Thomas Calculus 13th [Solutions]

1026 Chapter 14 Partial Derivatives
17.
18.
2 2
fx ( x , y ) 3 x 3 y 15 0 and
2
f y ( x , y ) 6 xy 3 y 15 0 critical points are (2, 1), ( 2, 1), 0, 5 ,
and 0, 5 ; for (2, 1) : fxx (2, 1) 6 x | (2,1) 12, f yy (2, 1) (6x 6 y)| (2,1) 18, fxy
(2, 1) 6 y| (2,1) 6
2
fxx f yy f xy 180 0 and f xx 0 local minimum of f (2, 1) 30; for ( 2, 1):
fxx ( 2, 1) 6 x| ( 2, 1) 12, f yy ( 2, 1) (6x 6 y)| ( 2, 1) 18, fxy
( 2, 1) 6 y| ( 2, 1) 6
2
fxx f yy f xy 180 0 and f xx 0 local maximum of f ( 2, 1) 30; for 0, 5 :
fxx 0, 5 6 x| 0, f 0, 5 (6 6 )| 6 5, 0, 5 6 | 6 5
0, 5 yy x y f
0, 5 xy y
0, 5
2
fxx f yy f xy 180 0 saddle point; for
0, 5 : fxx
0, 5 6 x| 0
0, 5
2
f yy 0, 5 (6x 6 y) | 6 5, f 0, 5 6 | 6 5 180 0
0, 5 xy y f
0, 5
xx f yy fxy
saddle point.
2
fx ( x, y) 6x 18x 0 6 x( x 3) 0 x 0 or x 3;
2
f y ( x, y) 6y 6y 12 0 6( y 2)( y 1) 0
y 2 or y 1 the critical points are (0, 2), (0,1), (3, 2), and (3, 1); fxx
( x, y) 12x
18,
f yy ( x, y) 12y 6, and fxy
( x, y ) 0; for (0, 2) : fxx (0, 2) 18, f yy (0, 2) 18, fxy
(0, 2) 0
2
fxx f yy f xy 324 0 and f xx 0 local maximum of f (0, 2) 20; for (0, 1) : f xx (0,1) 18,
2
f yy (0,1) 18, fxy (0, 1) 0 fxx f yy f xy 324 0 saddle point; for (3, 2) : f xx (3, 2) 18,
2
f yy (3, 2) 18, fxy (3, 2) 0 fxx f yy f xy 324 0 saddle point; for (3, 1) : f xx (3, 1) 18,
2
f yy (3,1) 18, fxy (3, 1) 0 fxx f yy f xy 324 0 and f xx 0 local minimum of f (3, 1) 34
19.
3
fx ( x , y ) 4 y 4 x 0 and
points are (0, 0), (1, 1), and ( 1,
3 2
f y ( x, y) 4x 4y 0 x y x 1 x 0 x 0, 1, 1 the critical
2 2
1); for (0, 0) : fxx
(0, 0) 12 x | (0, 0) 0, f yy (0, 0) 12 y | (0, 0) 0,
2
fxy (0, 0) 4 fxx f yy f xy 16 0 saddle point; for (1,1) : fxx (1, 1) 12, f yy (1, 1) 12, fxy
(1, 1) 4
2
fxx f yy f xy 128 0 and f xx 0 local maximum of f (1,1) 2; for ( 1, 1) : f xx ( 1, 1) 12,
2
f yy ( 1, 1) 12, fxy ( 1, 1) 4 fxx f yy f xy 128 0 and f xx 0 local maximum of f ( 1, 1) 2
20.
3
fx ( x , y ) 4 x 4 y 0 and
the critical points are (0, 0), (1,
3 3 2
f y ( x, y) 4y 4x 0 x y x x 0 x 1 x 0 x 0, 1, 1
2 2
1), and ( 1, 1); fxx
( x, y) 12 x , f yy ( x, y) 12 y , and fxy
( x, y) 4;
2
for (0, 0) : fxx (0, 0) 0, f yy (0, 0) 0, fxy (0, 0) 4 fxx f yy f xy 16 0 saddle point; for (1, 1) :
2
fxx (1, 1) 12, f yy (1, 1) 12, fxy (1, 1) 4 fxx f yy f xy 128 0 and f xx 0 local minimum of
f (1, 1) 2; for
2
( 1, 1) : fxx ( 1,1) 12, f yy ( 1, 1) 12, fxy ( 1, 1) 4 fxx f yy f xy 128 0 and
f xx 0 local minimum of f ( 1, 1) 2
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