Thomas Calculus 13th [Solutions]

1072 Chapter 14 Partial Derivatives
82. (i)
(ii) For the interior of the disk, fx
( x, y) 2x 0 and f y ( x, y) 6y 2 0 x 0 and y 1
0,
1
3 3
is an interior critical point with f 0,
1 1
. Therefore the absolute maximum of f on the disk is 5 at
3 3
(0, 1) and the absolute minimum of f on the disk is
1
at 0,
1
.
3 3
2 2
f ( x, y) x y 3x xy on
2 2
x y f x y y x
9 (2 3 ) i (2 ) j and g 2xi
2yj so that
f g (2x 3 y) i (2 y x) j (2xi 2 yj ) 2x 3 y 2x
and 2y x 2y
and
2 x(1 ) y 3 and
2 2
x y 3 y . Thus,
x 2 y (1 ) 0 1
x
and (2 x) x y 3 x y 3y
2 y
2y
2 2
2 2 2 2 2
9 x y y 3y y 2y 3y 9 0 (2y 3)( y 3) 0
y 3 or y 3
2 . For 2 2
2 2
y 3, x y 9 x 0 yielding the point (0, 3). For y 3
2 , x y 9
2 9 2 27 3 3
3 3 27 3
x 9 x x . Evaluations give f (0, 3) 9, f ,
3
9 20.691,
4 4 2
2 2 4
f
3 3 3 27 3
2 2 4
, 9 2.691.
(ii) For the interior of the disk, fx
( x, y) 2x 3 y 0 and f y ( x, y) 2y x 0 x 2 and
y 1 (2, 1) is an interior critical point of the disk with f (2, 1) 3. Therefore, the absolute maximum
of f on the disk is
27 3 3 3
9 at ,
3
and the absolute minimum of f on the disk is 3 at (2, 1).
4 2 2
83. f i j k and g 2xi 2yj 2zk so that f g i j k (2xi 2yj 2 zk) 1 2 x ,
84. Let
1 2 y ,1 2 z x y z 1
. Thus
1 1 1
3 3 3
2 2 2 2
1 3 1
1
3
x y z x x yielding the points
, , and
1
,
1
,
1
. Evaluations give the absolute maximum value of
1 1 1 3
3 3 3 3
3 3 3
f , , 3 and the absolute minimum value of f 1
,
1
,
1
3.
2 2 2
f ( x, y, z)
x y z be the square of the distance to the origin and
3 3 3
2
g( x, y, z) x zy 4. Then
f 2xi 2yj 2zk and g 2xi zj yk so that f g 2x 2 x, 2 y z , and 2z y
x 0 or 1.
CASE 1: x 0 zy 4 z 4
y
and y 4 2
4
z y
and 4
8 2
2 z y and
z
8 2 2 2
2
2
z y z y z . But y x z 4 leads to no solution, so y z z 4
z 2 yielding the points (0, 2, 2) and (0, 2, 2).
y
2
CASE 2: 1 2z y and 2y z 2y 4y y y 0 z 0 x 4 0
2
x 2 yielding the points ( 2, 0, 0) and (2, 0, 0).
Evaluations give f (0, 2, 2) f (0, 2, 2) 8 and f ( 2, 0, 0) f (2, 0, 0) 4. Thus the points ( 2, 0, 0) and
(2, 0, 0) on the surface are closest to the origin.
85. The cost is f ( x, y, z) 2axy 2bxz 2cyz subject to the constraint xyz V . Then f g
2ay 2 bz yz, 2ax 2 cz xz , and 2bx 2cy xy 2axy 2 bxz xyz, 2axy 2 cyz xyz , and
2 bxz 2 cyz xyz 2 axy 2 bxz 2 axy 2 cyz y b
.
c
x Also 2 axy 2 bxz 2 bxz 2 cyz z a x
c
.
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