Thomas Calculus 13th [Solutions]

1028 Chapter 14 Partial Derivatives
fxx ( 2, 0)
x 2 2 2 x 2
x
e x 4x 2 y , f ( 2, 0) 2 , ( 2, 0) 2
2 yy e f
2 xy
ye
( 2, 0) e
( 2, 0) e
( 2, 0)
0
fxx f yy
2
f 4
xy 0 and f 0
4
xx local maximum of f ( 2, 0) 4
2
e
e
29. f ( , ) 4 2
x x y 0 and f ( , ) 1 1 0
x
y x y critical point is 1 , 1 ; f 1 1
y
2 xx ,1 8, f ,1 1,
2 yy 2
1
2
fxy 2 , 1 0 fxx f yy f xy 8 0 and f xx 0 local maximum of f 1
2 , 1 3 2ln 2
30. f ( , ) 2 1
x x y x 0 and ( , ) 1 1 0
x y
f y x y x y
critical point is 1 , 3 ; f 1 3
2 2 xx , 1,
2 2
f yy
31. (i) On OA,
1 , 3 1,
2 2
1 3
2
fxy , 1 f 2 0
2 2
xx f yy f xy
saddle point
2
f ( x, y) f (0, y) y 4y 1 on
0 y 2; f (0, y) 2y 4 0 y 2;
f (0, 0) 1 and f (0, 2) 3
(ii) On AB,
2
f ( x, y) f ( x, 2) 2x 4x 3 on
0 x 1; f ( x, 2) 4x 4 0 x 1;
f (0, 2) 3 and f (1, 2) 5
(iii) On OB,
2
f ( x, y) f ( x, 2 x) 6x 12x 1 on 0 x 1; endpoint values have been found above;
f ( x, 2 x) 12x 12 0 x 1 and y 2, but (1, 2) is not an interior point of OB
(iv) For interior points of the triangular region, fx
( x, y) 4x 4 0 and f y ( x, y) 2y 4 0 x 1 and
y 2, but (1, 2) is not an interior point of the region. Therefore, the absolute maximum is 1 at (0, 0)
and the absolute minimum is 5 at (1, 2).
32. (i) On
(ii) On
2
OA, D( x, y) D(0, y) y 1 on 0 y 4;
D (0, y) 2y 0 y 0; D (0, 0) 1 and
D(0, 4) 17
2
AB, D( x, y) D( x, 4) x 4x 17 on
0 x 4; D ( x, 4) 2x 4 0 x 2 and
(2, 4) is an interior point of AB; D (2, 4) 13 and
D(4, 4) D(0, 4) 17
2
(iii) On OB, D( x, y) D( x, x) x 1 on 0 x 4; D ( x, x) 2x 0 x 0 and y 0, which is not an
interior point of OB; endpoint values have been found above
(iv) For interior points of the triangular region, fx
( x, y) 2x y 0 and f y ( x, y) x 2y 0 x 0 and
y 0, which is not an interior point of the region. Therefore, the absolute maximum is 17 at (0, 4) and
(4, 4), and the absolute minimum is 1 at (0, 0).
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