Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 199
43. S =6sh − 3 2 s2 cot θ +3s 2 √ 3
2 csc θ
(a) dS
dθ = 3 2 s2 csc 2 θ − 3s 2 √ 3
2 csc θ cot θ or 3 2 s2 csc θ csc θ − √ 3cotθ .
(b) dS
dθ =0when csc θ − √ 3cotθ =0 ⇒ 1
sin θ − √ 3 cos θ =0 ⇒ cos θ = √
1
sin θ 3
. The First Derivative Test shows
that the minimum surface area occurs when θ =cos −1 √
1
3
≈ 55 ◦ .
(c) If cos θ = 1 √
3
,thencot θ = 1 √
2
and csc θ = √ 3
√
2
, so the surface area is
S =6sh − 3 2 s2 1
√
2
+3s 2 √ √
3 3
2
=6sh + 6
2 √ 2 s2 =6s h + 1
2 √ s 2
√
2
=6sh − 3
2 √ 2 s2 + 9
2 √ 2 s2
√
x2 +25
45. Here T (x) =
6
+ 5 − x , 0 ≤ x ≤ 5 ⇒ T 0 (x) =
8
x
6 √ x 2 +25 − 1 8 =0 ⇔ 8x =6√ x 2 +25 ⇔
16x 2 =9(x 2 + 25) ⇔ x = 15 √
7
.But 15 √
7
> 5, soT has no critical number. Since T (0) ≈ 1.46 and T (5) ≈ 1.18,he
should row directly to B.
47. There are (6 − x) km over land and √ x 2 +4km under the river.
We need to minimize the cost C (measured in $100,000) of the pipeline.
C(x) =(6− x)(4) + √ x 2 +4 (8)
⇒
C 0 (x) =−4+8· 1
2 (x2 +4) −1/2 8x
(2x) =−4+ √
x2 +4 .
C 0 (x) =0 ⇒ 4=
8x
√
x2 +4
⇒ √ x 2 +4=2x ⇒ x 2 +4=4x 2 ⇒ 4=3x 2 ⇒ x 2 = 4 3
⇒
x =2/ √ 3 [0 ≤ x ≤ 6]. Compare the costs for x =0, 2/ √ 3,and6. C(0) = 24 + 16 = 40,
C 2/ √ 3 =24− 8/ √ 3+32/ √ 3=24+24/ √ 3 ≈ 37.9,andC(6) = 0 + 8 √ 40 ≈ 50.6. So the minimum cost is about
$3.79 million when P is 6 − 2/ √ 3 ≈ 4.85 km east of the refinery.
49. The total illumination is I(x) = 3k
x + k
, 0