Graphics Reference
In-Depth Information
Let us now restate our new extremum problem as follows:
The constrained extremum problem:
Find the extrema of a function f(x
1
,...,x
n
) subject
to a constraint g(x
1
,...,x
n
) = 0.
Note first that the set
Z
= g
-1
(0) of zeros of g is intuitively an (n - 1)-dimensional
space with unique normal line. Therefore, Propositions 4.5.7 and 4.5.8 suggest that
—f and —g must be parallel on
Z
since they are vectors that are both perpendicular to
all curves in
Z
. This observation leads to the following:
The method of Lagrange multipliers:
Find the extrema of f subject to a con-
straint g by solving the equation
(
)
=
—-
f
l
0.
for l.
We shall demonstrate the use of the Lagrange method with two examples.
Solution to Example 4.5.6.
If
(
)
(
)
=-
2
2
Fxy
,
xy
l
x
+-
y
1
,
then
∂
∂
F
x
∂
∂
F
y
=-
y
2
l
x
and
=-
xy
2
l
.
Setting ∂F/∂x to zero, implies l=y/(2x). If we substitute this l into the equation
∂F/∂y = 0 and solve for x, then we will get that x =±y. Now substitute ±y for x in the
constraint equation and solve for y. It follows that local extrema occur at
1
2
1
2
Ê
Ë
ˆ
¯
(
)
=±
xy
,
,
±
.
4.5.9. Example.
To show that of all the triangles inscribed in a fixed circle, the equi-
lateral triangle has the largest perimeter.
Solution.
See Figure 4.15. The length of the side subtended by angle a is 2R sin (a/2).
A similar formula holds for the other sides. Therefore,
a
b
g
Ê
Ë
ˆ
¯
(
)
=
f
abg
, ,
2
R
sin
+
sin
+
sin
2
2
2
is the formula for the perimeter with the constraint
(
)
=++- 20
g abg
,,
a b
g
p
.
