Jim Lambers
MAT 419/519
Summer Session 2011-12
Lecture 7 Notes
These notes correspond to Section 2.4 in the text.
Convexity and the Arithmetic-Geometric Mean Inequality
We now describe a highly useful application of convexity in the solution of constrained optimization
problems. Consider the function f (x) = − ln x on the domain D = (0, ∞). Because f 00 (x) = 1/x2 ,
this function is strictly convex on D. It follows
Pk that for positive real numbers x1 , x2 , . . . , xk and
positive real numbers δ1 , δ2 , . . . , δk such that i=1 δk = 1, we have
− ln
k
X
!
≤−
δ i xi
i=1
k
X
δi ln xi ,
i=1
with equality if and only if all of the xi are equal. Negating and then exponentiating both sides
yields
k
k
X
Y
δ i xi .
xδi i ≤
i=1
i=1
This yields the following theorem.
Theorem (Arithmetic-Geometric Mean Inequality or A-G Inequality) Let x1 , x2 , . . . , xn
be positive real numbers, and let δ1 , δ2 , . . . , δn be positive real numbers whose sum is one. Then
n
Y
xδi i ≤
n
X
δ i xi ,
i=1
i=1
with equality if and only if x1 = x2 = · · · = xn .
Example Let δi = 1/n for i = 1, 2, . . . , n. Then the A-G Inequality reduces to
√
n
x1 x2 · · · xn ≤
1
(x1 + x2 + · · · + xn ),
n
where the left side is the geometric mean of x1 , x2 , . . . , xn , and the right side is their arithmetic
mean. 2
We now demonstrate how the A-G Inequality can be used to find the maximum or minimum of
a function of n variables, subject to a constraint on the values of those variables.
1
Example Consider an open rectangular box with dimensions x, y and z. We wish to maximize
the objective function, the volume V = xyz, subject to the constraint that the surface area S =
xy + 2yz + 2xz is equal to a constant S0 . Note that the coefficient of xy is equal to 1, rather than
2, because the box is open at the top.
The idea is to apply the A-G Inequality with a function of the volume on the left side of the
inequality, and a function of the surface area on the right side, which will serve as an upper bound.
By forcing equality in the A-G Inequality, the maximizer of the volume can be found.
The challenge is to choose the numbers x1 , x2 , . . . , xn and the numbers δ1 , δ2 , . . . , δn , which must
sum to one, to apply to A-G Inequality. To that end, we choose x1 , x2 , x3 to be the three terms in
S, and δi = 1/3 for i = 1, 2, 3. The A-G Inequality yields
1
1
1
(xy)1/3 (2yz)1/3 (2xz)1/3 ≤ xy + 2yz + 2xz,
3
3
3
or
1
41/3 V ≤ S0 .
3
This inequality is an equality if xy = 2yz = 2xz = 13 S0 . It follows that x = y = 2z. From
xy = x2 = 13 S0 , we obtain
r
r
S0
1 S0
x=y=
, z=
,
3
2 3
and that the maximum volume is
3/2
S0
V = xyz =
.
2(33/2 )
2
Example We now solve the dual problem of the previous example, which means that the objective
function and constraint are interchanged, and the new objective function is to be minimized rather
than maximized. In this case, we wish to minimize the surface area S = xy + 2yz + 2xz, subject
to the constraint that the volume V = xyz is equal to a fixed constant V0 .
Applying the A-G Inequality with the same xi ’s and δi0 s yields
1
41/3 V0 ≤ S.
3
As before, this inequality is an equality if x = y = 2z. In this case, it follows from xyz = V0 that
4z 3 = V0 , which yields
1/3
1/3
1/3
V0
V0
V0
z=
, x=2
, y=2
.
4
4
4
We conclude that the minimum surface area is
S = xy + 2yz + 2xz = 12
2
V0
4
2/3
= 3(4V02 )1/3 .
2
Example We wish to maximize the volume V = πr2 h subject to the constraint that its cost
c0 = 2πr2 c1 + 2πrhc2
is fixed, where c1 is the cost per unit of area of the top and bottom, and c2 is the cost per unit of
area of the side.
In this problem, we cannot simply choose x1 and x2 to be the two terms in the cost and then
choosing δ1 = δ2 = 1/2, because the product xδ11 xδ22 would not be a function of the volume. Instead,
we must determine the proper choices of δ1 and δ2 that have this effect.
To that end, we let
γ1
γ2
δ1 =
, δ2 =
, γ1 , γ2 > 0,
γ1 + γ2
γ1 + γ2
which ensures that δ1 + δ2 = 1. We then set the xi ’s equal to each term in the cost, divided by one
of the γi ’s. The A-G Inequality then yields
c0 = 2πr2 c1 + 2πrhc2
2πr2 c1
2πrhc2
= γ1
+ γ2
γ1
γ2
2πr2 c1
2πrhc2
= (γ1 + γ2 ) δ1
+ δ2
γ1
γ2
δ
2πr2 c1 1 2πrhc2 δ2
≥ (γ1 + γ2 )
γ1
γ2
≥ (γ1 + γ2 )2π(c1 /γ1 )δ1 (c2 /γ2 )δ2 r2δ1 +δ2 hδ2
≥ (γ1 + γ2 )2π(c1 /γ1 )δ1 (c2 /γ2 )δ2 (r2γ1 +γ2 hγ2 )1/(γ1 +γ2 ) .
In order for the last expression above to be equal to a function of the volume V = πr2 h, we
must have
γ2
2γ1 + γ2
= ,
2
1
which yields γ1 = 1, γ2 = 2, and therefore
δ1 =
1
1
= ,
1+2
3
δ2 =
2
2
= .
1+2
3
We then have
1
2 2πrhc2
1/3
1/3 2/3
2
c0 = 3
2πr c1 +
≥ 3(2π)c1 (c2 /2)2/3 (r4 h2 )1/3 = 3(2π)1/3 c1 c2 V 2/3 .
3
3 2
3
Therefore, to maximize the volume, we must have x1 = x2 = c0 /(γ1 + γ2 ), which implies that
2πr2 c1
2πrhc2
c0
=
= .
γ1
γ2
3
We conclude that the volume is maximized when
r
r
c0
2rc1
2 c0 c1
r=
, h=
=
.
6πc1
c2
c2
6π
2
Example We wish to maximize the function
f (x, y, z) = xy 2 z,
x, y, z > 0,
subject to the constraint
x + y + z 2 = k,
k > 0.
To apply the A-G Inequality, we proceed as in the previous example, requiring
δi =
γi
,
γ1 + γ2 + γ3
x1 =
x
,
γ1
i = 1, 2, 3,
and
x2 =
y
,
γ2
x3 =
z2
.
γ3
From the A-G Inequality, we then obtain
k = x + y + z2
x
y
z2
= γ1 + γ2 + γ3
γ1
γ2
γ3
x
y
z2
= (γ1 + γ2 + γ3 ) δ1 + δ2 + δ3
γ1
γ2
γ3
≥ (γ1 + γ2 + γ3 )(x/γ1 )δ1 (y/γ2 )δ2 (z 2 /γ3 )δ3
≥ (γ1 + γ2 + γ3 )γ1−δ1 γ2−δ2 γ3−δ3 (xγ1 y γ2 z 2γ3 )1/(γ1 +γ2 +γ3 ) .
In order to make the lower end of this inequality a function of xy 2 z, we must have
γ2
2γ3
γ1
=
=
1
2
1
which is satisfied by choosing γ1 = 2, γ2 = 4 and γ3 = 1, which yields
2
δ1 = ,
7
4
δ2 = ,
7
4
1
δ3 = .
7
The A-G inequality then yields
k ≥ 7(1/2)2/7 (1/4)4/7 (xy 2 z)2/7 .
Equality holds when x1 = x2 = x3 = k/(γ1 + γ2 + γ3 ), or
x
y
z2
k
= =
= .
2
4
1
7
Substituting these relations into x + y + z 2 = k yields
2k
,
x=
7
4k
y=
,
7
r
z=
k
.
7
2
The A-G Inequality can also be used for unconstrained minimization. In this case, the idea is
to make the lower end of the A-G Inequality equal to a constant.
Example We wish to find the minimium of the function
f (x, y) = 4x +
x
4y
+ .
2
y
x
This can be quite difficult to solve by examination of the Hessian Hf (x, y). Instead, we try to use
the A-G Inequality as before.
We set
4x
x
4y
x1 =
, x2 =
, x3 =
,
2
δ1
δ2 y
δ3 x
where
δ1 + δ2 + δ3 = 1.
The A-G Inequality then yields
f (x, y) = δ1 x1 + δ2 x2 + δ3 x3
≥ xδ11 xδ22 xδ33
δ1 x δ2 4y δ3
4x
≥
δ1
δ2 y 2
δ3 x
δ1 δ2 δ3
4
1
4
≥
xδ1 +δ2 −δ3 y δ3 −2δ2 .
δ1
δ2
δ3
To make the lower end of the inequality a constant, we must have
δ1 + δ2 − δ3 = 0,
5
δ3 − 2δ2 = 0,
along with δ1 + δ2 + δ3 = 1. These equations are satisfied by δ1 = δ2 =
Inequality then yields
f (x, y, z) ≥ 4(41/4 )(21/2 ) ≥ 8.
1
4,
δ3 =
1
2.
The A-G
We conclude that the minimum value of f (x, y, z) is 8, and it is achieved when x1 = x2 = x3 = 8,
or
4y
4x
x
=
=
= 8.
1/4
(1/4)y 2
(1/2)x
This yields the minimizer (x∗ , y ∗ ) = (1/2, 1/2). 2
Exercises
1. Chapter 2, Exercise 15ad
2. Chapter 2, Exercise 16
3. Chapter 2, Exercise 17
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