⑤
28
First
of
all
cannot be
then
let's note
,
cube
a
⑤
:
x-
2530L
=
Assume that
①
-
①
-
②
z
③
¥5
¥
⑤
④
x
=
x
50
-
-
y
since if
2-
=
x
-
CZ
f y
)
x
-
so
2> +
a
-
-
cy
at
-
a ez
€5
.
)
a - 2- ⑥
)
a
x
-
.
x
=y⑦
⑧
za
-
5013
X # 2-
,
=
x)
=
x- y- Z
impossible
750
cy
y
,
that the box
7-so
⑨
x=
22=750-232
2a
too
a
3 a?
•
2
=
-
422--750
100
a t
750
5045M£ ¥
×
=
⑥y
⑦
or
-
-
22
=
0
.
So
:
50-13050-1
2-
=
50 t 5
To
-
3
29
a
•
=
5035kg
s
x=
Soto
3
⑥
¥3 y
i
2-
50+5%3
.
f-(50-130%2 5otg# 50+512)
,
,
( 87500
2 Soo
:
-
t)
=
← win
-
f
(
507,0kg 50-53%-150-552 )
,
¥
=
( 8750 Ot
of
then
x
,
by
me
we
wax
consider that instead
,
is the
y or Z
the
To ) ←
-
Reina If
:
2500
=
distinct side ,
symmetry of fcx.y.tl
get the
same
result
.
=
Xy2-