Section 12.8: Lagrange Multipliers
In many applied problems, a function of three variables, f (x, y, z), must be optimized
subject to a constraint of the form g(x, y, z) = c.
Theorem: (Lagrange’s Theorem)
Suppose that f and g are functions with continuous first-order partial derivatives and f has
an extremum at (x0 , y0 , z0 ) on the smooth curve g(x, y, z) = c. If ∇g(x0 , y0 , z0 ) 6= ~0, then
there is a number λ such that
∇f (x0 , y0 , z0 ) = λg(x0 , y0 , z0 ).
The number λ is called a Lagrange multiplier.
Method of Lagrange Multipliers:
To find the extreme values of f (x, y, z) subject to the constraint g(x, y, z) = c,
1. Find all values of x, y, z, λ such that
∇f (x, y, z) = λ∇g(x, y, z),
g(x, y, z) = c.
2. Evaluate f at each point (x, y, z) found in step 1. The largest of these values is the
maximum value of f and the smallest is the minimum value of f .
Example: Find the maximum and minimum values of f (x, y) = 6x+8y on the circle x2 +y 2 =
25.
Let g(x, y) = x2 + y 2 . The gradient vectors of f and g are
∇f (x, y) = h6, 8i
and
∇g(x, y) = h2x, 2yi.
By Lagrange’s Theorem, there is a number λ such that
h6, 8i = λh2x, 2yi = h2λx, 2λyi.
Therefore, we consider the system
6 = 2λx,
8 = 2λy,
2
x + y 2 = 25.
Clearly λ 6= 0. It follows that x =
4
3
and y = . Using the constraint,
λ
λ
x2 + y 2
9
16
+ 2
2
λ
λ
25
λ2
λ2
λ
= 25
= 25
= 25
= 1
= ±1.
The points of interest are (x, y) = ±(3, 4). Therefore, the maximum and mimimum values
are
f (3, 4) = 50
and
f (−3, −4) = −50.
Example: Find the maximum and minimum values of f (x, y, z) = x + 3y + 5z on the sphere
x2 + y 2 + z 2 = 1.
Let g(x, y, z) = x2 + y 2 + z 2 . The gradient vectors of f and g are
∇f (x, y, z) = h1, 3, 5i
and
∇g(x, y, z) = h2x, 2y, 2zi.
Consider the system
1
3
5
2
2
x + y + z2
Clearly λ 6= 0. Thus, x =
=
=
=
=
2λx,
2λy,
2λz,
1.
3
5
1
,y=
,z=
, and
2λ
2λ
2λ
x2 + y 2 + z 2 = 1
1
9
25
+ 2+ 2 = 1
2
4λ
4λ
4λ
35
= 1
4λ2
35
λ2 =
4√
λ = ±
35
.
2
It follows that the points of interest are
(x, y, z) = ±
1
3
5
√ ,√ ,√
35 35 35
.
Therefore, the maximum and minimum values are
√
√
1
3
5
3
5
1
f √ ,√ ,√
= 35
= − 35.
and
f −√ , −√ , −√
35 35 35
35
35
35
Example: Find the maximum and minimum values of the function f (x, y, z) = xyz on the
ellipsoid x2 + 2y 2 + 3z 2 = 6.
Let g(x, y, z) = x2 + 2y 2 + 3z 2 . The gradient vectors of f and g are
∇f (x, y, z) = hyz, xz, xyi
∇g(x, y, z) = h2x, 4y, 6zi.
and
Consider the system
yz
xz
xy
2
2
x + 2y + 3z 2
=
=
=
=
2λx,
4λy,
6λz,
6.
xyz
xyz
xyz
2
2
x + 2y + 3z 2
=
=
=
=
2λx2 ,
4λy 2 ,
6λz 2 ,
6.
This is equivalent to the system
Clearly λ 6= 0. (Otherwise, x = y = z = 0 which does not satisfy the constraint.) Thus, the
first three equations yield
x2 = 2y 2 = 3z 2 .
Using the constraint,
x2 + 2y 2 + 3z 2
3x2
x2
x
=
=
=
=
6
6
2
√
± 2.
The points of interest are
r !
2
.
3
√
Therefore, the maximum and minimum values are ±2/ 3.
√
(x, y, z) = ±
2, 1,
Example: A rectangular box without a lid is to be made of 12 square meters of cardboard.
Find the maximum volume of such a box.
Let x, y, and z denote the dimensions of the box (in meters). We would like to maximize
the volume of the box
f (x, y, z) = xyz
subject to the constraint
g(x, y, z) = xy + 2xz + 2yz = 12.
The gradient vectors of f and g are
∇f (x, y, z) = hyz, xz, xyi
and
∇g(x, y, z) = hy + 2z, x + 2z, 2x + 2yi.
Consider the system
yz
xz
xy
xy + 2xz + 2yz
=
=
=
=
λ(y + 2z),
λ(x + 2z),
λ(2x + 2y),
12.
This is equivalent to the system
xyz
xyz
xyz
xy + 2xz + 2yz
=
=
=
=
λ(xy + 2xz),
λ(xy + 2yz),
λ(2xz + 2yz),
12.
Clearly λ 6= 0. (Otherwise, x = y = z = 0 which does not satisfy the constraint). The first
two equations imply
xy + 2xz = xy + 2yz
2xz = 2yz
x = y.
Similarly, the second and third equations yield
xy + 2yz = 2xz + 2yz
xy = 2xz
y = 2z.
Using the constraint,
xy + 2xz + 2yz = 4z 2 + 4z 2 + 4z 2 = 12z 2 = 12.
Then z = 1 which implies that x = y = 2. The maximum volume of the box is 4 cubic
meters.
Note: This method can be extended to functions of more than three variables as well as
situations with multiple constraints.
Suppose we want to find the extreme values of a function f (x, y, z) subject to the two
constraints g(x, y, z) = c1 and h(x, y, z) = c2 .
Theorem: If an extreme value of f occurs at (x0 , y0 , z0 ), then there exist numbers λ and µ
such that
∇f (x0 , y0 , z0 ) = λ∇g(x0 , y0 , z0 ) + µ∇h(x0 , y0 , z0 ).
Example: Find the maximum and minimum values of f (x, y, z) = x + 2y subject to the
constraints x + y + z = 1 and y 2 + z 2 = 4.
Let g(x, y, z) = x + y + z and h(x, y, z) = y 2 + z 2 . The gradient vectors for f , g, and h are
∇f (x, y, z) = h1, 2, 0i
∇g(x, y, z) = h1, 1, 1i
∇h(x, y, z) = h0, 2y, 2zi.
Consider the system
1
2
0
x+y+z
y2 + z2
=
=
=
=
=
λ,
λ + 2µy,
λ + 2µz,
1,
4.
Since λ = 1, this is equivalent to
2µy
2µz
x+y+z
y2 + z2
Clearly µ 6= 0. Thus, y =
=
=
=
=
1,
−1,
1,
4.
1
1
and z = − . Since y = −z, the third equation implies x = 1.
2µ
2µ
By the last equation,
y2 + z2 = 4
2
= 4
4µ2
1
µ2 =
8
1
µ = ± √ .
2 2
√
√ √
√
The points of interest are (1, 2, − 2) and (1, − 2, 2). Therefore, the maximum and
minimum values are
√
√
√
√ √
√
f (1, 2, − 2) = 1 + 2 2
and
f (1, − 2, 2) = 1 − 2 2.