Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 REVIEW ¤ 223
53. By similar triangles, y x = r
√ , so the area of the triangle is
x2 − 2rx
A(x) = 1 2 (2y)x = xy = rx 2
√
x2 − 2rx
⇒
A 0 (x) = 2rx √ x 2 − 2rx − rx 2 (x − r)/ √ x 2 − 2rx
x 2 − 2rx
when x =3r.
= rx2 (x − 3r)
=0
(x 2 3/2
− 2rx)
A 0 (x) < 0 when 2r 3r. Sox =3r gives a minimum and A(3r) = r(9r2 )
√
3 r
=3 √ 3 r 2 .
55. We minimize L(x) =|PA| + |PB| + |PC| =2 √ x 2 +16+(5− x),
57. v = K
L
C + C L ⇒ dv
dL =
0 ≤ x ≤ 5. L 0 (x) =2x √ x 2 +16 − 1=0 ⇔ 2x = √ x 2 +16 ⇔
4x 2 = x 2 +16 ⇔ x = √ 4
3
. L(0) = 13, L
minimum occurs when x = 4 √
3
≈ 2.3.
√
4
3
≈ 11.9, L(5) ≈ 12.8,sothe
K 1
2 (L/C)+(C/L) C − C
=0 ⇔ 1 L 2 C = C ⇔ L 2 = C 2 ⇔ L = C.
L 2
This gives the minimum velocity since v 0 < 0 for 0 C.
59. Let x denote the number of $1 decreases in ticket price. Then the ticket price is $12 − $1(x), and the average attendance is
11,000 + 1000(x). Nowtherevenuepergameis
R(x) =(price per person) × (number of people per game)
=(12− x)(11,000 + 1000x) =−1000x 2 +1000x +132,000
for 0 ≤ x ≤ 4 [since the seating capacity is 15,000] ⇒ R 0 (x) =−2000x +1000=0 ⇔ x =0.5. Thisisa
maximum since R 00 (x) =−2000 < 0 for all x. Now we must check the value of R(x) =(12− x)(11,000 + 1000x) at
x =0.5 and at the endpoints of the domain to see which value of x gives the maximum value of R.
R(0) = (12)(11,000) = 132,000, R(0.5) = (11.5)(11,500) = 132,250,andR(4) = (8)(15,000) = 120,000. Thus,the
maximum revenue of $132,250 per game occurs when the average attendance is 11,500 and the ticket price is $11.50.
61. f(x) =x 5 − x 4 +3x 2 − 3x − 2 ⇒ f 0 (x) =5x 4 − 4x 3 +6x − 3,sox n+1 = x n − x5 n − x 4 n +3x 2 n − 3x n − 2
.
5x 4 n − 4x 3 n +6x n − 3
Now x 1 =1 ⇒ x 2 =1.5 ⇒ x 3 ≈ 1.343860 ⇒ x 4 ≈ 1.300320 ⇒ x 5 ≈ 1.297396 ⇒
x 6 ≈ 1.297383 ≈ x 7 ,sotherootin[1, 2] is 1.297383, to six decimal places.