Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 4 REVIEW ¤ 225
79. (a) The cross-sectional area of the rectangular beam is
A =2x · 2y =4xy =4x √ 100 − x 2 , 0 ≤ x ≤ 10,so
dA
dx =4x
1
2 (100 − x 2 ) −1/2 (−2x)+(100− x 2 ) 1/2 · 4
−4x 2
=
(100 − x 2 ) + 4(100 − 1/2 x2 ) 1/2 = 4[−x2 + 100 − x 2 ]
.
(100 − x 2 ) 1/2
dA
dx =0when −x2 + 100 − x 2 =0 ⇒ x 2 =50 ⇒ x = √
50 ≈ 7.07 ⇒ y = 100 − √ 50 2 √
= 50.
Since A(0) = A(10) = 0, the rectangle of maximum area is a square.
(b)
The cross-sectional area of each rectangular plank (shaded in the figure) is
A =2x y − √ 50 =2x √ 100 − x 2 − √ 50 , 0 ≤ x ≤ √ 50,so
dA
dx =2√ 100 − x 2 − √ 50 +2x 1
2
(100 − x 2 ) −1/2 (−2x)
=2(100− x 2 ) 1/2 − 2 √ 50 −
2x 2
(100 − x 2 ) 1/2
Set dA
dx =0: (100 − x2 ) − √ 50 (100 − x 2 ) 1/2 − x 2 =0 ⇒ 100 − 2x 2 = √ 50 (100 − x 2 ) 1/2 ⇒
10,000 − 400x 2 +4x 4 = 50(100 − x 2 ) ⇒ 4x 4 − 350x 2 + 5000 = 0 ⇒ 2x 4 − 175x 2 +2500=0 ⇒
x 2 = 175 ± √ 10,625
4
≈ 69.52 or 17.98 ⇒ x ≈ 8.34 or 4.24. But8.34 > √ 50, sox 1 ≈ 4.24 ⇒
y − √ 50 = 100 − x 2 1 − √ 50 ≈ 1.99. Each plank should have dimensions about 8 1 inches by 2 inches.
2
(c) From the figure in part (a), the width is 2x and the depth is 2y, so the strength is
S = k(2x)(2y) 2 =8kxy 2 =8kx(100 − x 2 )=800kx − 8kx 3 , 0 ≤ x ≤ 10. dS/dx =800k − 24kx 2 =0when
24kx 2 =800k ⇒ x 2 = 100
3
⇒ x = √ 10
200
3
⇒ y =
3
= 10 √ √
2
3
= √ 2 x. SinceS(0) = S(10) = 0, the
maximum strength occurs when x = √ 10
3
. The dimensions should be √ 20
3
≈ 11.55 inches by 20 √ √
2
3
≈ 16.33 inches.
81. We first show that
x
1+x < 2 tan−1 x for x>0. Letf(x) =tan −1 x − x
1+x .Then 2
f 0 (x) = 1
1+x − 1(1 + x2 ) − x(2x)
= (1 + x2 ) − (1 − x 2 ) 2x 2
= > 0 for x>0. Sof(x) is increasing
2 (1 + x 2 ) 2 (1 + x 2 ) 2 (1 + x 2 )
2
on (0, ∞). Hence, 0