LECTURE 9: LAGRANGE MULTIPLIERS
MINGFENG ZHAO
January 23, 2015
Polar coordinates
For the point (x, y) 6= (0, 0) in the xy-plane, let r = |(x, y)| and θ be the angle between x-axis and the vector from
(0, 0) to (x, y), then we have
x = r cos(θ),
and y = r sin(θ),
which is called the Polar coordinates.
Figure 1. Polar coordinates in the plane
Remark 1. For the ellipse
x2
y2
+
= 1, we can use the following polar coordinates:
a2
b2
x = a cos(θ),
and y = b sin(θ),
Gradient
1
for all 0 ≤ θ ≤ 2π.
2
MINGFENG ZHAO
Definition 1. The gradient of a function f (x, y, z) at point (a, b, c), denoted by ∇f (x, y, z) is defined as:
∇f (a, b, c) = hfx (a, b, c), fy (a, b, c), fz (a, b, c)i.
Remark 2. A critical point is just a point such that the gradient at this points equals zero vector.
Example 1. Let f (x, y) = ln(xy + sin(y)), compute the gradient of f (x, y).
In fact, we have
fx (x, y)
1
· y By the chain rule
xy + sin(y)
y
xy + sin(y)
1
· cos(y) By the chain rule
xy + sin(y)
=
=
fy (x, y)
=
cos(y)
.
xy + sin(y)
=
So we have
∇f (x, y) =
y
cos(y)
,
xy + sin(y) xy + sin(y)
.
Example 2. Let f (x, y, z) = x2 + 2y 2 + 4z 2 − 1, compute the gradient of f (x, y, z) at point (0, 1, 2).
In fact, we have
fx = 2x,
fy = 4y,
and fz = 8z.
Then
∇f (x, y, z) = h2x, 4y, 8zi.
So we get
∇f (0, 1, 2) = h0, 4, 16i.
Lagrange multipliers
Consider the optimization problem:
max / min
f (x, y) −→
g(x, y) = 0
Method of Lagrange Multipliers:
−→
objective function
constraint.
LECTURE 9: LAGRANGE MULTIPLIERS
3
If ∇g(x, y) 6= 0 for all (x, y) such that g(x, y) = 0, to find the maximum and minimum values of f (x, y)
subject to the constraint g(x, y) = 0, we have two steps:
Step 1. Find the values of x, y and λ such that
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Step 2. Among the points (x, y) found in Step 1, select the largest and smallest corresponding function
values. These values are the maximum and minimum values of f (x, y) subject to the constraint
g(x, y) = 0.
Remark 3. For the above Step 1, we can define the function h(x, y, λ) = f (x, y) − λg(x, y) which is a function of three
variables x, y and λ, to find values in Step 1, we only need to find critical points of h(x, y, λ), that is, we need to solve
the following system:
∇h(x, y, λ) = 0.
Remark 4. For solving ∇h(x, y, λ) = 0, fist, you should check whether λ = 0 or not. Usually, λ 6= 0
Example 3. Find the maximum and minimum values of the objective function f (x, y) = 2x2 + y 2 + 2, where x and y
lie on the ellipse C given by g(x, y) = x2 + 4y 2 − 4 = 0.
Step 1: Let’s compute the gradients of f and g, we have
∇f (x, y) = h4x, 2yi,
and ∇g(x, y) = h2x, 8yi.
Let’s solve the system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
That is,
(1)
4x = λ · (2x) = 2λx
(2)
2y = λ · (8y) = 8λy
(3)
x2 + 4y 2 − 4 = 0
It’s easy to see that λ 6= 0, otherwise, by (1) and (2), we have x = y = 0, which does not satisfy (3). By (1) and (2),
we have
4x(8λy) = 2λx(2y).
4
MINGFENG ZHAO
Since λ 6= 0, then 32xy = 4xy, that is, xy = 0. So either x = 0 or y = 0.
When x = 0, plug x = 0 into (3), we get 4y 2 − 4 = 0, that is, y 2 = 1, so y = 1 or y = −1. By (2), we have λ = 41 . So
we have two solutions
1
(x, y, λ) = 0, 1,
,
4
and
(x, y, λ) =
1
0, −1,
4
.
When y = 0, plug y = 0 into (3), we get x2 − 4 = 0, then x = 2 or x = −2. By (1), we have λ = 2. So we have two
solutions:
(x, y, λ) = (2, 0, 2) ,
and
In summary, we have four solutions:
1
1
, (x, y, λ) = 0, −1,
,
(x, y, λ) = 0, 1,
4
4
(x, y, λ) = (−2, 0, 2) .
(x, y, λ) = (2, 0, 2) ,
and
(x, y, λ) = (−2, 0, 2) .
Step 2: Compute the values of f (x, y) on the above points, we have
f (0, 1)
f (0, −1)
f (2, 0)
f (−2, 0)
=
2 · 02 + 12 + 2
=
3
=
2 · 02 + (−1)2 + 2
=
3
=
2 · 22 + 02 + 2
=
10
=
2 · (−2)2 + 02 + 2
=
10.
In summary, the minimum values of f (x, y) subject to the constraint g(x, y) = 0 is 3 = f (0, 1) = f (0, −1), and the
maximum values of f (x, y) subject to the constraint g(x, y) = 0 is 10 = f (2, 0) = f (−2, 0).
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca