Thomas Calculus 13th [Solutions]

338 Chapter 4 Applications of Derivatives
2
17. The surface area of the cylinder is S 2 r 2 rh.
From the diagram we have r H h h RH rH
R H R
and S( r) 2 r( r h)
2 r r H r H
R
2
2 1 H r 2 Hr,
where 0 .
R r R
Case 1: H R S( r ) is a quadratic equation containing the origin and concave upward S( r ) is maximum
at r R.
Case 2: H R S( r ) is a linear equation containing the origin with a positive slope S( r ) is maximum
at r R.
Case 3: H R S( r ) is a quadratic equation containing the origin and concave downward.
Then dS 4 1 H
dr R r 2 H and dS 0 4 1 H
dr R r RH
2( H R)
. we let r* RH .
2( H R)
(a) If R H 2 R , then 0 H 2R H 2( ) * RH
2( H R)
R . the right endpoint R of the interval 0 r R because S( r ) is an increasing function of r.
2
(b) If H 2 R , then r * 2R
R S
2R
r ) is maximum at r R.
(c) If H 2 R , then 2R H 2H H 2( H R)
H 1 RH R r* R.
Therefore, S( r ) is
2( H R)
2( H R)
a maximum at r r* RH .
2( H R)
Conclusion: If H (0, 2 R ], then the maximum surface area is at r R . If H (2 R , ), then the maximum is
at r r* RH .
2( H R)
18. f ( x) mx 1 1
x
f ( x)
m 1 and f ( x ) 2 0 when x 0. Then f ( x) 0 x 1 yields a minimum.
2
3
x
x
m
If f 1 0, then m 1 m 2 m 1 0 m 1 . Thus the smallest acceptable value for m is 1
m
4
4 .
19. (a) lim 2sin(5 x) 2sin(5 ) 10 sin(5 ) 10 10
3 lim x lim x
3
0 0 (5 ) 3 (5 ) 3 1
x
x
x x
3
5 x 0
x
sin(5 ) cos(3 ) 3sin(5 )sin(3 ) 5cos(5 )cos(3 )
(b) lim sin(5 ) cot(3 ) lim x x 5
0 0
sin(3 ) lim x x x x
x x
x x
x
x 0
3cos(3 x) 3
(c)
x
x
1
2x
2 2sin 2x
cos 2x
2
2x
2x
lim 2 1 2
x csc 2 x lim lim lim lim lim 1 1
x 0 x 0 sin 2x x 0 x 0 sin 2 2x x 0 cos 2 2x x 0 cos 2 2x
2 2
(d) lim (sec x tan x ) lim 1 sin x
cos
cos
lim x
x
sin x
0
x x x
3
(e) lim x sin x lim 1 cos x lim 1 cos x lim cos x 1 lim sin x lim sin x lim cos x 1
tan 2 2 2 2 2sin x
x
x x
x 1 sec x x 5 tan x x tan x x 2 tan xsec
x x x
2 2
(f)
(g)
(h)
2 2 2 2 2
lim sin( x ) 2 cos( ) (2 )sin( ) 2cos( ) 2
sin lim x x
cos sin lim x x x
x x x x x xsin x 2cos x 2
1
x x x
lim 3 2
sec x 1 lim sec x tan x sec tan sec 1 0 1
2
2 lim x x x
x x x
x
x
2 2 2
lim
3 2
( 2)( 2 4)
2
x 8 lim x x x
2 4 4 4 4
2
4 ( 2)( 2) lim x x
x x x 2 4
3
x x x x
cos
3
x
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