110 √
=
Chapter 3: Application of Differentiation The function is increasing on to the left of d = 200 and decreasing to the right of d = 200 therefore there is a relative maximum at d = 200. We find h(200)
2
We find f ( 3π 4) 3π ) 4
f(
= =
h(200) = = = 86.6 −0.002(200)2 + 0.8(200) + 6.6 −80 + 160 + 6.6
3π ( ) 3 cos( ) −cos33π 2 4 4 √ 2 −
We find f(π)
There is a relative maximum at (200,86.6) fπ()
= =
80
( )4 3 cosπ () −cos33π 2 −1
60
We find f ( 5π 4) f(
5π 4
)
= =
5π 3 cos( ) −cos35π 2 ( ) 4 4 √ − 2
40
20
We find f ( 7π 4) f(
7π 4
100
0
)
= =
s37π 7π 3 cos( ) − 2co ( ) 4 4 √2
√ 4, 2), ( π,√ −1) and Thereisarelativemaximumat(π √ 3π − (7π 2) and 4, 2) and a relative minimum at (4 , − √ (5π 4, 2).Wesketchthegraph
.
T (t) = −0.1t2 + 1.2t + 98.6, 0 ≤ t ≤ 12
95. T ′(t) = −0.2t + 1.2 T′(t) exists for all real numbers. Solve T′(t) = 0. −0.2t+1.2 = 0 −0.2t = t =
3 y2 1 1
2
–1
3 x
4
5
6
400
94 Answers vary.
4
0
300
200 d
−1.2 6
The only critical point is at t = 6. We use it to divide the interval [0,12] (the domain of T(t)) into two intervals, A: [0, 6) and B: (6, 12].
–2
A: Test 0, T′(0) =
–3
B: Test 7, T′(7) = Since T(t) is increasing on [0, 6) and decreasing on (6, 12], there is a relative maximum at x = 6. T (6) = −0.1(6)2 + 1.2(6) + 98.6 = 102.2 There is a relative maximum at (6, 102.2◦). We sketch the graph.
–4
47. - 92. Left to the student . 93. 2h(d) . = −0.002d + 0.8d + 6 6 h′ (d) = −0.004d + 0.8
−0.2(0) −0.2(7) + + 1.2 1.2 = = 1.2 −0.2><00
102
Solve h′(d) = 0.
101.5
−0.004d+0.8 = 0 − 0.004d =
d =
A: Test 100, f′(50) = 0.4 > 0
101
−0.8 20
100.5 100 99.5 99
B: Test 300, f′(50) = 0.4−< 0
0
2
4
6 t
8
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