Thomas Calculus 13th [Solutions]

1030 Chapter 14 Partial Derivatives
(v) For interior points of the rectangular region, Tx
( x, y) 2x y 6 0 and Ty
( x, y) x 2y 0 x 4
and y 2, an interior critical point with T (4, 2) 10. Therefore the absolute maximum is 11 at
(0, 3) and the absolute minimum is 10 at (4, 2).
36. (i) On OA, f ( x, y) f (0, y) 24y on 0 y 1;
(ii) On
(iii) On
f (0, y) 48y 0 y 0 and x 0, but
(0, 0) is not an interior point of OA; f (0, 0) 0
and f (0, 1) 24
3
AB, f ( x, y) f ( x,1) 48x 32x 24 on
2
x f x x x 1
and
2
0 1; ( , 1) 48 96 0
2
y 1, or x 1
and y 1, but
1
,1 is not in the interior of AB; f , 1 16 2 24 and
2
2
12
f (1, 1) 8
2
BC, f ( x, y) f (1, y) 48y 32 24y on 0 y 1; f (1, y) 48 48y 0 y 1 and x 1, but
(1,1) is not an interior point of BC; f (1, 0) 32 and f (1, 1) 8
(iv) On OC, f ( x, y) f ( x, 0) 32x on
not an interior point of OC; f (0, 0) 0 and f (1, 0) 32
(v) For interior points of the rectangular region,
3
2
0 x 1; f ( x, 0) 96x 0 x 0 and y 0, but (0, 0) is
x
2
f ( x, y) 48y 96x 0 and f ( x, y) 48x 48y
0
x 0 and y 0, or x 1
and
1
2
y 2 , but (0, 0) is not an interior point of the region; f 1
,
1
2.
2 2
Therefore the absolute maximum is 2 at
1
,
1
and the absolute minimum is 32 at (1, 0).
2 2
37. (i) On AB, f ( x, y) f (1, y) 3cos y on
x
f
y f (1, y ) 3sin y 0 y 0 and
4 4 ;
1;
3 2
4 2
f (1, 0) 3, f 1, , and
3 2
1,
4 2
y
(ii) On CD, f ( x, y) f (3, y) 3cos y on
(iii) On
(iv) On
3 2
f (3, 0) 3, f 3,
and f
4 2
y f (3, y) 3sin y 0 y 0 and x 3;
4 4 ;
3 2
3,
4 2
2 2
BC, f ( x, y) f x, 4x x on
4 2
3 2
f 2, 2 2, f 1, , and f
4 4 2
1 x 3; f x, 2(2 x) 0 x 2 and
4
3 2
3,
4 2
2 2
AD, f ( x, y) f x, 4x x on
4 2
y 4 ;
3 2
f 2, 2 2, f 1, , and f
4 4 2
1 x 3; f x, 2(2 x) 0 x 2 and
4
3 2
4 2
(v) For interior points of the region, fx
( x, y) (4 2 x)cos y 0 and
3,
2
f y ( x , y ) 4 x x sin y 0
x 2 and y 0, which is an interior critical point with f (2, 0) 4. Therefore the absolute maximum
is 4 at (2, 0) and the absolute minimum is 3 2
2
at
3, , 3, ,
4 4
1, , and
4
1, .
4
y
4 ;
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