Thomas Calculus 13th [Solutions]

Chapter 14 Additional and Advanced Exercises 1079
12.
(2x 3 y) (2x 3 y)
f ( x, y) 6 xye fx
( x, y) 6 y(1 2 x) e 0 and
(2x
3 y)
f y ( x, y) 6 x(1 3 y) e 0 x 0
and y 0, or x 1 and y 1
2
3 . The value f (0, 0) 0 is on the boundary, and f 1 , 1 1 . On the positive
2 3 2
e
y -axis, f (0, y ) 0, and on the positive x -axis, f ( x , 0) 0. As x or y we see that f ( x, y) 0.
Thus the absolute maximum of f in the closed first quadrant is 1
2
e
at the point 1 , 1 .
2 3
13. Let
2 2 2
y
2 2 y
f ( x, y , z) x z 1 f x i j 2z
k an equation of the plane tangent at the point
2 2 2 2 2 2
a b c a b c
P0 ( x0, y0, y 0 ) is
intercepts of the plane are
x0 y0 z0 2
x0 2
y0 2
z0
2 2 2 2 2 2
2 2 2 2 2 2
x0 y0 z0
x y z 2 or x y z 1. The
2 2 2
a b c a b c
a b c
2 2
a , 0, 0 , 0, b , 0 and
x
y
0 0
2
0, 0, c
z
. The volume of the tetrahedron formed by the
2 2 2
plane and the coordinate planes is V 1 1 a b c we need to maximize
3 2 x y z
2
( abc) 1
V ( x, y, z) ( xyz ) subject to the constraint
6
2
( abc) 1 2x
6 2
x yz
2
a
,
2
( abc) 1 2 y
6 2
xy z
2
b
0 0 0
,
0
2 2 2
y
f ( x, y, z ) x z 1. Thus,
2 2 2
a b c
and
2
( abc) 1 2z
6
2
xyz
2
c
. Multiply the first equation by
2
2
2
2 2 2 2
a yz , the second by b xz , and the third by c xy . Then equate the first and second a y b x
b
2 2 2 2
y x, x 0; equate the first and third a z c x z c x, x 0; substitute into
a
a
3
f ( x , y , z ) 0 x a y b z c V abc
3 3 3 2
.
14. 2( x u) , 2( y v) , 2( x u ) , and
2( y v) 2 v x u v y, x u , and
2
y v v x u v v 1 or 0.
2 2
CASE 1: 0 x u, y v , and 0; then y x 1 v u 1 and
2 2
v u v v
2 1 1 4
v v 1 0 v no real solution.
2
CASE 2: 1
2
v and u v u 1 ; x 1 1 y and y x 1 x 1 x 1 2x 1 x 1
2
4 4 2
4 2 4 8
y 7
8 .
2 2 2
Then f 1 , 7 , 1 , 1 1 1 7 1 2 3 the minimum distance is 3 2. (Notice that f has no
8 8 4 2 8 4 8 2 8
8
maximum value.)
15. Let ( x0 , y 0)
be any point in R. We must show lim f ( x, y) f ( x0 , y 0)
or, equivalently that
( x, y) ( x0 , y0)
lim f ( x0 h, y0 k) f ( x0 , y 0) 0. Consider f ( x0 h, y0 k) f ( x0, y0)
( h, k) (0, 0)
f ( x0 h, y0 k) f ( x0, y0 k) f ( x0, y0 k) f ( x0, y 0 ) . Let F( x) f ( x, y0
k ) and apply the
Mean Value Theorem: there exists with x0 x0
h such that F ( ) h F( x0 h) F( x0
)
hfx ( , y0 k ) f ( x0 h , y0 k ) f ( x0 , y0
k ). Similarly, k f y ( x0 , ) f ( x0 , y0 k ) f ( x0 , y 0 ) for
some with y0 y0 k . Then f ( x0 h, y0 k) f ( x0 , y0 ) hfx
( , y0 k) + k f y ( x 0, ) . If M, N are
positive real numbers such that | fx|
M and f y N for all ( x, y ) in the
xy -plane, then
1
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