Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 203
69. From the figure, tan α = 5 2
and tan β =
x 3 − x .Since
5
α + β + θ = 180 ◦ = π, θ = π − tan −1 x
2
− 5
1
−
x 2
dθ
dx = − 1
5
1+
x
Now dθ
dx =0 ⇒ 5
x 2 +25 = 2
x 2 − 6x +13
= x2
x 2 +25 ·
1+
5 (3 − x)2
−
x2 (3 − x) 2 +4 ·
2
3 − x
2
− tan −1 3 − x
2
2
(3 − x) 2
2
(3 − x) 2 .
⇒ 2x 2 +50=5x 2 − 30x +65 ⇒
⇒
3x 2 − 30x +15=0 ⇒ x 2 − 10x +5=0 ⇒ x =5± 2 √ 5. We reject the root with the + sign, since it is larger
than 3.
dθ/dx > 0 for x5 − 2 √ 5,soθ is maximized when
|AP | = x =5− 2 √ 5 ≈ 0.53.
71. In the small triangle with sides a and c and hypotenuse W , sin θ = a W and
cos θ = c W . In the triangle with sides b and d and hypotenuse L, sin θ = d L and
cos θ = b .Thus,a = W sin θ, c = W cos θ, d = L sin θ,andb = L cos θ,sothe
L
area of the circumscribed rectangle is
A(θ)=(a + b)(c + d) =(W sin θ + L cos θ)(W cos θ + L sin θ)
= W 2 sin θ cos θ + WLsin 2 θ + LW cos 2 θ + L 2 sin θ cos θ
= LW sin 2 θ + LW cos 2 θ +(L 2 + W 2 )sinθ cos θ
= LW (sin 2 θ +cos 2 θ)+(L 2 + W 2 ) · 1
2 · 2sinθ cos θ = LW + 1 2 (L2 + W 2 )sin2θ, 0 ≤ θ ≤ π 2
This expression shows, without calculus, that the maximum value of A(θ) occurs when sin 2θ =1 ⇔ 2θ = π 2
⇒
θ = π . So the maximum area is A
π
4 4 = LW +
1
2 (L2 + W 2 )= 1 2 (L2 +2LW + W 2 )= 1 (L + W 2 )2 .
73. (a) If k = energy/km over land, then energy/km over water =1.4k.
So the total energy is E =1.4k √ 25 + x 2 + k(13 − x), 0 ≤ x ≤ 13,
and so dE
dx =
1.4kx
− k.
(25 + x 2 1/2
)
Set dE
dx =0: 1.4kx = k(25 + x2 ) 1/2 ⇒ 1.96x 2 = x 2 +25 ⇒ 0.96x 2 =25 ⇒ x = 5 √
0.96
≈ 5.1.
Testing against the value of E at the endpoints: E(0) = 1.4k(5) + 13k =20k, E(5.1) ≈ 17.9k, E(13) ≈ 19.5k.
Thus, to minimize energy, the bird should fly to a point about 5.1 km from B.