Part II: The Answers
290
Answers
401–500
445.
5
36π
ft/min
The volume of a cone is V
r h
= 1
3
2
π . The problem tells you that dV
dt
= 20 cubic feet per
minute and that d = 2h. Because the diameter is twice the radius, you have 2r = d = 2h
so that r = h; therefore, the volume becomes
V
h h
V
h
=
=
1
3
1
3
2
3
π
π
Take the derivative with respect to time:
dV
dt
h dh
dt
= 1
3
3
2
π
( ) ( )
Substitute in the given information and solve for dh
dt
:
20
1
3
3 12
20
1
3
3 12
5
36
2
2
=
=
=
π
π
π
( ) ( )
( ) ( )
( )
( )
dh
dt
dh
dt
When the pile is 12 feet high, the height of the pile is increasing at a rate of 5
36π
(about 0.04) feet per minute.
446.
no absolute maximum; absolute minimum: y = –1; no local maxima; local minimum: (1, –1)
There’s no absolute maximum because the graph doesn’t attain a largest y value. There
are also no local maxima. The absolute minimum is y = –1. The point (1, –1) is a local
minimum.
447.
absolute maximum: y = 4; absolute minimum: y = 0; local maximum: (5, 4); local
minima: (1, 0), (7, 2)
The absolute maximum value is 4, which the graph attains at the point (5, 4). The absolute
minimum is 0, which is attained at the point (1, 0). The point (5, 4) also corresponds to a
local maximum, and the local minima occur at (1, 0) and (7, 2).
290
Answers
401–500
445.
5
36π
ft/min
The volume of a cone is V
r h
= 1
3
2
π . The problem tells you that dV
dt
= 20 cubic feet per
minute and that d = 2h. Because the diameter is twice the radius, you have 2r = d = 2h
so that r = h; therefore, the volume becomes
V
h h
V
h
=
=
1
3
1
3
2
3
π
π
Take the derivative with respect to time:
dV
dt
h dh
dt
= 1
3
3
2
π
( ) ( )
Substitute in the given information and solve for dh
dt
:
20
1
3
3 12
20
1
3
3 12
5
36
2
2
=
=
=
π
π
π
( ) ( )
( ) ( )
( )
( )
dh
dt
dh
dt
When the pile is 12 feet high, the height of the pile is increasing at a rate of 5
36π
(about 0.04) feet per minute.
446.
no absolute maximum; absolute minimum: y = –1; no local maxima; local minimum: (1, –1)
There’s no absolute maximum because the graph doesn’t attain a largest y value. There
are also no local maxima. The absolute minimum is y = –1. The point (1, –1) is a local
minimum.
447.
absolute maximum: y = 4; absolute minimum: y = 0; local maximum: (5, 4); local
minima: (1, 0), (7, 2)
The absolute maximum value is 4, which the graph attains at the point (5, 4). The absolute
minimum is 0, which is attained at the point (1, 0). The point (5, 4) also corresponds to a
local maximum, and the local minima occur at (1, 0) and (7, 2).
