Thomas Calculus 13th [Solutions]

292 Chapter 4 Applications of Derivatives
2
$10/ft $0.67 $18
15 ft 3 3
ft yd
fabrication and $750 for the excavation.
. The total cost of the least expensive tank is $3375, which is the sum of $2625 for
11. The area of the printing is ( y 4)( x 8) 50.
Consequently, y
50
4. The area of the paper is
x 8
A( x) x
50
4 , where 8 x . Then
x 8
2
50 50 4( x 8) 400
A ( x) 4 x
0
x 8 2
2
( x 8) ( x 8)
the critical points are 2 and 18, but 2 is not in the
domain. Thus A (18) 0 at x 18 we have a
minimum. Therefore the dimensions 18 by 9 inches
minimize the amount of paper.
1
2
12. The volume of the cone is V r h , where r x 9 y and h y 3 (from the figure in the text). Thus,
3
2 2 3
2
V ( y) (9 y )( y 3) (27 9y 32 y y ) V ( y) (9 6y 3 y ) (1 y)(3 y ). The critical
3 3
3
points are 3 and 1, but 3 is not in the domain. Thus V (1) ( 6 6(1)) 0 at y 1 we have a maximum
3
volume of V (1) (8)(4)
32
cubic units.
3 3
ab sin
13. The area of the triangle is A ( ) , where
2
ab cos
0 . Solving A ( ) 0 0 .
2 2
ab sin
Since A ( ) A 0, there is a
2 2
maximum at
2 .
2 1000
14. A volume V r h 100 h . The amount
2
r
of material is the surface area given by the sides and
2 2
bottom of the can S 2 rh r
2000
r ,
r
3
0 r . Then dS 2000
2 r 0 r 1000
0.
dr 2 2
r r
The critical points are 0 and 10 3
, but 0 is not in the
2
domain. Since d s 4000
2 0, we have a
2 3
dr r
minimum surface area when r 10 cm and
3
h
1000 10 cm. Comparing this result to the result
2 3
r
found in Example 2, if we include both ends of the
can, then we have a minimum surface area when
the can is shorterspecifically, when the height of
the can is the same as its diameter.
2 ,
1000 .
15. With a volume of 1000 cm 3 and V r h then h The amount of aluminum used per can is
2
r
2 2 3
A 8r 2 rh 8 r
2000
. Then
r
A ( r ) 16 r 2000
0
8r
1000
0 the critical points are 0 and 5, but
2 2
r
r
r 0 results in no can. Since A ( r ) 16
1000
0 we have a minimum at r 5 h
40
and h: r 8: .
3
r
2
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