Math 21a Handout on Lagrange Multipliers  Fall 2000
The principal purpose of this handout is to supply some additional examples of the Lagrange multiplier
method for solving constrained equations for three unknowns.
In this regard, remember that the basic
problem is to find the maxima, minima and/or stationary points of some given function
f
(
x
,
y
,
z
) on
R
3
subject to the constraint that
g
(
x
,
y
,
z
) = 0 where
g
is another function.
Note that these constrained problems typically arise in one of two ways: First, they can arise when
looking for the extreme points of a function in some region of space; for in this case a search must be made
for extreme points both inside the region and on its boundary.
In particular, the problem of finding the
extreme points on the boundary can be phrased as a constrained extremal problem.
For example, imagine
that the region in question is the solid Earth, and the function in question gives the temperature at each
point.
Then, a search for extreme points must look for interior extreme points and also extreme points on
the surface.
Constrained extremal problems also arise when looking for the stationary points of a given function’s
restriction to some given surface in
R
3
. For example, take the surface in question to be the surface of the
earth, and consider finding the extreme points of the function whose values measure a year’s average of
sun light intensity at each point.
Anyway, what follows are some sample constrained problems.
Example 1:
Model the surface of the Earth as the sphere where
x
2
+
y
2
+
z
2
= 1.
Then, suppose that
the average temperature at each point on the surface is given by the values of the function
T =
x
2
+ 2
xy
+
y
2
 2
z
2
.
Find the points with the maximum and also the minimum temperature.
Here is the solution: The constraint function is
g
(
x
,
y
,
z
)
≡
x
2
+
y
2
+
z
2
 1, and the extreme points occur
where
∇
T =
λ∇
g
for some constant
λ
.
A computation finds:
∇
g
= (2
x
, 2
y
, 2
z
)
∇
T = (2
x
+ 2
y
, 2
y
+ 2
x
,  4
z
)
(1)
Thus, the equation
∇
T =
λ∇
g
holds if and only if
2
x
+ 2
y
=
λ
2
x
2
y
+ 2
x
=
λ
2
y
4
z
=
λ
2
z
(2)
The last point in (2) can hold only if
λ
= 2 or
z
= 0.
I’ll treat these two possibilities in turn.
Consider the case
λ
= 2.
Then, the first and second points in (2) read
2
x
+ 2
y
= 4
x
2
y
+ 2
x
= 4
y
(3)
The latter points can be rephrased to read
6
x
= 2
y
6
y
= 2
x
(4)
This first line says that
x
= 
y
/3, while the second says that
x
= 3
y
.
Clearly, these two assertions are
compatible only if
x
=
y
= 0.
Finally, this last forces
z
= ±1 since
x
2
+
y
2
+
z
2
= 1.
Thus, I have found two
stationary points, (0, 0, 1) and (0, 0, 1).
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 Fall '11
 GregoryJ.Galloway
 Addition, Equations, Multivariable Calculus, Optimization, Fermat's theorem, stationary points

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