5128_ch04ansTE_pp652-661
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Additional Answers <strong>661</strong><br />
(c) Maximum volume 66.019 in 3 when x 1.962 in.<br />
(d) V(x) 6x 2 – 50 x 75 which is zero at x 25 57<br />
1.962.<br />
6<br />
(b)<br />
y<br />
2<br />
70. (a) The only x-value for which f has a relative maximum is x – 2. That<br />
is the only place where the derivative of f goes from positive to<br />
negative.<br />
(b) The only x-value for which f has a relative minimum is x 0. That is<br />
the only place where the derivative of f goes from negative to positive.<br />
(c) The graph of f is concave up on (1, 1) and on (2, 3). Those are the intervals<br />
on which the derivative of f is increasing.<br />
(d)<br />
y<br />
LRAM 1.25<br />
2<br />
x<br />
6. (a)<br />
2<br />
y<br />
–3 3<br />
x<br />
72. (a) V a 2 b, and b 60 2a<br />
15 a 4<br />
2 , so V 15pa2 p a 3 .<br />
2<br />
Thus d V<br />
30pa 3p a 2<br />
3 pa(20 a). The domain of consideration<br />
da<br />
2 2<br />
for a in this problem is (0, 30), so a 20 is the only critical number.<br />
The cylinder of maximum volume is formed when a 20 and b 5.<br />
(b) The sign graph for the derivative d V<br />
3 pa(20 – a) on the interval<br />
da<br />
2<br />
(0, 30) is as follows:<br />
<br />
x<br />
0 20 30<br />
By the First Derivative Test, there is a maximum at x 20.<br />
(b)<br />
2<br />
y<br />
RRAM 1.25<br />
2<br />
x<br />
CHAPTER 5<br />
2<br />
x<br />
Section 5.1<br />
Exercises 5.1<br />
5. (a) y<br />
2<br />
7.<br />
MRAM 1.375<br />
n LRAM n MRAM n RRAM n<br />
10 1.32 1.34 1.32<br />
50 1.3328 1.3336 1.3328<br />
100 1.3332 1.3334 1.3332<br />
500 1.333328 1.333336 1.333328<br />
R<br />
2<br />
x<br />
13.<br />
14.<br />
n<br />
MRAM<br />
10 526.21677<br />
20 524.25327<br />
40 523.76240<br />
80 523.63968<br />
160 523.60900<br />
n error % error<br />
10 2.61799 0.5<br />
20 0.65450 0.125<br />
40 0.16362 3.12 10 2<br />
80 0.04091 7.8 10 3<br />
160 0.01023 2 10 3