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Lagrange Multipliers

Lagrange Multipliers
In the previous section we optimized (i
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found the absolute extrema) a function on a region that
contained its boundary
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However, as we saw in the examples finding potential optimal points on the boundary was often a
fairly long and messy process
...
The constraint(s) may be the equation(s) that describe the boundary of a
region although in this section we won’t concentrate on those types of problems since this method
just requires a general constraint and doesn’t really care where the constraint came from
...
We want to optimize (find the minimum and maximum) of a function,
f ( x, y, z ) , subject to the constraint g ( x, y, z ) = k
...
The process is actually fairly simple,
although the work can still be a little overwhelming at times
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Solve the following system of equations
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Plug in all solutions, ( x, y, z ) , from the first step into f ( x, y, z ) and identify the
minimum and maximum values, provided they exist
...

Notice that the system of equations actually has four equations, we just wrote the system in a
simpler form
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fx , f y , fz = λ gx , g y , gz = λ gx , λ g y , λ gz
In order for these two vectors to be equal the individual components must also be equal
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fx = λ gx

fy = λgy

fz = λ gz

These three equations along with the constraint, g ( x, y, z ) = c , give four equations with four
unknowns x, y, z, and λ
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As a final note we also need to be careful with the fact that in some cases minimums and
maximums won’t exist even though the method will seem to imply that they do
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Let’s work a couple of examples
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Solution
Before we start the process here note that we also saw a way to solve this kind of problem in
Calculus I, except in those problems we required a condition that related one of the sides of the
box to the other sides so that we could get down to a volume and surface area function that only
involved two variables
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Now, let’s get on to solving the problem
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Let’s set the length of the box to be x, the width of the box
to be y and the height of the box to be z
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We want to find the largest volume and so the function that we want to optimize is given by,

f ( x, y, z ) = xyz

Next we know that the surface area of the box must be a constant 64
...

The surface area of a box is simply the sum of the areas of each of the sides so the constraint is
given by,

2 xy + 2 xz + 2 yz = 64

⇒

xy + xz + yz = 32

Note that we divided the constraint by 2 to simplify the equation a little
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g ( x, y, z ) = xy + xz + yz
Here are the four equations that we need to solve
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We’ll solve it in the following way
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This gives,

xyz = λ x ( y + z )

(5)

xyz = λ y ( x + z )

(6)

xyz = λ z ( x + y )

(7)

Now notice that we can set equations (5) and (6) equal
...
The first, λ = 0 is not possible since if this was the case equation (1)
would reduce to

yz = 0

⇒

y = 0 or z = 0

Since we are talking about the dimensions of a box neither of these are possible so we can
discount λ = 0
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xz = yz

Since we know that z ≠ 0 (again since we are talking about the dimensions of a box) we can
cancel the z from both sides
...
Doing this gives,

λ y ( x + z) = λz ( x + y)

λ ( yx + yz − zx − zy ) = 0
λ ( yx − zx ) = 0
λ = 0 or yx = zx
⇒
As already discussed we know that λ = 0 won’t work and so this leaves,
yx = zx

We can also say that x ≠ 0 since we are dealing with the dimensions of a box so we must have,

z=y

(9)

Plugging equations (8) and (9) into equation (4) we get,

y + y + y = 3 y = 32
2

2

2

2

32
y=±
= ± 3
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Therefore the only solution that makes physical sense here is

x = y = z = 3
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We should be a little careful here
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The method of Lagrange
Multipliers will give a set of points that will either maximize or minimize a given function subject
to the constraint, provided there actually are minimums or maximums
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The only real restriction that we’ve got is that all the
variables must be positive
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The function will not have a maximum if all the variables are allowed to increase without bound
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So, if one of the variables gets very large, say x, then because each of the products
must be less than 32 both y and z must be very small to make sure the first two terms are less than
32
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This isn’t a rigorous proof that the function will have a maximum, but it should help to visualize
that in fact it should have a maximum and so we can say that we will get a maximum volume if
the dimensions are : x = y = z = 3
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Notice that we never actually found values for λ in the above example
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The value of λ isn’t really important to determining if the point is a
maximum or a minimum so often we will not bother with finding a value for it
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Example 2 Find the maximum and minimum of f ( x, y ) = 5x − 3y subject to the constraint

x 2 + y 2 = 136
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Also,
note that it’s clear from the constraint that region of possible solutions lies on a disk of radius

136 with is a closed and bounded region and hence by the Extreme Value Theorem we know
that a minimum and maximum value must exist
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5 = 2λ x
−3 = 2λ y
x 2 + y 2 = 136

Notice that, as with the last example, we can’t have λ = 0 since that would not satisfy the first
two equations
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This gives,

5
x=
2λ
Plugging these into the constraint gives,

3
y=−
2λ

25
9
17
+ 2 = 2 = 136
2
4λ
4λ
2λ
We can solve this for λ
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1
If λ = − we get,
4
1
and if λ = we get,
4

x = −10

y=6

x = 10

y = −6

To determine if we have maximums or minimums we just need to plug these into the function
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Here are the minimum and maximum values of the function
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Do not always expect this to happen
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Let’s take a look at another example
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Assume that x, y, z ≥ 0
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Therefore it is clear that our solution will fall in the range 0 ≤ x, y, z ≤ 1
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Here is the system of equation that we need to solve
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So, let’s start off by setting equations (10) and (11) equal
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Let’s start off with by assuming that z = 0
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From equation (12) we
see that this means that xy = 0
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So, we’ve got two possible cases to deal with there
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Once we know this we can plug into the constraint, equation (13), to find the remaining
value
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Now let’s go back and take a look at the other possibility, y = x
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This first case is x = y = 0
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The second case is x = y ≠ 0
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xz = xy

⇒

x ( z − y) = 0

⇒

x = 0 or z = y

Now, we’ve already assumed that x ≠ 0 and so the only possibility is that z = y
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⎝3 3 3⎠
We got four solutions by setting the first two equations equal
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Doing this gives,

yz = xy

⇒

y ( z − x) = 0

⇒

y = 0 or z = x

xz = xy

⇒

x ( z − y) = 0

⇒

x = 0 or z = y

Both of these are very similar to the first situation that we looked at and we’ll leave it up to you to
show that in each of these cases we arrive back at the four solutions that we already found
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f ( 0, 0,1) = 0

f ( 0,1, 0 ) = 0

f (1, 0, 0 ) = 0

⎛1 1 1⎞ 1
f ⎜ , , ⎟=
⎝ 3 3 3 ⎠ 27

All Minimums
Maximum

So, in this case the maximum occurs only once while the minimum occurs three times
...
This
assumption is here mostly to make sure that we really do have a maximum and a minimum of the
function
...
For example
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However, all of these examples required negative values of x, y
and/or z to make sure we satisfy the constraint
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To this point we’ve only looked at constraints that were equations
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The process for these types of problems is nearly identical to what we’ve
been doing in this section to this point
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Let’s work an example to see how these kinds of problems work
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Solution
Note that the constraint here is the inequality for the disk
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The first step is to find all the critical points that are in the disk (i
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satisfy the constraint)
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Here are the two first order partial derivatives
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At this point we proceed with Lagrange Multipliers and we treat the constraint as an equality
instead of the inequality
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So, here is the system of equations that we need to solve
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If we have λ = 4 the second equation gives us,

20 y = 8 y
The constraint then tells us that x = ± 2
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So, Lagrange Multipliers gives us four points to check : ( 0, 2 ) , ( 0, −2 ) , ( 2, 0 ) , and ( −2, 0 )
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f ( 0, 0 ) = 0

Minimum

f ( 2, 0 ) = f ( −2, 0 ) = 16
f ( 0, 2 ) = f ( 0, −2 ) = 40

Maximum

In this case, the minimum was interior to the disk and the maximum was on the boundary of the
disk
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We will look only at two constraints, but we can naturally extend the work here to
more than two constraints
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The system that we need to solve in this case is,

∇f ( x, y, z ) = λ∇g ( x, y, z ) + μ∇h ( x, y, z )
g ( x, y , z ) = c
h ( x, y , z ) = k

So, in this case we get two Lagrange Multipliers
...
Let’s see an example of this kind of optimization
problem
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Solution
Verifying that we will have a minimum and maximum value here is a little trickier
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With this in mind there must
also be a set of limits on z in order to make sure that the first constraint is met
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We won’t do that here
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Here is the system of equations that we need to solve
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Plugging this into equation (14) and
equation (15) and solving for x and y respectively gives,

0 = 4 + 2μ x
4 = −2 + 2 μ y

⇒

x=−

⇒

3

y=

2

μ

μ

Now, plug these into equation (18)
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First, let’s see what we get when μ = 13
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Let’s now see what we get if we take μ = − 13
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Now all that we need to is check the two solutions in the function to see which is the maximum
and which is the minimum
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2111
13
26
= −3
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