3.1 Lagrangian Multipliers

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A function may be given not on R2 but on some curve in R2. (More generally

on some hypersurface is Rn).

Finding critical points in this case is rather different.

Example 3.1.1. Define f : R2 ! R by

_

x

y

_

  x2 − y2.

Find the maxima and minima of f on the unit circle

S1 =

__

x

y

_

2 R2 : x2 + y2 = 1

_

.

Plotting a few points we see f is zero at x = ±y, negative and a minimum

at x = 0y = ±1, positive and a maximum at x = ±1y = 0. Better yet we

use Mathematica and type in:

ParametricPlot3D[{Cos[t],Sin[t],s*Sin[2t]},{t,0,2_},{s,0,1},PlotPoints->50]

It is obvious that Df =

h

@f

@x , @f

@y

i

is zero only at the origin - which is not

much help since the origin isn’t on the circle.

Q 3.1.2 How can we solve this problem? (Other than drawing the graph)

17

18 CHAPTER 3. CONSTRAINED OPTIMISATION

Figure 3.1: Graph of x2 − y2 over unit circle

3.1. LAGRANGIAN MULTIPLIERS 19

Answer 1 Express the circle parametrically by ’drawing’ the curve

eg

x = cos t

y = sin t

_

t 2 [0, 2_)

Then f composed with the map

_

x

y

_

: [0, 2_) −! R2

t 

_

cos t

sin t

_

is

f _

_

x

y

_

(t) = cos2 t − sin2 t

= cos 2t

which has maxima at 2t = 0 and 2t = 2_ i.e. at t = 0, _ and minima at

2t = _, 3_ i.e. t = _/2, 3_/2.

It is interesting to observe that the function x2 − y2 restricted to the circle

is just cos 2t.

Answer 2 (Lagrange Multipliers)

Observe that if f has a critical point on the circle, then

rf =

_ @f

@x

@f

@y

_

(= (Df)T )

which points in the direction where f is increasing most rapidly, must be

normal to the circle. (Since if there was a component tangent to the circle,

f couldn’t have a critical point there, it would be increasing or decreasing in

their direction.)

Now if g

_

x

y

_

= x2 + y2 − 1 we have rg

_

x

y

_

is normal to the circle from

first semester.

So for f to have a critical point on S1, rf

_

x

y

_

and rg

_

x

y

_

must be

pointing in the same direction (or maybe the exactly opposite direction), ie

9_ 2 R, rf

_

x

y

_

= _rg

_

x

y

_

.

20 CHAPTER 3. CONSTRAINED OPTIMISATION

Well, rg

_

x

y

_

=

_ @g

@x

@g

@y

_

=

_

2x

2y

_

and rf

_

x

y

_

=

_ @f

@x

@f

@y

_

=

_

2x

−2y

_

So rf

_

x

y

_

= _

_

rg

_

x

y

__

9_ 2 R

,

_

2x

2y

_

= _

_

2x

−2y

_

9_ 2 R

if x = 0, _ = −1 is possible and on the unit circle, x = 0 ) y = ±1 so

_

0

1

_

,

_

0

−1

_

are two possibilities.

If x 6= 0 then 2x = 2_x ) _ = 1 whereupon y = −y ) y = 0 and on

S1, y = 0 ) x = ±1. So _

1

0

_ _

−1

0

_

are the other two.

It is easy to work out which are the maxima and which the minima by

plugging in values.

Remark 3.1.1. The same idea generalises for maps

f : Rn ! R

and constraints

g1 : Rn ! R

g2 : Rn ! R

...

gn : Rn ! R

with the problem:

Find the critical points of f subject to

gi(x) = 0 8i : 1 _ i _ k

3.1. LAGRANGIAN MULTIPLIERS 21

We have two ways of proceeding: one is to parametise the hypersurface given

by the {gi} by finding

M : Rn−k ! Rn

such that every y 2 Rn such that gi(y) = 0 for all i comes from some

p 2 Rn−k, preferably only one.

Then f _M : Rn−k ! R can have its maxima and minima found as usual by

setting

D(f _M) = [|0, 0,{.z. . , 0}

n−k

]

Or we can find rf(x) and rgi(x).

Then:

Proposition 3.1.1. x is a critical point of f restricted to the set

{gi(x) = 0; 1 _ i _ k}

provided 9 _1, _2, . . . _k 2 R

rf(x) =

X1

i=1,K

_irgi(x)

“Proof” rgi(x) is normal to the ith constraint and the hypersurface is

given by k constraints which (we hope) are all independent. If you think of

k tangents at the point x, you can see them as intersecting hyperplanes, the

dimension of the intersection being, therefore, n − k.

rf cannot have a component along any of the hyperplanes at a critical point,

ie rf(x) is in the span of the k normal vectors rgi(x) : 1 _ i _ k. _

Example 3.1.2. Find the critical points of f

_

x

y

_

= x + y2 subject to the

constraint x2 + y2 = 4.

Solution We have rf

_

x

y

_

=

_ @f

@x

@f

@y

_

=

_

1

2y

_

and

g_ : R2 ! R

x

y

_

  x2 + y2 − 4

22 CHAPTER 3. CONSTRAINED OPTIMISATION

Figure 3.2: Graph of x + y2

has S =

__

x

y

_

2 R2 : g

_

x

y

_

= 0

_

as the constrained set, with

rg =

_

2x

2y

_

Then rf

_

x

y

_

= _rg

_

x

y

_

9 _ 2 R

) 9_ 2 R,

_

1

2y

_

= _

_

2x

2y

_

) _ = 1 and x = 1/2 ) y = ±

q

15

4

or y = 0 and x = ±2 and _ = ±1

2

so the critical points are at

_

1/2 p

15

2

_ _

1/2

−

p

15

2

_ _

2

0

_ _

−2

0

_

If you think about f

_

x

y

_

= x + y2 you see it is a parabolic trough, as in

figure 3.2.

Looking at the part over the circle x2 + y2 = 4 it looks like figure 3.3. Both

pictures were drawn by Mathematica.

3.1. LAGRANGIAN MULTIPLIERS 23

Figure 3.3: Graph of x + y2 over unit circle

So a minimum occurs at

x = −2

y = 0

a local minimum at

x = 2

y = 0

and maxima

at

_

x = 1/4

y =

±p

15

2

_

. _

24 CHAPTER 3. CONSTRAINED OPTIMISATION

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