Thomas Calculus 13th [Solutions]

Section 14.7 Extreme Values and Saddle Points 1027
21. ( , ) 2x
2 y
fx
x y 0 and f
2 2
2
y ( x, y) 0 x 0 and y 0 the critical points is (0, 0);
2 2
2
x y 1
x y 1
2 2 2 2
4x 2 y 2 2x 4 y 2 8xy
fxx , f
3 yy , f ;
3 xy
f
3 xx (0, 0) 2, f yy (0, 0) 2, fxy
(0, 0) 0
2 2 2 2 2 2
x y 1 x y 1 x y 1
2
fxx f yy f xy 4 0 and f xx 0 local maximum of f (0, 0) 1
22. f 1
x ( x , y ) y 0 and f 1
2
y ( x , y ) x 0 x 1 and y 1 the critical point is (1,1);
2
x
y
f 2 2
xx , f , 1;
3 yy f
3 xy
x y
local minimum of f (1,1) 3
2
fxx (1, 1) 2, f yy (1,1) 2, fxy (1,1) 1 fxx f yy f xy 3 0 and f xx 2
23. fx
( x, y) y cos x 0 and f y ( x, y) sin x 0 x n , n an integer, and y 0 the critical points are
( n , 0), n an integer (Note: cos x and sin x cannot both be 0 for the same x, so sin x must be 0 and y 0 );
fxx y sin x, f yy 0, fxy
cos x ; fxx ( n , 0) 0, f yy ( n , 0) 0, fxy
( n , 0) 1 if n is even and
fxy ( n , 0) 1 if n is odd
2
fxx f yy f xy 1 0 saddle point.
24.
2x
fx
( x, y) 2e cos y 0 and
2x
2x
f y ( x, y) e sin y 0 no solution since e 0 for any x and the
functions cos y and sin y cannot equal 0 for the same y no critical points no extrema and no saddle
points
25.
2 2 4
2 2 4
x y x
x y x
fx
( x, y) (2x 4) e 0 and f y ( x, y) 2ye 0 critical point is (2, 0); f (2, 0) 2
xx ,
4
e
2
fxy (2, 0) 0, f 2 4
yy (2, 0) f 0
4 xx f yy f xy and f
8
xx 0 local minimum of f (2, 0) 1
4
e
e
e
x
y x
26. fx
( x, y) ye 0 and f y ( x, y) e e 0 critical point is (0, 0); fxx
(2, 0) 0, fxy
(2, 0) 1,
2
f yy (2, 0) 1 fxx f yy f xy 1 0 saddle point
y
27. fx
( x, y) 2xe 0 and
y y 2 2
f y ( x, y) 2ye e x y 0 critical points are (0, 0) and (0, 2);
y y y y 2 2
for (0, 0) : fxx
(0, 0) 2e 2, f yy (0, 0) 2e 4ye e x y 2,
(0, 0) (0, 0)
y
2
fxy (0, 0) 2 xe | (0, 0) 0, fxx f yy f xy 4 0 and f xx 0 local minimum of f (0, 0) 0;
y
for 2 y y y 2 2
(0, 2) : f (0, 2) 2 , (0, 2) 2 4 2
xx e f ,
2 yy e ye e x y
2
(0,2) e
(0, 2) e
y
2
f 4
xy (0, 2) 2xe 0 fxx f yy f xy 0 saddle point
4
(0, 2)
e
28.
x 2 2
x
fx
( x, y) e x 2x y 0 and f y ( x, y) 2ye 0 critical points are (0, 0) and ( 2, 0); for
x 2 2
x x
(0, 0) : fxx (0, 0) e x 4x 2 y 2, f yy (0, 0) 2e 2, fxy
(0, 0) 2ye
0
(0, 0)
(0, 0) (0, 0)
2
fxx f yy f xy 4 0 and f xx 0 saddle point; for ( 2, 0):
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