The solution of
V
(
x
) = 0 in the domain of
V
is
x
= 10.
Because
V
(10) = 1000
> V
(1) = 149
.
5
>
V
(
10
√
3
)
= 0, this shows that
x
= 10 maximizes
V
and that the maximum value of
V
is 1000 cm
3
.
3.6.18:
Let
x
denote the radius of the cylinder and
y
its height. Then its total surface area is
πx
2
+2
πxy
=
300
π
, so
x
2
+ 2
xy
= 300. We are to maximize its volume
V
=
πx
2
y
. Because
y
=
300
−
x
2
2
x
,
it follows that
V
=
V
(
x
) =
π
2
(300
x
−
x
3
)
,
0
x
10
√
3
.
It is then easy to show that
x
= 10 maximizes
V
(
x
), that
y
=
x
= 10 as well, and thus that the maximum
possible volume of the can is 1000
π
in.
3
3.6.19:
Let
x
be the length of the edge of each of the twelve small squares. Then each of the three cross-
shaped pieces will form boxes with base length 1
−
2
x
and height
x
, so each of the three will have volume
x
(1
−
2
x
)
2
. Both of the two cubical boxes will have edge
x
and thus volume
x
3
. So the total volume of all
five boxes will be
V
(
x
) = 3
x
(1
−
2
x
)
2
+ 2
x
3
= 14
x
3
−
12
x
2
+ 3
x,
0
x
1
2
.
Now
V
(
x
) = 42
x
2
−
24
x
+ 3;
V
(
x
) = 0 when 14
x
2
−
8
x
−
1 = 0.
The quadratic formula gives the two
solutions
x
=
1
14
(
4
±
√
2
)
. These are approximately 0
.
3867 and 0
.
1847, and both lie in the domain of
V
.
Finally,
V
(0) = 0,
V
(0
.
1847)
≈
0
.
2329,
V
(0
.
3867)
≈
0
.
1752, and
V
(0
.
5) = 0
.
25. Therefore, to maximize
V
,
one must cut each of the three large squares into four smaller squares of side length
1
2
each and form the
resulting twelve squares into two cubes. At maximum volume there will be only two boxes, not five.
3.6.20:
Let
x
be the length of each edge of the square base of the box and let
h
denote its height. Then its
volume is
V
=
x
2
h
. The total cost of the box is $144, hence
4
xh
+
x
2
+ 2
x
2
= 144
and thus
h
=
144
−
3
x
2
4
x
.
Therefore
V
=
V
(
x
) =
x
4
(
144
−
3
x
2
)
= 36
x
−
3
4
x
3
.
172

The natural domain of
V
is the open interval
(
0
,
4
√
3
)
, but we may adjoin the endpoints as usual to obtain
a closed interval. Also
V
(
x
) = 36
−
9
4
x
2
,
so
V
(
x
) always exists and is zero only at
x
= 4 (reject the other root
x
=
−
4). Finally,
V
(
x
) = 0 at the
endpoints of its domain, so
V
(4) = 96 (ft
3
) is the maximum volume of such a box. The dimensions of the
largest box are 4 ft square on the base by 6 ft high.
3.6.21:
Let
x
denote the edge length of one square and
y
that of the other. Then 4
x
+4
y
= 80, so
y
= 20
−
x
.
The total area of the two squares is
A
=
x
2
+
y
2
, so
A
=
A
(
x
) =
x
2
+ (20
−
x
)
2
= 2
x
2
−
40
x
+ 400
,
with domain (0
,
20); adjoin the endpoints as usual. Then
A
(
x
) = 4
x
−
40, which always exists and which
vanishes when
x
= 10. Now
A
(0) = 400 =
A
(20), whereas
A
(10) = 200. So to minimize the total area of the
two squares, make two equal squares. To maximize it, make only one square.
3.6.22:
Let
r
be the radius of the circle and
x
the edge of the square.
We are to maximize total area
A
=
πr
2
+
x
2
given the side condition 2
πr
+ 4
x
= 100. From the last equation we infer that
x
=
100
−
2
πr
4
=
50
−
πr
2
.
So
A
=
A
(
r
) =
πr
2
+
1
4
(50
−
πr
)
2
=
π
+
1
4
π
2
r
2
−
25
πr
+ 625
for 0
r
50
/π
(because
x
0). Now
A
(
r
) = 2
π
+
1
4
π
2
r
−
25
π
;
A
(
r
) = 0
when
r
=
25
2 +
π
2
=
50
π
+ 4
;
that is, when
r
≈
7. Finally,
A
(0) = 625
,
A
50
π
≈
795
.
77
and
A
50
π
+ 4
≈
350
.
06
.