Lesson 17: The Method of Lagrange Multipliers

Section 11.8
Lagrange Multipliers

Math 21a

March 14, 2008

Announcements
◮ Midterm is graded
◮ Ofﬁce hours Tuesday, Wednesday 2–4pm SC 323
◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b

. .
Image: Flickr user Tashland
. . . . . .

Announcements

◮ Midterm is graded
◮ Ofﬁce hours Tuesday, Wednesday 2–4pm SC 323
◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b

. . . . . .

Happy Pi Day!

3:14 PM Digit recitation contest! Recite all the digits you know of π (in
order, please). Please let us know in advance if you’ll recite π in
a base other than 10 (the usual choice), 2, or 16. Only positive
integer bases allowed – no fair to memorize π in base
π /(π − 2)...
4 PM — Pi(e) eating contest! Cornbread are square; pie are round.
You have 3 minutes and 14 seconds to stuff yourself with as
much pie as you can. The leftovers will be weighed to calculate
how much pie you have eaten.
Contests take place in the fourth ﬂoor lounge of the Math
Department. .

.
Image: Flickr user Paul Adam Smith
. . . . . .

Outline

Introduction

The Method of Lagrange Multipliers

Examples

For those who really must know all

. . . . . .

The problem

Last time we learned how to ﬁnd the critical points of a function of
two variables: look for where ∇f = 0. That is,
∂f ∂f
= =0
∂x ∂y

Then the Hessian tells us what kind of critical point it is.

. . . . . .

The problem

Last time we learned how to ﬁnd the critical points of a function of
two variables: look for where ∇f = 0. That is,
∂f ∂f
= =0
∂x ∂y

Then the Hessian tells us what kind of critical point it is. Sometimes,
however, we have a constraint which restricts us from choosing
variables freely:
◮ Maximize volume subject to limited material costs
◮ Minimize surface area subject to ﬁxed volume
◮ Maximize utility subject to limited income

. . . . . .

Example
Maximize the function
√
f( x , y ) = xy
subject to the constraint

g(x, y) = 20x + 10y = 200.

. . . . . .

√
Maximize the function f(x, y) = xy subject to the constraint
20x + 10y = 200.

. . . . . .

√
20x + 10y = 200.
Solution
Solve the constraint for y and make f a single-variable function:
2x + y = 20, so y = 20 − 2x. Thus
√ √
f(x) = x(20 − 2x) = 20x − 2x2
1 10 − 2x
f ′ (x ) = √ (20 − 4x) = √ .
2 20x − 2x 2 20x − 2x2

Then f′ (x) = 0 when 10 − 2x = 0, or x = 5. Since y = 20 − 2x, y = 10.
√
f(5, 10) = 50.

. . . . . .

Checking maximality: Closed Interval Method
Cf. Section 4.2

Once the function is restricted to the line 20x + 10y = 200, we can’t
plug in negative numbers for f(x). Since
√
f(x) = x(20 − 2x)

we have a restricted domain of 0 ≤ x ≤ 10. We only need to check f
on these two endpoints and its critical√
point to ﬁnd the maximum
value. But f(0) = f(10) = 0 so f(5) = 50 is the maximum value.

. . . . . .

Checking maximality: First Derivative Test
Cf. Section 4.3

We have
10 − 2x
f ′ (x ) = √
20x − 2x2
The denominator is always positive, so the fraction is positive exactly
when the numerator is positive. So f′ (x) < 0 if x < 5 and f′ (x) > 0 if
x > 5. This means f changes from increasing to decreasing at 5. So 5
is the global maximum point.

. . . . . .

Checking maximality: Second Derivative Test
Cf. Section 4.3

We have
100
f′′ (x) = −
(20x − 2x2 )3/2
So f′′ (5) < 0, which means f has a local maximum at 5. Since there
are no other critical points, this is the global maximum.

. . . . . .

Example
Find the maximum and minimum values of

f (x, y) = x2 + y2 − 2x − 2y + 14.


g (x, y) = x2 + y2 − 16 ≡ 0.

. . . . . .

Example

f (x, y) = x2 + y2 − 2x − 2y + 14.


g (x, y) = x2 + y2 − 16 ≡ 0.

This one’s harder. Solving for y in terms of x involves the square root,
of which there’s two choices.

. . . . . .

Example

f (x, y) = x2 + y2 − 2x − 2y + 14.


g (x, y) = x2 + y2 − 16 ≡ 0.

This one’s harder. Solving for y in terms of x involves the square root,
of which there’s two choices.
There’s a better way!

. . . . . .

Consider a path that moves across a hilly terrain. Where are the
critical points of elevation along your path?

..
. . . . . .

Simpliﬁed map

l
.evel curves of f .evel curve g = 0
l

- .9. . . . - - -
. 10 876 5 4. 3. 2 . 1
-. - - - -
- .

. . . . . .

Simpliﬁed map

l
.evel curves of f .evel curve g = 0
l

.
At the constrained
critical point, the
. tangents to the
- .9. . . . - - -
. 10 876 5 4. 3. 2 . 1
-. - - - -
-
level curves of f
and g are in the
same direction!

. . . . . .

The slopes of the tangent lines to these level curves are
( ) ( )
dy f dy g
= − x and =− x
dx f fy dx g gy

So they are equal when

fx g f fy
= x ⇐⇒ x =
fy gy gx gy

If λ is the common ratio on the right, we have

fx g
= x =λ
fy gy

So

f x = λg x
f y = λg y

This principle works with any number of variables.
. . . . . .

Theorem (The Method of Lagrange Multipliers)
Let f(x1 , x2 , . . . , xn ) and g(x1 , x2 , . . . , xn ) be functions of several
variables. The critical points of the function f restricted to the set g = 0
are solutions to the equations:

∂f ∂g
(x1 , x2 , . . . , xn ) = λ (x1 , x2 , . . . , xn ) for each i = 1, . . . , n
∂ xi ∂ xi
g (x 1 , x 2 , . . . , x n ) = 0 .

Note that this is n + 1 equations in the n + 1 variables x1 , . . . , xn , λ.

. . . . . .

Example
√
20x + 10y = 200.

. . . . . .

Let’s set g(x, y) = 20x + 10y − 200. We have
√
∂f 1 y ∂g
= = 20
∂x 2 x ∂x
√
∂f 1 x ∂g
= = 10
∂y 2 y ∂y

So the equations we need to solve are
√ √
1 y 1 x
= 20λ = 10λ
2 x 2 y
20x + 10y = 200.

. . . . . .

Solution (Continued)
Dividing the ﬁrst by the second gives us
y
= 2,
x
which means y = 2x. We plug this into the equation of constraint to get

20x + 10(2x) = 200 =⇒ x = 5 =⇒ y = 10.

. . . . . .

Caution

When dividing equations, one must take care that the equation we
divide by is not equal to zero. So we should verify that there is no
solution where √
1 x
= 10λ = 0
2 y
If this were true, then λ = 0. Since y = 800λ2 x, we get y = 0. Since
x = 200λ2 y, we get x = 0. But then the equation of constraint is not
satisﬁed. So we’re safe.
Make sure you account for these because you can lose solutions!

. . . . . .

Example

f (x, y) = x2 + y2 − 2x − 6y + 14.


g (x, y) = x2 + y2 − 16 ≡ 0.

. . . . . .

Example

f (x, y) = x2 + y2 − 2x − 6y + 14.


g (x, y) = x2 + y2 − 16 ≡ 0.

Solution
We have the two equations

2x − 2 = λ(2x)
2y − 6 = λ(2y).

as well as the third
x2 + y2 = 16.

. . . . . .

Solving both of these for λ and equating them gives

x−1 y−3
= .
x y

Cross multiplying,

xy − y = xy − 3x =⇒ y = 3x.

Plugging this in the equation of constraint gives

x2 + (3x)2 = 16,
√ √
which gives x = ± 8/5, and y = ±3 8/5.

. . . . . .

Looking at the function

f (x, y) = x2 + y2 − 2x − 2y + 14

We see that
( ) √
√ √ 94 + 10 5
f −2 2/5, −6 2/5 =
5
is the maximum and
( √ √
√ ) 94 − 10 5
f 2 2/5, 6 2/5 =
5
is the minimum value of the constrained function.

. . . . . .

Contour Plot

4

2

The green curve is the
0 constraint, and the two
green points are the
constrained max and min.
2

4

4 2 0 2 4

. . . . . .

Compare and Contrast

Elimination Lagrange Multipliers
◮ solve, then differentiate ◮ differentiate, then solve
◮ messier (usually) ◮ nicer (usually) equations
equations ◮ more equations
◮ fewer equations ◮ adaptable to more than
◮ more complex with more one constraint
constraints ◮ second derivative test
◮ second derivative test is (won’t do) is harder
easier ◮ multipliers have
contextual meaning

. . . . . .

Example
A rectangular box is to be constructed of materials such that the
base of the box costs twice as much per unit area as does the sides
and top. If there are D dollars allocated to spend on the box, how
should these be allocated so that the box contains the maximum
possible value?

. . . . . .

Example
A rectangular box is to be constructed of materials such that the
base of the box costs twice as much per unit area as does the sides
and top. If there are D dollars allocated to spend on the box, how
should these be allocated so that the box contains the maximum
possible value?

Answer
√ √
1 D 1 D
x=y= z=
3 c 2 c
where c is the cost per unit area of the sides and top.

. . . . . .

Solution

Let the sides of the box be x, y, and z. Let the cost per unit area of
the sides and top be c; so the cost per unit area of the bottom is 2c.
If x and y are the dimensions of the bottom of the box, then we want
to maximize V = xyz subject to the constraint that
2cyz + 2cxz + 3cxy − D = 0. Thus

yz = λc(2z + 3y)
xz = λc(3x + 2z)
xy = λc(2x + 2y)

. . . . . .

Before dividing, check that none of x, y, z, or λ can be zero. Each of
those possibilities eventually leads to a contradiction to the
constraint equation.
Dividing the ﬁrst two gives

y 2z + 3y
= =⇒ y(3x + 2z) = x(2z + 3y) =⇒ 2yz = 2xz
x 3x + 2z
Since z ̸= 0, we have x = y.

. . . . . .

The last equation now becomes x2 = 4λcx. Dividing the second
equation by this gives

z 3x + 2z
= =⇒ z = 3 x.
2
x 4x
Putting these into the equation of constraint we have

D = 3cxy + 2cyz + 2xz = 3cx2 + 3cx2 + 3cx2 = 9cx2 .

So √ √
1 D 1 D
x=y= z=
3 c 2 c
It also follows that √
x 1 D
λ= =
4c 12 c3

. . . . . .

Interpretation of λ

Let V∗ be the maximum volume found by solving the Lagrange
multiplier equations. Then
( √ )( √ )( √ ) √
1 D 1 D 1 D 1 D3
V∗ = =
3 c 3 c 2 c 18 c3

Now √ √
dV∗ 3 1 D 1 D
= 3
= =λ
dD 2 18 c 12 c3
This is true in general; the multiplier is the derivative of the extreme
value with respect to the constraint.

. . . . . .

The second derivative test for constrained optimization

Constrained extrema of f subject to g = 0 are unconstrained critical
points of the Lagrangian function

L(x, y, λ) = f(x, y) − λg(x, y)

The hessian at a critical point is
 
0 gx g y
HL = gx fxx fxy 
gy fxy fyy

For (x, y, λ) to be minimal, we need det(HL) < 0, and for (x, y, λ) to
be maximal, we need det(HL) > 0.

. . . . . .

Lesson 17: The Method of Lagrange Multipliers

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