Unformatted text preview: en x = 0, or when
dx
2
dx
6
26
√
√
S
S
S
S
,x=
. If x = 0,
,
, then
6x = S − πx2 , 6x2 = S − πx2 , x2 =
6+π
6+π
6+π
π
S 3/2
S 3/2
S 3/2
S
S
, √ so that V is smallest when x =
, and hence when y =
,
V = 3/2 , √
6+π
6+π
6
6 6+π 6 π
thus x = y . (b) From part (a), the sum of the volumes is greatest when there is no cube.
33. 34. Let h and r be the dimensions shown in the ﬁgure, then the volume
1
is V = πr2 h. But r2 + h2 = L2 thus r2 = L2 − h2 so
3
1
1
V = π (L2 − h2 )h = π (L2 h − h3 ) for 0 ≤ h ≤ L.
3
3
√
√
dV
1
dV
2
2
= π (L − 3h ).
= 0 when h = L/ 3. If h = 0, L/ 3, 0
dh
3
dh
2π 3
then V = 0, √ L , 0 so the volume is as large as possible when
93
√
h = L/ 3 and r = 2/3L. Let r and h be the radius and height of the cone (see ﬁgure). The
slant height of any such cone will be R, the radius of the circular
sheet. Refer to the solution of Exercise 33 to ﬁnd that the largest
2π
volume is √ R3 .
93 h L r h R r Exercise Set 6.2 35. 36. 37. 38. 39. 188 √
The area of the paper is A = πrL = πr r2 + h2 , but
1
V = πr2 h = 10 thus h = 30/(πr2 ) so A = πr r2 + 900/(π 2 r4 ).
3
To simplify the computations let S = A2 ,
900
900
S = π 2 r2 r2 + 2 4 = π 2 r4 + 2 for r > 0,
πr
r
1800
4(π 2 r6 − 450)
dS
23
= 4π r − 3 =
, dS/dr = 0 when
dr
r
r3
r = 6 450/π 2 , d2 S/dr2 > 0, so S and hence A is least when
30 3 2
π /450.
r = 6 450/π 2 , h =
π
1
hb. By similar triangles (see ﬁgure)
2
R
b/2
2Rh
Rh2
=√
,b= √
so A = √
for h > 2R,
h
h2 − 2Rh
h2 − 2Rh
h2 − 2Rh
2
Rh (h − 3R) dA
dA
=2
= 0 for h > 2R when h = 3R, by the ﬁrst
,
dh
(h − 2Rh)3/2 dh
derivative test A is minimum when h = 3R. If h = 3R then
√
b = 2 3R (the triangle is equilateral). r The area of the triangle is A = 1
The volume of the cone is V = πr2 h. By similar triangles (see
3
R
r
Rh
ﬁgure) = √
,r=√
so
2 − 2Rh
2 − 2Rh
h
h
h
1
h3
h2
1
= πR2
for h > 2R,
V = πR2 2
3
h − 2Rh
3
h − 2R
1
dV
h(h − 4R) dV
= πR2
= 0 for h > 2R when h = 4R, by the
,
dh
3
(h − 2R)2 dh
ﬁrst √
derivative test V is minimum when h = 4R. If h = 4R then
r = 2R. h−R dA/dθ = 0 for 0 < θ < π/2 when cos θ = 1/2, θ = π/3. If
√
θ = 0, π/3, π/2 then A = 0, 75 3/4, 25 so the crosssectional area is
greatest when θ = π/3. h2 − 2Rh h
R
R
b /2 b h−R h2 − 2Rh h
R
R r The area is (see ﬁgure)
1
A = (2 sin θ)(4 + 4 cos θ)
2
= 4(sin θ + sin θ cos θ)
for 0 ≤ θ ≤ π/2;
dA/dθ = 4(cos θ − sin2 θ + cos2 θ)
= 4(cos θ − [1 − cos2 θ] + cos2 θ)
= 4(2 cos2 θ + cos θ − 1)
= 4(2 cos θ − 1)(cos θ + 1)
dA/dθ = 0 when θ = π/3 for 0 < θ < π/2. If θ = 0, π/3, π/2 then
√
√
A = 0, 3 3, 4 so the maximum area is 3 3.
Let b and h be the dimensions shown in the ﬁgure, then the
1
crosssectional area is A = h(5 + b). But h = 5 sin θ and
2
5
b = 5 + 2(5 cos θ) = 5 + 10 cos θ so A = sin θ(10 + 10 cos θ)
2
= 25 sin θ(1 + cos θ) for 0 ≤ θ ≤ π/2.
dA/dθ = −25 sin2 θ + 25 cos θ(1 + cos θ)
= 25(− sin2 θ + cos θ + cos2 θ)
= 25(−1 + cos2 θ + cos θ + cos2 θ)
= 25(2 cos2 θ + cos θ − 1) = 25(2 cos θ − 1)(cos θ + 1). L h 4 cos θ
2 sin θ 2
θ 4 5 cos θ 5 2 cos θ b h = 5 sin θ θ 5 5 189 Chapter 6 , k the constant of proportionality. If h is the height of the lamp above the table then
√
r2 − 2h2
h
h
dI
dI
cos φ = h/ and = h2 + r2 so I = k 3 = k 2
=k 2
=0
for h > 0,
,
2 )3/2
2 )5/2 dh
dh
(h + r
(h + r
√
√
when h = r/ 2, by the ﬁrst derivative test I is maximum when h = r/ 2. 41. Let L, L1 , and L2 be as shown in the ﬁgure, then
L = L1 + L2 = 8 csc θ + sec θ,
dL
= −8 csc cot θ + sec θ tan θ, 0 < θ < π/2
dθ
sin θ
8 cos θ
−8 cos3 θ + sin3 θ
+
;
=−
=
2
2θ
cos
sin θ
sin2 θ cos2 θ
dL
= 0 if sin3 θ = 8 cos3 θ, tan3 θ = 8, tan θ = 2 which gives the
dθ
absolute minimum for L because lim L = lim − L = +∞. If
θ→0+
√θ→π/2
√
tan θ = 2, then csc θ = √ 5/2 and sec θ = 5 so
√
√
L = 8( 5/2) + 5 = 5 5 ft. 42. I=k cos φ 40. 2 L2
L
L1
8
θ 1 Let x = number of steers per acre
w = average market weight per steer
T = total market weight per acre
then T = xw where w = 2000 − 50(x − 20) = 3000 − 50x
so
T = x(3000 − 50x) = 3000x − 50x2 for 0 ≤ x ≤ 60,
dT /dx = 3000 − 100x and dT /dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so the total
market weight per acre is largest when 30 steers per acre are allowed. 43. The daily proﬁt is
P = (revenue) − (production cost) = 100x − (100, 000 + 50x + 0.0025x2 )
= −100, 000 + 50x − 0.0025x2
for 0 ≤ x ≤ 7000, so dP/dx = 50 − 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000
is not in the interval [0, 7000], the maximum proﬁt must occur at an endpoint. When x = 0,
P = −100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and sold
daily. (b) Yes, because dP/dx > 0 when x = 7000 so proﬁt is increasing at this production level. (a) R(x) = px but p = 1000 − x so R(x) = (1000 −...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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