Thomas Calculus 13th [Solutions]

38. (i) On OA, f ( x, y) f (0, y) 2y 1 on 0 y 1;
f (0, y ) 2 no interior critical points;
f (0, 0) 1 and f (0,1) 3
(ii) On OB, f ( x, y) f ( x, 0) 4x 1 on 0 x 1;
(iii) On
f ( x , 0) 4 no interior critical points;
f (1, 0) 5
2
AB, f ( x, y) f ( x, x 1) 8x 6x 3 on 0 1;
5 3 5 15
8 8 8 8
y ; f , , f (0,1) 3, and f (1, 0) 5
Section 14.7 Extreme Values and Saddle Points 1031
x f ( x, x 1) 16x 6 0 x 3
and
8
(iv) For interior points of the triangular region, fx
( x, y) 4 8y 0 and f 1
y ( x , y ) 8 x 2 0 y and
1
4
x which is an interior critical point with f 1
,
1
2. Therefore the absolute maximum is 5 at (1,0)
and the absolute minimum is 1 at (0, 0).
4 2
2
39. Let
b
2
a
F( a, b) 6 x x dx where a b . The boundary of the domain of F is the line a b in the abplane,
and F( a, a ) 0, so F is identically 0 on the boundary of its domain. For interior critical points we have:
F
a
2
a a a and
6 0 3, 2
interior critical point ( 3, 2) and
2
F
b
2
6 b b 0 b 3, 2. Since a b , there is only one
2 2
F( 3, 2) 6 x x dx gives the area under the parabola
y 6 x x that is above the x -axis. Therefore, a 3 and b 2.
3
40. Let
b
2
1/3
F( a, b) 24 2x x dx where a b . The boundary of the domain of F is the line a b and on
a
this line F is identically 0. For interior critical points we have:
F
b
2
1/3
F
a
2
1/3
a a a and
24 2 0 4, 6
24 2b b 0 b 4, 6. Since a b , there is only one critical point ( 6, 4) and
4 2
1/3
F( 6, 4) 24 2x x dx gives the area under the curve
6
x -axis. Therefore, a 6 and b 4.
2
1/3
y 24 2x x that is above the
41. Tx
( x, y) 2x 1 0 and T 1
y ( x , y ) 4 y 0 x and y 0 with T 1
, 0
1
; on the boundary
2
2 4
2 2 2
x y 1: T ( x, y) x x 2 for 1 x 1 T ( x, y) 2x 1 0 x 1
and
1 3 9 1 3 9
2 2 4 2 2 4
T , , T , , T ( 1, 0) 2, and T (1, 0) 0 the hottest is 2
1
at
2
4
y
3
2 ;
1 3
2 2
, and
1 3
2 2
, ; the coldest is
1
4
at
1
2 , 0 .
42. f ( , ) 2
2
x x y y 0 and f ( , )
1
0
1
y x y x x and y 2; f 1 , 2 2
xx
8,
2
x
2 2
y 1
4 2
, 2
2
Copyright
y
1 , 2 1 1 , 1 , 2 1 2
f yy fxy fxx f yy f xy 1 0 and f xx 0 a local minimum of
f
1 , 2 2 ln 1 2 ln 2
2 2
2
2014 Pearson Education, Inc.
2
x
1
2
, 2