Thomas Calculus 13th [Solutions]

Section 14.7 Extreme Values and Saddle Points 1033
parallel to i 2j k which is normal to the plane if x 1 and y 1. Thus the point 1 , 1, 1 1 10 or
2
2 4
1 , 1, 45 is the point on the surface
2 4
2 2
z x y 10 nearest the plane x 2y z 0.
51.
2 2 2
d( x, y, z) ( x 0) ( y 0) ( z 0) we can minimize d( x, y, z ) by minimizing
2 2 2
D( x, y, z) x y z ;
2 2 2
3x 2y z 6 z 6 3x 2 y D( x, y) x y 6 3x 2y
Dx ( x , y ) 2 x 6(6 3 x 2 y ) 0 and Dy
( x, y) 2y 4(6 3x 2 y ) 0 critical point is
9 6 3
2
, z ; D 9 6 9 6 9 6
7 7 7 xx , 20, D
7 7 yy , 10, D , 12 56 0
7 7 xy D
7 7
xx Dyy D xy and D xx 0
local minimum of d
9 6 3 3 14
, ,
7 7 7 7
52.
2 2 2
d( x, y, z) ( x 2) ( y 1) ( z 1) we can minimize d( x, y, z ) by minimizing
2 2 2
D( x, y, z) ( x 2) ( y 1) ( z 1) ; x y z 2 z x y 2
2 2 2
D( x, y) ( x 2) ( y 1) ( x y 3) Dx
( x, y) 2( x 2) 2( x y 3) 0 and
Dy ( x, y) 2( y 1) 2( x y 3) 0 critical point is 8 , 1 z 1 ; D 8 1 8 1
3 3 3 xx , 4, D , 4
3 3 yy
,
3 3
8 1
2
Dxy , 2 D 12 0
3 3
xxDyy D xy and D xx 0 local minimum of d 8 , 1 , 1 2
3 3 3 3
53.
2 2 2
s( x, y, z) x y z ;
2 2 2
x y z 9 z 9 x y s( x, y) x y (9 x y)
sy
x y y x y critical point is (3, 3) z 3;
2
s xx local minimum of
sx ( x , y ) 2 x 2(9 x y ) 0 and ( , ) 2 2(9 ) 0
sxx (3, 3) 4, syy (3, 3) 4, sxy (3, 3) 2 sxxsyy s xy 12 0 and 0
s(3, 3, 3) 27
54. p( x, y, z) xyz;
2
px ( x , y ) 3 y 2 xy y 0 and
and (1,1); for (0, 0) z 3;
point; for (0, 3) z 0;
point; for (3, 0) z 0;
point; for (1, 1) z 1;
x y z 3 z 3 x y p( x, y) xy(3 x y) 3xy 2
x y
2
xy
py ( x , y ) 3 x
2
x 2 xy 0 critical points are (0, 0), (0, 3), (3, 0),
pxx (0, 0) 0, pyy (0, 0) 0, pxy (0, 0) 3 pxx pyy 2
p xy 9 0 saddle
pxx (0, 3) 6, pyy (0, 3) 0, pxy (0, 3) 3 pxx pyy 2
p xy 9 0 saddle
pxx (3, 0) 0, pyy (3, 0) 6, pxy (3, 0) 3 pxx pyy 2
p xy 9 0 saddle
pxx (1, 1) 2, pyy (1,1) 2, pxy (1, 1) 1 pxx pyy 2
p xy 3 0 and p xx 0
local maximum of p(1, 1,1) 1
55. s( x, y, z) xy yz xz ; x y z 6 z 6 x y s( x, y) xy y(6 x y) x(6 x y)
2 2
6x 6 y xy x y sx
( x, y) 6 2x y 0 and sy
( x, y) 6 x 2y 0 critical point is (2, 2)
2
z 2; sxx (2, 2) 2, syy (2, 2) 2, sxy (2, 2) 1 sxxsyy s xy 3 0 and s xx 0 local maximum
of s(2, 2, 2) 12
56.
2 2 2
d( x, y, z) ( x 6) ( y 4) ( z 0) we can minimize d( x, y, z ) by minimizing
2 2 2
D( x, y, z) ( x 6) ( y 4) z ;
2 2 2 2 2 2
z x y D( x, y) ( x 6) ( y 4) x y
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