Thomas Calculus 13th [Solutions]

1042 Chapter 14 Partial Derivatives
9.
2 2 2 2
V r h 16 r h 16 r h g( r, h) r h 16;
2
g 2rhi
r j so that
2
S 2 rh 2 r S (2 h 4 r) i 2 rj and
2
S g (2 rh 4 r) i 2 rj 2rhi r j 2 rh 4 r 2rh
and
2
2 r r r 0 or 2 . But r 0 gives no physical can, so r 0 2 2 h 4 r 2rh
2
r
r
r
2
2r h 16 r (2 r) r 2 h 4; thus r 2 cm and h 4 cm give the only extreme surface area of
2
24 cm . Since r 4 cm and
then
3
h 1 cm V 16 cm and
2
24 cm must be the minimum surface area.
2
S 40 cm , which is a larger surface area,
10. For a cylinder of radius r and height h we want to maximize the surface area S 2 rh subject to the constraint
2
2
2
g( r, h) r h a 0. Thus S 2 hi
2 rj and g 2ri h j so that S g 2 h 2 r and
2
2
2 h h
2 2 2 2 2 2
r and 2 r h h 4r h h 2r r 4r
a 2r a r a
2 r
r 2 4 2
2
h a 2 S 2 a a 2 2 a .
2
2
11. A (2 x)(2 y) 4xy subject to
2
2 y
g( x, y) x 1 0; A 4yi
4xj and
16 9
2
4 4 x y
4 x
2 y
32 y
A g yi xj i j y and 4x and 4x
8 9 8
9
x
2
3
2
x
4 2
2 y
g x i j so that
8 9
2 y 32 y
9 x
y 3 x x 1 x 8 x 2 2. We use x 2 2 since x represents distance. Then
4 16 9
3 3 2
y 2 2 , so the length is 2x 4 2 and the width is 2y
3 2.
4 2
12. P 4x 4y subject to
2 2
y
2 2
g( x, y) x 1 0; P 4i 4j and 2 2 y
g x i j so that P g
2 2
a b
a b
2
2
b 2
x
4 2x
2 y
2
and 4
2a
2 y 2 2 2 2
2 2 2
and 4 2a a
y b x x 1 x b x 1
2
2
a
b
x
2 x
2 2 2 2 4
b a a b a a
2 2 2 4
2
2 2
2
a b x a x a , since x 0 y b x b width 2x 2a
and
2 2
2 2 2
a b
a 2 2
a b
a b
2
2 2
height 2 2b
2 2
y perimeter is P 4x 4y 4a
4b
4 a b
2 2
2 2
a b
a b
13. f 2xi 2yj and g (2x 2) i (2y
4) j so that f g 2xi 2 yj (2x 2) i (2y
4) j
2 x (2x 2) and 2 y (2y 4) x , and y 2 1 y 2x
1
1
2 2
x 2 x (2 x) 4(2 x) 0 x 0 and y 0, or x 2 and y 4. f (0,0) 0 is the minimum value and
f (2, 4) 20 is the maximum value. (Note that 1 gives 2x 2x 2 or 0 2, which is impossible.)
14. f 3i j and g 2xi 2yj so that f g 3 2 x and 1 2 y 3 and 1 2 3
2x
2x
2
2
2
y x x x 4 10x 36 x 6 x 6 and
3 3 y 2 , or x 6 and
10 10
10 10
y 2 . Therefore f 6 , 2 20 6 2 10 6 12.325 is the maximum value, and f 6 , 2
10 10 10 10
10 10
2 10 6 0.325 is the minimum value.
y
Copyright
2014 Pearson Education, Inc.