Thomas Calculus 13th [Solutions]

294 Chapter 4 Applications of Derivatives
2
2 (200 3 )
r r
. The critical point for 0 r 10 occurs at r
200
10
2.
Since V ( r ) 0 for 0 r 10
2
2
100 r
3 3
3
and V ( r ) 0 for 10
2
r 10, the critical point corresponds to the maximum volume. The dimensions are
3
r 10
2
8.16 cm and h
20
11.55 cm, and the volume is
4000
3
2418.40 cm .
3
3
3 3
20. (a) From the diagram we have 4x 108 and
V x
2 . The volume of the box is
2
V ( x) x (108 4 x ), where 0 x 27. Then
2
V ( x) 216x 12x 12 x(18 x ) 0 the
critical points are 0 and 18, but x 0 results in
no box. Since V ( x) 216 24x 0 at x 18
we have a maximum. The dimensions of the
box are 18 18 36 in.
2 108
4
(b) In terms of length, V ( ) x . The
graph indicates that the maximum volume
occurs near 36, which is consistent with the
result of part (a).
2
21. (a) From the diagram we have 3h 2w 108 and
2 2 3 2 3 3
V h w V ( h) h 54 h 54 h h .
(b)
2 2
9 2 9
2 2
Then V ( h) 108 h h h(24 h) 0
h 0 or h 24, but h 0 results in no box.
Since V ( h) 108 9h 0 at h 24, we
have a maximum volume at h 24 and
w 54 h 18.
3
2
22. From the diagram the perimeter is P 2r 2 h r,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light
2
transmitted proportional to A 2rh 1
r
4
1 2 2 3 2
4 4
3 2P
2 8 3
4P
2 P (4 ) P
8 3 8 3 8 3
r( P 2 r r) r rP 2 r r . Then
dA P 4r r 0 r
dr
2 h P . Therefore,
2r
8
gives the proportions that admit the most
h 4
2
light since d A 4
3
0.
2
dr 2
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