14.8: LAGRANGE MULTIPLIERS
KIAM HEONG KWA
1.
Method of Lagrange Multipliers
Let
f
(
x,y,z
) and
g
(
x,y,z
) be continuously diﬀerentiable in the sense
that all their ﬁrst partial derivatives exist and are continuous.
If
f
(
x
0
,y
0
,z
0
) is a local extreme value of
f
on the level surface
S
de-
scribed by
g
(
x,y,z
) =
k
and
(1.1)
∇
g
(
x
0
,y
0
,z
0
)
6
=
0
,
then
(1.2)
∇
f
(
x
0
,y
0
,z
0
) =
λ
∇
g
(
x
0
,y
0
,z
0
)
for a real number
λ
that we refer to as a Lagrange multiplier
. In
other words, at the point (
x
0
,y
0
,z
0
), the vectors
∇
f
(
x
0
,y
0
,z
0
) and
∇
g
(
x
0
,y
0
,z
0
) are parallel.
The idea of a proof is this: if
f
(
x
0
,y
0
,z
0
) is a local extreme value
of
f
on the surface
S
, then it is also a local extreme value of
f
when
f
is evaluated along any smooth curve on
S
that passes through the
point (
x
0
,y
0
,z
0
). To make this precise, let
r
(
t
) =
h
x
(
t
)
,y
(
t
)
,z
(
t
)
i
be a
smooth curve that lies on
S
and passes through (
x
0
,y
0
,z
0
) when
t
=
t
0
.
Then the function
F
(
t
) =
f
(
r
(
t
)) =
f
(
x
(
t
)
,y
(
t
)
,z
(
t
))
gives the value of the function
f
along the curve
r
(
t
) and has a local
extreme value when
t
=
t
0
. Furthermore,
F
is diﬀerentiable since so
are
f
and
r
(
t
). Thanks to this, we can apply the chain rule to obtain
F
0
(
t
) =
f
x
(
x
(
t
)
,y
(
t
)
,z
(
t
))
x
0
(
t
) +
f
y
(
x
(
t
)
,y
(
t
)
,z
(
t
))
y
0
(
t
)
+
f
z
(
x
(
t
)
,y
(
t
)
,z
(
t
))
z
0
(
t
)
=
∇
f
(
r
(
t
))
·
r
0
(
t
)
.
Date
: October 2, 2010.
1