It follows that sec
θ
= 3 or sec
θ
=
−
1. We reject the latter as before, and find that sec
θ
= 3, so
θ
≈
1
.
23095
(radians). The resulting minimum volume of the cone is
8
3
πR
3
, twice the volume of the sphere!
r
Θ
Θ
h
R
3.7.82:
Let
L
be the length of the crease. Then the right triangle of which
L
is the hypotenuse has sides
L
cos
θ
and
L
sin
θ
. Now 20 =
L
sin
θ
+
L
sin
θ
cos 2
θ
, so
L
=
L
(
θ
) =
20
(sin
θ
)(1 + cos 2
θ
)
,
0
< θ
π
4
.
Next,
dL
dθ
= 0 when
(cos
θ
)(1 + cos 2
θ
) = (sin
θ
)(2 sin 2
θ
);
(cos
θ
)(2 cos
2
θ
) = 4 sin
2
θ
cos
θ
;
203

so cos
θ
= 0 (which is impossible given the domain of
L
) or
cos
2
θ
= 2 sin
2
θ
= 2
−
2 cos
2
θ
;
cos
2
θ
=
2
3
.
This implies that cos
θ
=
1
3
√
6 and sin
θ
=
1
3
√
3. Because
L
→
+
∞
as
θ
→
0
+
, we have a minimum either
at the horizontal tangent just found or at the endpoint
θ
=
1
4
π
. The value of
L
at
1
4
π
is 20
√
2
≈
28
.
28 and
at the horizontal tangent we have
L
= 15
√
3
≈
25
.
98. So the shortest crease is obtained when cos
θ
=
1
3
√
6;
that is, for
θ
approximately 35
◦
15 52 . The bottom of the crease should be one-quarter of the way across
the page from the lower left-hand corner.
3.7.83:
Set up coordinates so the diameter is on the
x
-axis and the equation of the circle is
x
2
+
y
2
= 1; let
(
x , y
) denote the northwest corner of the trapezoid. The chord from (1
,
0) to (
x, y
) forms a right triangle
with hypotenuse 2, side
z
opposite angle
θ
, and side
w
; moreover,
z
= 2 sin
θ
and
w
= 2 cos
θ
. It follows that
y
=
w
sin
θ
= 2 sin
θ
cos
θ
and
−
x
= 1
−
w
cos
θ
=
−
cos 2
θ.
Now
A
=
y
(1
−
x
) = (2 sin
θ
cos
θ
)(1
−
cos 2
θ
) = 4 sin
θ
cos
θ
sin
2
θ,
and therefore
A
=
A
(
θ
) = 4 sin
3
θ
cos
θ,
π
4
θ
π
2
.
A
(
θ
) = 12 sin
2
θ
cos
2
θ
−
4 sin
4
θ
= (4 sin
2
θ
)(3 cos
2
θ
−
sin
2
θ
)
.
To solve
A
(
θ
) = 0, we note that sin
θ
= 0, so we must have 3 cos
2
θ
= sin
2
θ
; that is tan
2
θ
= 3. It follows
that
θ
=
1
3
π
. The value of
A
here exceeds its value at the endpoints, so we have found the maximum value
of the area—it is
3
4
√
3.
3.7.84:
Let
θ
=
α/
2 (see Fig. 3.7.18 of the text) and denote the radius of the circular log by
r
. Using the
technique of the solution of Problem 82, we find that the area of the hexagon is
A
=
A
(
θ
) = 8
r
2
sin
3
θ
cos
θ,
0
θ
π
2
.
After some simplifications we also find that
dA
dθ
= 8
r
2
(sin
2
θ
)(4 cos
2
θ
−
1)
.
204

Now
dA/dθ
= 0 when sin
θ
= 0 and when cos
θ
=
1
2
.
When sin
θ
= 0,
A
= 0; also,
A
(0) = 0 =
A
(
1
2
π
)
.
Therefore
A
is maximal when cos
θ
=
1
2
:
θ
=
1
3
π
.
When this happens, we find that
α
=
2
3
π
and that
β
=
π
−
θ
=
2
3
π
. Therefore the figure of maximal area is a regular hexagon.
3.7.85:
The area in question is the area of the sector minus the area of the triangle in Fig. 3.7.19 and turns
out to be
A
=
1
2
r
2
θ
−
r
2
cos
θ
2
sin
θ
2
=
1
2
r
2
(
θ
−
sin
θ
) =
s
2
(
θ
−
sin
θ
)
2
θ
2
because
s
=
rθ
. Now
dA
dθ
=
s
2
(2 sin
θ
−
θ
cos
θ
−
θ
)
2
θ
3
,
so
dA/dθ
= 0 when
θ
(1 + cos
θ
) = 2 sin
θ
. Let
θ
= 2
x
; note that 0
< x
π
because 0
< θ
2
π
. So the
condition that
dA/dθ
= 0 becomes
x
=
sin
θ
1 + cos
θ
= tan
x.