lesson 27: lagrange multipliers i
DESCRIPTION
The method of Lagrange mutipliers allows easy solution to a constrained optimization problem.TRANSCRIPT
Lesson 27 (Chapter 18.1–2)Constrained Optimization I: Lagrange Multipliers
Math 20
November 26, 2007
Announcements
I Problem Set 10 on the website. Due November 28.
I next OH: Today 1–2, tomorrow 3–4 (SC 323)
I Midterm II Review: Tuesday 12/4, 7:30–9:00pm in Hall E
I Midterm II: Thursday, 12/6, 7–8:30pm in Hall A
Outline
Motivation
The Method of Lagrange Multipliers
Examples
The problem
We know how to find the critical points of a function of twovariables: look for where ∇f = 0. That is,
∂f
∂x=∂f
∂y= 0
Sometimes, however, we have a constraint which restricts us fromchoosing variables freely:
I Maximize volume subject to limited material costs
I Minimize surface area subject to fixed volume
I Maximize utility subject to limited income
The problem
We know how to find the critical points of a function of twovariables: look for where ∇f = 0. That is,
∂f
∂x=∂f
∂y= 0
Sometimes, however, we have a constraint which restricts us fromchoosing variables freely:
I Maximize volume subject to limited material costs
I Minimize surface area subject to fixed volume
I 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 constraint20x + 10y = 200.
SolutionSolve 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
f ′(x) =1
2√
20x − 2x2(20− 4x) =
10− 2x√20x − 2x2
.
Then f ′(x) = 0 when 10− 2x = 0, or x = 5. Since y = 20− 2x,y = 10. f (5, 10) =
√50.
Maximize the function f (x , y) =√
xy subject to the constraint20x + 10y = 200.
SolutionSolve 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
f ′(x) =1
2√
20x − 2x2(20− 4x) =
10− 2x√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 MethodCf. Section 9.3
Once the function is restricted to the line 20x + 10y = 200, wecan’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 checkf on these two endpoints and its critical point to find themaximum value. But f (0) = f (10) = 0 so f (5) =
√50 is the
maximum value.
Checking maximality: First Derivative TestCf. Section 9.2
We have
f ′(x) =10− 2x√20x − 2x2
The denominator is always positive, so the fraction is positiveexactly when the numerator is positive. So f ′(x) < 0 if x < 5 andf ′(x) > 0 if x > 5. This means f changes from increasing todecreasing at 5. So 5 is the global maximum point.
Checking maximality: Second Derivative TestCf. Section 9.4
We have
f ′′(x) = − 100
(20x − 2x2)3/2
So f ′′(5) < 0, which means f has a local maximum at 5. Sincethere 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.
subject to the constraint
g (x , y) = x2 + y2 − 16 ≡ 0.
This one’s harder. Solving for y in terms of x involves the squareroot, of which there’s two choices.There’s a better way!
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 squareroot, of which there’s two choices.
There’s a better way!
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 squareroot, of which there’s two choices.There’s a better way!
Outline
Motivation
The Method of Lagrange Multipliers
Examples
Consider a path that moves across a hilly terrain. Where are thecritical points of elevation along your path?
Simplified map
-1-2-3-4-5-6-7-8-9-10
level curves of f level curve g = 0
At the constrainedcritical point, thetangents to thelevel curves of fand g are in thesame direction!
The slopes of the tangent lines to these level curves are(dy
dx
)f
= − f ′xf ′y
and
(dy
dx
)g
= −g ′xg ′y
So they are equal when
f ′xf ′y
=g ′xg ′y⇐⇒ f ′x
g ′x=
f ′yg ′y
If λ is the common ratio on the right, we have
f ′xf ′y
=g ′xg ′y
= λ
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 severalvariables. The critical points of the function f restricted to the setg = 0 are solutions to the equations:
∂f
∂xi(x1, x2, . . . , xn) = λ
∂g
∂xi(x1, x2, . . . , xn) for each i = 1, . . . , n
g(x1, x2, . . . , xn) = 0.
Note that this is n + 1 equations in the n + 1 variables x1, . . . , xn, λ.
Outline
Motivation
The Method of Lagrange Multipliers
Examples
Example
Maximize the function f (x , y) =√
xy subject to the constraint20x + 10y = 200.
Solution
Let’s set g(x , y) = 20x + 10y − 200. We have
∂f
∂x=
1
2
√y
x
∂g
∂x= 20
∂f
∂y=
1
2
√x
y
∂g
∂y= 10
So the equations we need to solve are
1
2
√y
x= 20λ
1
2
√x
y= 10λ
20x + 10y = 200.
Solution (Continued)
Dividing the first by the second gives us
y
x= 2,
which means y = 2x. We plug this into the equation of constraintto get
20x + 10(2x) = 200 =⇒ x = 5 =⇒ y = 10.
Caution
When dividing equations, one must take care that the equation wedivide by is not equal to zero. So we should verify that there is nosolution where
1
2
√x
y= 10λ = 0
If this were true, then λ = 0. Since y = 800λ2x , we get y = 0.Since x = 200λ2y , we get x = 0. But then the equation ofconstraint is not satisfied. So we’re safe.Make sure you account for these because you can lose solutions!
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.
SolutionWe have the two equations
2x − 2 = λ(2x)
2y − 6 = λ(2y).
as well as the thirdx2 + y2 = 16.
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.
SolutionWe have the two equations
2x − 2 = λ(2x)
2y − 6 = λ(2y).
as well as the thirdx2 + y2 = 16.
Solution (Continued)
Solving both of these for λ and equating them gives
x − 1
x=
y − 3
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.
Solution (Continued)
Looking at the function
f (x , y) = x2 + y2 − 2x − 2y + 14
We see that
f(−2√
2/5,−6√
2/5)
=94 + 10
√5
5
is the maximum and
f(
2√
2/5, 6√
2/5)
s =94− 10
√5
5
is the minimum value of the constrained function.
Contour Plot
-4 -2 0 2 4
-4
-2
0
2
4
The green curve is theconstraint, and the twogreen points are theconstrained max andmin.
Compare and Contrast
Elimination
I solve, then differentiate
I messier (usually)equations
I fewer equations
I adaptable to more thanone constraint
I second derivative test iseasier
Lagrange Multipliers
I differentiate, then solve
I nicer (usually) equations
I more equations
I adaptable to more thanone constraint
I second derivative test(later) is harder
I multipliers havecontextual meaning
Another argument for Lagrange multipliers
To find the critical points of f subject to the constraint thatg = 0, create the lagrangian
L = f (x1, x2, . . . , xn)− λg(x1, x2, . . . , xn)
If L is restricted to the set g = 0, L = f and so the constrainedcritical points are unconstrained critical points of L . So for each i ,
∂L
∂xi= 0 =⇒ ∂f
∂xi= λ
∂g
∂xi.
But also,∂L
∂λ= 0 =⇒ g(x1, x2, . . . , xn) = 0.
Example
A rectangular box is to be constructed of materials such that thebase of the box costs twice as much per unit area as does the sidesand top. If there are D dollars allocated to spend on the box, howshould these be allocated so that the box contains the maximumpossible value?
Answer.
x = y =1
3
√D
cz =
1
2
√D
c
where c is the cost per unit area of the sides and top.
Example
A rectangular box is to be constructed of materials such that thebase of the box costs twice as much per unit area as does the sidesand top. If there are D dollars allocated to spend on the box, howshould these be allocated so that the box contains the maximumpossible value?
Answer.
x = y =1
3
√D
cz =
1
2
√D
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 areaof the sides and top be c ; so the cost per unit area of the bottomis 2c . If x and y are the dimensions of the bottom of the box, thenwe want to maximize V = xyz subject to the constraint that2cyz + 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. Eachof those possibilities eventually leads to a contradiction to theconstraint equation.Dividing the first two gives
y
x=
2z + 3y
3x + 2z=⇒ y(3x + 2z) = x(2z + 3y) =⇒ 2yz = 2xz
Since z 6= 0, we have x = y .
The last equation now becomes x2 = 4λcx . Dividing the secondequation by this gives
z
x=
3x + 2z
4x=⇒ z = 3
2x .
Putting these into the equation of constraint we have
D = 3cxy + 2cyz + 2xz = 3cx2 + 3cx2 + 3cx2 = 9cx2.
So
x = y =1
3
√D
cz =
1
2
√D
c
It also follows that
λ =x
4c=
1
12
√D
c3
Interpretation of λ
Let V ∗ be the maximum volume found by solving the Lagrangemultiplier equations. Then
V ∗ =
(1
3
√D
c
)(1
3
√D
c
)(1
2
√D
c
)=
1
18
√D3
c3
NowdV ∗
dD=
3
2
1
18
√D
c3=
1
12
√D
c3= λ
This is true in general; the multiplier is the derivative of theextreme value with respect to the constraint.