2.3. LAGRANGE MULTIPLIERS 57
2.3 LAGRANGE MULTIPLIERS
is section introduces constrained optimization using a technique called Lagrange mul-
tipliers. e basic idea is that we want to optimize a function f .x; y/ at those points where
g.x; y/ D c. e function g.x; y/ is called the constraint.
e proof that Lagrange multipliers work is beyond the scope of this text, so we will begin by
just stating the technique.
Knowledge Box 2.7
e method of LaGrange Multipliers with two variables
Suppose that z D f .x; y/ defines a surface and that g.x; y/ D c spec-
ifies points of interest. en the optima of f .x; y/, subject to the con-
straint that g.x; y/ D c, occur at solutions to the system of equations
f
x
.x; y/ D g
x
.x; y/
f
y
.x; y/ D g
y
.x; y/
g.x; y/ D c
where is an auxiliary variable.
e variable is new and strange – it is the “multiplier” – and, as we will see, correct solutions
to the system of equations that arise from Lagrange multipliers typically use in a fashion that
causes it to drop out.
If the constraint g.x; y/ D c is thought of as the boundary of the domain of optimization, then
using Lagrange multipliers gives us a tool for resolving the boundary as a source of optima as
per the extreme value theorem.
With that context, let’s practice our Lagrange multipliers.
58 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
Example 2.38 Find the maxima and minima of
f .x; y/ D x C 4y 2
on the ellipse 2x
2
C 3y
2
D 36.
Solution:
e equations arising from the Lagrange multiplier technique are:
1 D 4x
4 D 6y
2x
2
C 3y
2
D 36
Solving:
x D
1
4
y D
2
3
Plug into the constraint and we get:
2
16
2
C
12
9
2
D 36
1
8
C
4
3
D 36
2
35=24 D 36
2
35=864 D
2
or
D ˙
p
35=864
2.3. LAGRANGE MULTIPLIERS 59
Which yields candidate points:
x D ˙
1
4
r
864
35
D ˙
r
54
35
y D ˙
2
3
r
864
35
D ˙
r
384
35
Since the equation of f .x; y/ is a plane that gets larger as x and y get larger, we see that the
maximum is
f
p
54=35;
p
384=35
Š 12:5
and the minimum is
f
p
54=35;
p
384=35
Š 16:5
˙
Notice that in the calculations in Example
2.38, the auxiliary variable
served as an informa-
tional conduit that let us discover the candidate points. e next example is much simpler.
Example 2.39 Find the point of closest approach of the line 2x C5y D 3 to the origin (0,0).
Solution:
is problem is like the closest approach of a plane to the origin problems, but in a lower dimen-
sion. e tricky part of this problem is phrasing it as a function to be minimized and a constraint.
Since we are minimizing distance we get that the function is the distance of a point
.x; y/ from the origin:
d.x; y/ D
p
.x 0/
2
.y 0/
2
D
p
x
2
C y
2
As per the monotone function shortcut, we can instead optimize the function
d
2
.x; y/ D x
2
C y
2
:
e constraint function is the line. We can phrase the line as a constraint by saying:
g.x; y/ D 2x C 5y D 3
60 2. MULTIVARIATE AND CONSTRAINED OPTIMIZATION
With the parts in place, we can extract the Lagrange multiplier equations.
2x D 2
2y D 5
2x C 5y D 3
Solve:
x D
y D
5
2
2./ C 5
5
2
D 3
29
2
D 3
D
6
29
Yielding: gx D
6
29
and y D
15
29
is makes the point on 2x C 5y D 3 closest to the origin
6
29
;
15
29
.
˙
Example 2.40 Find the maximum value of
f .x; y/ D x
2
C y
3
subject to the constraint x
2
C y
2
D 9.
2.3. LAGRANGE MULTIPLIERS 61
Solution:
is problem is already in the correct form for Lagrange multipliers, so we may immedi-
ately derive the system of equations.
2x D 2x
2y D 3y
2
x
2
C y
2
D 9
e first equation yields the useful information that D 1 but that x may take on any value,
unless x D 0 in which case may be anything.
Given that D 1, the second equation tells us y D 0 or y D 2=3. If is free and x D 0
then y may be anything.
Plugging the values for y into the constraint that says the points lie on a circle, we can retrieve
values for x: when y D 0, x D ˙3; when y D 2=3, x D ˙
p
9 4=9 D ˙
p
77=9.
If x D 0 and y is free the constraint yields y D ˙3.
is gives us the candidate points .˙3; 0/, .0; ˙3/, .2=3;
p
77=3/, and .2=3;
p
77=3/.
e sign of the x-coordinate is unimportant because f .x; y/ depends on x
2
; since f .x; y/
depends on y
3
, positive values yield larger values of f , and negative ones yield smaller values
of f .
Since f .3; 0/ D 9, f .2=3;
p
77=3/ Š 25:47, and f .0; 3/ D 27, we get that the maximum value
of the function on the circle is:
f .0; 3/ D 27
˙
In Section 2.2 we found the minimum of a line on a quadratic surface that opens upward (Ex-
ample 2.23). e next example lets us try a problem like this using the formalism of Lagrange
multipliers.
Example 2.41 Find the minimum of f .x; y/ D x
2
xy Cy
2
on the line 5x C 7y D 18.
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