Thomas Calculus 13th [Solutions]

Chapter 14 Practice Exercises 1073
Then
b a 3
x x x V x c V
width
c c ab
2
2 1/3
c V
ab ,
x Depth
2 1/3 2 1/3
b c V b V
c ab ac ,
y and
Height
z
2 1/3 2 1/3
a c V a V
c ab bc .
86. The volume of the pyramid in the first octant formed by the plane is V ( a, b, c) 1 1 ab c 1 abc . The point
(2, 1, 2) on the plane
2 1 2
1.
a b c
bc ac ab
6 6 6
bc
ac
6 6
abc
(2bc 2 ab abc ), and
6
6
Thus,
3 2 6
We want to minimize V subject to the constraint 2bc ac 2 ab abc.
V i j k and g ( c 2 b bc) i (2c 2 a ac) j (2 b a ab)
k so that V g
ab
( c 2 b bc), (2c 2 a ac ), and 6
(2 b a ab) 6
( ac 2 ab abc),
abc
(2 bc ac abc) ac 2 bc and 2 ab 2 bc . Now 0 since
a 0, b 0, and c 0 ac 2bc and ab bc a 2 b c . Substituting into the constraint equation gives
2 2 2
1 a 6 b 3 and c 6. Therefore the desired plane is
x z
a a a
6 3 6
1 or x 2y z 6.
87. f ( y z) i xj xk, g 2xi 2 yj , and h zi xk so that f g h ( y z)
i xj xk
(2xi 2 yj) ( zi xk ) y z 2 x z, x 2 y, x x x 0 or 1.
CASE1: x 0 which is impossible since xz 1.
CASE 2: 1 y z 2 x z y 2 x and x 2 y y (2 )(2 y) y 0 or
then
2
x 1 x 1 so with xz 1 we obtain the points (1, 0, 1) and ( 1, 0, 1). If
abc
y
2
4 1. If y 0,
2
4 1,
then
1
2 . For 1
2 , y x so 2 2 2
x y 1 x 1 x 1
with xz 1 z 2,
2 2
and we obtain the points
1
,
1
, 2 and
1
,
1
, 2 . For
1
2 2
2 2
2 , y x
2
x 1 x 1
with xz 1 z 2, and we obtain the points
1
,
1
, 2 and
2 2
1
,
1
, 2 .
2 2
2 2
Evaluations give f (1, 0, 1) 1, f ( 1, 0, 1) 1, f 1 , 1 , 2 1 , f 1 , 1 , 2 1
,
1 1
2 2
3
2
2 2 2 2 2 2
f , , 2 , and f 1 , 1
, 2 3
. Therefore the absolute maximum is 3 2 at 1
,
1
, 2
and
1 1
2 2
2 2
2
, , 2 , and the absolute minimum is 1 2 at 1
,
1
, 2 and
1
,
1
, 2 .
2 2 2
2 2
2 2
88. Let f ( x, y, z)
x y z be the square of the distance to the origin. Then f 2xi 2yj 2 zk,
g i j k , and h 4xi 4yj 2zk so that f g h 2x 4 x , 2y 4 y , and
2z 2z 2 x(1 2 ) 2 y(1 2 ) 2 z(1 2 ) x y or
1
2 .
2 2
CASE 1: x y z 4x z 2x so that x y z 1 x x 2x 1 or x x 2x 1 (impossible and
CASE 2:
1
2
1 1
4 4
x y and z 1
yielding the point
1
,
1
,
1
.
2
0 0 2 z(1 1) z 0 so that
fails to satisfy the first constraint x y z 1.
4 4 2
Therefore, the point
1
,
1
,
1
on the curve of intersection is closest to the origin.
4 4 2
2 2
2 2
2x 2y 0 x y 0. But the origin (0, 0, 0)
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