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√
dC/dx = 0 when x = 3 3375 = 15, d2 C/dx2 > 0 so C is least when x = 15, y = 10. 21. Let x = length of each edge of base, y = height, k = $/cm2 for the sides. The cost is
C = (2k )(2x2 ) + (k )(4xy ) = 4k (x2 + xy ), but V = x2 y = 2000 thus y = 2000/x2 so
C = 4k (x2 + 2000/x) for x > 0 dC/dx = 4k (2x − 2000/x2 ), dC/dx = 0 when
√
x = 3 1000 = 10, d2 C/dx2 > 0 so C is least when x = 10, y = 20. 22. Let x and y be the dimensions shown in the ﬁgure and V the
volume, then V = x2 y . The amount of material is to be 1000 ft2 ,
thus (area of base) + (area of sides) = 1000, x2 + 4xy = 1000,
1
1000 − x2
1000 − x2
so V = x2
= (1000x − x3 ) for
y=
4x √
4x
4
0 < x ≤ 10 10.
dV
1
dV
= (1000 − 3x2 ),
= 0 when x = 1000/3 = 10 10/3.
dx
4
dx
√
5000
10/3, 0;
If x = 0, 10 10/3, 10 10 then V = 0,
3
the volume is greatest for x = 10 10/3 ft and y = 5 10/3 ft. y x x 23. Let x = height and width, y = length. The surface area is S = 2x2 + 3xy where x2 y = V , so y = V /x2
and S = 2x2 + 3V /x for x > 0; dS/dx = 4x − 3V /x2 , dS/dx = 0 when x = 3 3V /4, d2 S/dx2 > 0 so S
4 3 3V
3 3V
,y=
.
is minimum when x =
4
3
4 24. Let r and h be the dimensions shown in the ﬁgure, then the volume
of the inscribed cylinder is V = πr2 h. But
2
h
h2
r2 +
= R2 thus r2 = R2 −
2
4
h
h2
h3
R
h
2
2
2
h=π R h−
so V = π R −
4
4
dV
dV
3
r
= π R 2 − h2 ,
=0
for 0 ≤ h ≤ 2R.
dh
4
dh
√
√
4π
when h = 2R/ 3. If h = 0, 2R/ 3, 2R then V = 0, √ R3 , 0 so the volume is largest when
33
√
h = 2R/ 3 and r = 2/3R. Exercise Set 6.2 186 25. Let r and h be the dimensions shown in the ﬁgure, then the surface
area is S = 2πrh + 2πr2 .
2
√
h
But r2 +
= R2 thus h = 2 R2 − r2 so
h
√2
S = 4πr R2 − r2 + 2πr2 for 0 ≤ r ≤ R,
R
h
2
4π (R2 − 2r2 )
dS
dS
=√
= 0 when
+ 4πr;
dr
dr
R2 − r 2
r
R2 − 2r2
√
= −r
(i)
R2 − r 2
√
R2 − 2r2 = −r R2 − r2
R4 − 4R2 r2 + 4r4 = r2 (R2 − r2 )
5r4 − 5R2 r2 + R4 = 0
√
√
√
5± 5
5R2 ± 25R4 − 20R4
5± 5 2
2
and using the quadratic formula r =
=
R,r=
R, of which
10
10
10
√
√
√
5+ 5
5+ 5
R satisﬁes (i). If r = 0,
R, 0 then S = 0, (5 + 5)πR2 , 2πR2 so the surface
only r =
10
10
√
√
5+ 5
5− 5
R and, from h = 2 R2 − r2 , h = 2
R.
area is greatest when r =
10
10 26. Let R and H be the radius and height of the cone, and r and h the
radius and height of the cylinder (see ﬁgure), then the volume of the
r
H −h
cylinder is V = πr 2 h. By similar triangles (see ﬁgure)
=
H
R
thus
H
H
H
h = (R − r) so V = π (R − r)r2 = π (Rr2 − r3 ) for 0 ≤ r ≤ R.
R
R
R
dV
H
dV
H
= π (2Rr − 3r2 ) = π r(2R − 3r),
= 0 for 0 < r < R
dr
R
R
dr
when r = 2R/3. If r = 0, 2R/3, R then V = 0, 4πR2 H/27, 0 so the
4
41 2
4πR2 H
=
πR H = · (volume of cone).
maximum volume is
27
93
9 r
H
h R 27. From (13), S = 2πr2 + 2πrh. But V = πr2 h thus h = V /(πr2 ) and so S = 2πr2 + 2V /r for r > 0.
dS/dr = 4πr − 2V /r2 , dS/dr = 0 if r = 3 V /(2π ). Since d2 S/dr2 = 4π + 4V /r3 > 0, the minimum
surface area is achieved when r = 3 V /2π and so h = V /(πr2 ) = [V /(πr3 )]r = 2r. 28. V = πr2 h where S = 2πr2 + 2πrh so h = 29. The surface area is S = πr2 + 2πrh where V = πr2 h = 500 so
h = 500/(πr2 ) and S = πr2 + 1000/r for r > 0;
dS/dr = 2πr − 1000/r2 = (2πr3 − 1000)/r2 , dS/dr = 0 when
r = 3 500/π , d2 S/dr2 > 0 for r > 0 so S is minimum when
500
500
500
3
500/π
r = 3 500/π and h = 2 = 3 r =
πr
πr
π (500/π ) 1
S − 2πr2
, V = (Sr − 2πr3 ) for r > 0.
2πr
2
1
d2 V
dV
= (S − 6πr2 ) = 0 if r = S/(6π ),
= −6πr < 0 so V is maximum when
dr
2
dr2
S − 2πr2
S − 2πr2
S − S/3
r = S/(6π ) and h =
=
r = 2r, thus the height is equal to the
r=
2πr
2πr2
S/3
diameter of the base. = 3 500/π. r h 187 30. Chapter 6 The total area of material used is
A = Atop + Abottom + Aside = (2r)2 + (2r)2 + 2πrh = 8r2 + 2πrh.
The volume is V = πr2 h thus h = V /(πr2 ) so A = 8r2 + 2V /r for r > 0,
√
dA/dr = 16r − 2V /r2 = 2(8r3 − V )/r2 , dA/dr = 0 when r = 3 V /2. This is the only critical point,
√
r
r
π3
=
=
r and, for
d2 A/dr2 > 0 there so the least material is used when r = 3 V /2,
h
V /(πr2 )
V
√
πV
π
r
3
=.
r = V /2, =
h
V8
8 31. Let x be the length of each side of the squares and y the height of the frame, then the volume is V = x2 y .
The total length of the wire is L thus 8x +4y = L, y = (L − 8x)/4 so V = x2 (L − 8x)/4 = (Lx2 − 8x3 )/4
for 0 ≤ x ≤ L/8. dV /dx = (2Lx − 24x2 )/4, dV /dx = 0 for 0 < x < L/8 when x = L/12. If
x = 0, L/12, L/8 then V = 0, L3 /1728, 0 so the volume is greatest when x = L/12 and y = L/12.
32. (a) Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume is
1
(S − πx2 )1/2
V = πx3 + y 3 and the surface area is S = πx2 + 6y 2 = constant. Thus y =
and
6
61/2
2 3/2
S
(S − πx )
π
;
for 0 ≤ x ≤
V = x3 +
6
π
63/2
√
dV
π
3π
π
dV
= x2 − 3/2 x(S − πx2 )1/2 = √ x( 6x − S − πx2 ).
= 0 wh...
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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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