The Arithmetic-Geometric Mean Inequality in
Precalculus
Elsie M. Campbell and Dionne T. Bailey
Angelo State University

March 1, 2013

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Arithmetic-Geometric Mean Inequality

If a1 , a2 , a3 , . . . , an ≥ 0, then
√
a1 + a2 + a3 + · · · + an
≥ n a1 a2 a3 · · · an
n
with equality if and only if a1 = a2 = a3 = · · · = an .

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a+b √
≥ ab
2

Figure: AB = a and BC = b
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Generalization of AM-GM Inequality

If x1 , x2 , x3 , . . . , xn > 0, w1 , w2 , w3 , . . . , wn > 0, and
w1 + w2 + w3 + · · · + wn = 1, then
w2
w3
wn
1
xw
1 · x 2 · x 3 · · · x n ≤ w1 x 1 + w2 x 2 + w3 x 3 + · · · + wn x n

with equality if and only if x1 = x2 = x3 = · · · = xn .

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Some Optimization Problems
1

Find the rectangle of perimeter P that encloses the largest area.

2

Find the dimensions of the closed box of volume V that requires the
least amount of material to build.

3

Find the dimensions of the closed cylindrical can of volume V that
requires the least amount of material to build.

4

Find the sides of the triangle of perimeter P that encloses the largest
area.

5

Find the cylinder of maximum volume that can be placed inside a
cone of height H and base radius R.

6

Find the dimensions of the rectangle of largest area which can be
situated so that two adjacent sides lie on the positive coordinate axes
and the remaining vertex lies on the line 2x + 3y = 12.

7

Find the largest possible volume of a right circular cylinder inscribed
in a sphere of radius R.

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Problem 1: Find the rectangle of perimeter P that
encloses the largest area.
Let x and y be the length and width of the rectangle, respectively.
Maximize A = xy subject to
2x + 2y = P

⇒

x+y =

P
.
2

By the Arithmetic-Geometric Mean Inequality,
A = xy


x+y 2
≤
2
 2
P
=
4
2
P
=
,
16
P
with equality if and only if x = y = .
4
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Problem 2: Find the dimensions of the closed box of
volume V that requires the least amount of material to
build.
Let x, y, and z be the length, width, and height of the closed box,
respectively. Minimize S.A. = 2xy + 2xz + 2yz subject to xyz = V.
By the Arithmetic-Geometric Mean Inequality,
S.A. = 2xy + 2xz + 2yz
p
≥ 3 3 (2xy)(2xz)(2yz)
p
= 6 3 x2 y 2 z 2
2

= 6(xyz) /3
= 6V

2/3

,

with equality if and only if x = y = z = V 1/3 .
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Problem 3: Find the dimensions of the closed cylindrical
can of volume V that requires the least amount of material
to build.
Let r and h be the radius and the height of the cylindrical can,
respectively. Minimize S.A. = 2πr2 + 2πrh subject to πr2 h = V.
By the Arithmetic-Geometric Mean Inequality,
S.A. = 2πr2 + 2πrh
= 2πr2 + πrh + πrh
p
≥ 3 3 (2πr2 )(πrh)(πrh)
p
= 3 3 2π(πr2 h)2
√
3
= 3 2πV 2
√
2
3
= 3 2πV /3 ,
3
with q
equality if and only
q if h = 2r, or equivalently V = 2πr . Thus,
V
V
r = 3 2π
and h = 2 3 2π
.
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Problem 4: Find the sides of the triangle of perimeter P
that encloses the largest area.
Let a, b, and c p
be the length of the three sides of the triangle, respectively.
Maximize A = s(s − a)(s − b)(s − c) subject to a + b + c = P and
s = 12 P .
By the Arithmetic-Geometric Mean Inequality,
p
A = s(s − a)(s − b)(s − c)
q
= 12 P ( 21 P − a)( 12 P − b)( 12 P − c)
s
 1
3
1
1
( 2 P −a)+( 2 P −b)+( 2 P −c)
1
≤ 2P
3
r
=
=

1
( 2 P )4
27

2
P√
,
12 3

with equality if and only if a = b = c =
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P
3.

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Problem 5: Find the cylinder of maximum volume that can
be placed inside a cone of height H and base radius R.
Maximize Vcyl = πh2 h subject to 13 πR2 H = Vcone .
Using similar triangles, h =

Campbell and Bailey (ASU)

H(R−r)
.
R

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By the Arithmetic-Geometric Mean Inequality,
Vcyl = πr2 h


2 H(R − r)
= πr
R
πH
=
(2r2 (R − r))
2R
πH
=
(r(r)(2R − 2r))
2R


πH r + r + (2R − 2r) 3
≤
2R
3
 3
πH 2R
=
2R
3
4
= Vcone ,
9
with equality if and only if r = 2R − 2r, or equivalently r =
h = H3 .
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2R
3

and

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Problem 6: Find the dimensions of the rectangle of largest
area which can be situated so that two adjacent sides lie
on the positive coordinate axes and the remaining vertex
lies on the line 2x + 3y = 12.
Maximize A = xy subject to 2x + 3y = 12.
Using similar triangles,

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4
6

=

y
6−x .

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By the Arithmetic-Geometric Mean Inequality,
A = xy
4
= x(6 − x)
6


4 x + (6 − x) 2
≤
6
2
= 6,
with equality if and only if x = 6 − x, or equivalently x = 3 and y = 2.

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Problem 7: Find the largest possible volume of a right
circular cylinder inscribed in a sphere of radius R.
Maximize Vcyl = πr2 h subject to 34 πR3 = Vs .
Using the Pythagorean Theorem, r2 = R2 − h2

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By the Arithmetic-Geometric Mean Inequality,
(Vcyl )2 = (πr2 (2h))2
= 4π 2 h2 (R2 − h2 )2

⇒ Vcyl

= 2π 2 (2h2 )(R2 − h2 )(R2 − h2 )
 2
3
2
2
2
2
2 2h + (R − h ) + (R − h )
≤ 2π
3
 2 3
2R
= 2π 2
3
1 4
= ( πR3 )2
3 3
1
= Vs2 ,
3
1
≤ √ Vs ,
3

with equality if and only if h =
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R
√
3

√

and r =

6
3 R.

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