The Method of Lagrange multipliers allows us to find constrained extrema. It's more equations, more variables, but less algebra.

1. 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
. . . . . .

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

3. 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
. . . . . .

4. Outline
Introduction
The Method of Lagrange Multipliers
Examples
For those who really must know all
. . . . . .

5. 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.
. . . . . .

6. 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
. . . . . .

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

8. √
Maximize the function f(x, y) = xy subject to the constraint
20x + 10y = 200.
. . . . . .

9. √
Maximize the function f(x, y) = xy subject to the constraint
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.
. . . . . .

10. 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.
. . . . . .

11. 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.
. . . . . .

12. 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.
. . . . . .

13. Example
Find the maximum and minimum values of
f (x, y) = x2 + y2 − 2x − 2y + 14.
subject to the constraint
g (x, y) = x2 + y2 − 16 ≡ 0.
. . . . . .

14. Example
Find the maximum and minimum values of
f (x, y) = x2 + y2 − 2x − 2y + 14.
subject to the constraint
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.
. . . . . .

15. Example
Find the maximum and minimum values of
f (x, y) = x2 + y2 − 2x − 2y + 14.
subject to the constraint
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!
. . . . . .

16. Outline
Introduction
The Method of Lagrange Multipliers
Examples
For those who really must know all
. . . . . .

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

18. Simplified map
l
.evel curves of f .evel curve g = 0
l
- .9. . . . - - -
. 10 876 5 4. 3. 2 . 1
-. - - - -
- .
. . . . . .

19. Simplified map
l
.evel curves of f .evel curve g = 0
l
- .9. . . . - - -
. 10 876 5 4. 3. 2 . 1
-. - - - -
- .
. . . . . .

20. 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!
. . . . . .

21. 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.
. . . . . .

22. 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.
. . . . . .

23. Outline
Introduction
The Method of Lagrange Multipliers
Examples
For those who really must know all
. . . . . .

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

25. 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.
. . . . . .

26. 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.
. . . . . .

27. 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!
. . . . . .

28. Example
Find the maximum and minimum values of
f (x, y) = x2 + y2 − 2x − 6y + 14.
subject to the constraint
g (x, y) = x2 + y2 − 16 ≡ 0.
. . . . . .

29. Example
Find the maximum and minimum values of
f (x, y) = x2 + y2 − 2x − 6y + 14.
subject to the constraint
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.
. . . . . .

30. Solution (Continued)
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.
. . . . . .

31. Solution (Continued)
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.
. . . . . .

32. 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
. . . . . .

33. 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
. . . . . .

34. 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?
. . . . . .

35. 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.
. . . . . .

36. 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)
. . . . . .

37. 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.
. . . . . .

38. 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
. . . . . .

39. 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.
. . . . . .

40. Outline
Introduction
The Method of Lagrange Multipliers
Examples
For those who really must know all
. . . . . .

41. 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.
. . . . . .