Solução_Calculo_Stewart_6e

F.
198 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
35. The volume is V = πr 2 h and the surface area is
V
S(r) =πr 2 +2πrh = πr 2 +2πr = πr 2 + 2V πr 2 r .
S 0 (r) =2πr − 2V
r 2 =0 ⇒ 2πr3 =2V ⇒ r = 3
V
π cm.
This gives an absolute minimum since S 0 (r) < 0 for 0 0 for r> 3
V
π .
When r = 3
V
π , h =
V
πr = V V 2 π(V/π) = 3 2/3 π cm.
37. h 2 + r 2 = R 2 ⇒ V = π 3 r2 h = π 3 (R2 − h 2 )h = π 3 (R2 h − h 3 ).
V 0 (h) = π 3 (R2 − 3h 2 )=0when h = 1 √
3
R. This gives an absolute maximum, since
V 0 (h) > 0 for 0 1 √
3
R. The maximum volume is
V
√
1
3
R
= π 3
√
1
3
R 3 − 1
3 √ 3 R3
39. By similar triangles, H R = H − h
r
so we’ll solve (1) for h.
h = H − Hr
R
=
HR − Hr
R
Hr
R = H − h
= 2
9 √ 3 πR3 .
(1). The volume of the inner cone is V = 1 3 πr2 h,
⇒
= H (R − r) (2).
R
Thus, V(r) = π 3 r2 · H πH
(R − r) =
R 3R (Rr2 − r 3 )
V 0 (r) = πH
3R (2Rr − 3r2 )= πH r(2R − 3r).
3R
V 0 (r) =0 ⇒ r =0or 2R =3r ⇒ r = 2 R and from (2), h = H
3 R −
2
R
R = H 1
3
R R = 1 H.
3 3
V 0 (r) changes from positive to negative at r = 2 3
R, so the inner cone has a maximum volume of
V = 1 3 πr2 h = 1 3 π 2
3 R 2 1
3 H = 4 27 · 1
3 πR2 H, which is approximately 15% of the volume of the larger cone.
41. P (R) = E2 R
(R + r) 2 ⇒
P 0 (R)= (R + r)2 · E 2 − E 2 R · 2(R + r)
[(R + r) 2 ] 2
⇒
= (R2 +2Rr + r 2 )E 2 − 2E 2 R 2 − 2E 2 Rr
(R + r) 4
= E2 r 2 − E 2 R 2
= E2 (r 2 − R 2 )
= E2 (r + R)(r − R)
= E2 (r − R)
(R + r) 4 (R + r) 4 (R + r) 4 (R + r) 3
P 0 (R) =0 ⇒ R = r ⇒ P (r) = E2 r
(r + r) = E2 r
2 4r = E2
2 4r .
The expression for P 0 (R) shows that P 0 (R) > 0 for Rr. Thus, the maximum value of the
power is E 2 /(4r), and this occurs when R = r.