Thomas Calculus 13th [Solutions]

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33. (i) On OA, f ( x, y) f (0, y)
y on 0 y 2;
f (0, y) 2y 0 y 0 and x 0; f (0, 0) 0
and f (0, 2) 4
2
(ii) On OB, f ( x, y) f ( x, 0) x on 0 x 1;
(iii) On
f ( x, 0) 2x 0 x 0 and y 0; f (0, 0) 0
and f (1, 0) 1
Section 14.7 Extreme Values and Saddle Points 1029
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AB, f ( x, y) f ( x, 2x 2) 5x 8x 4 on 0 x 1; f ( x, 2x 2) 10x 8 0 x 4 and
5
y 2 ; f 4 , 2 4 ; endpoint values have been found above.
5 5 5 5
(iv) For interior points of the triangular region, fx
( x, y) 2x 0 and f y ( x, y) 2y 0 x 0 and y 0,
but (0, 0) is not an interior point of the region. Therefore the absolute maximum is 4 at (0, 2) and the
absolute minimum is 0 at (0, 0).
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34. (i) On AB, T ( x, y) T (0, y)
y on 3 y 3;
T (0, y) 2y 0 y 0 and x 0; T (0, 0) 0,
T (0, 3) 9, and T (0, 3) 9
2
(ii) On BC, T ( x, y) T ( x, 3) x 3x 9 on
0 x 5; T ( x, 3) 2x 3 0 x 3 and
2
y 3; T 3 , 3 27 and T (5, 3) 19
2 4
2
(iii) On CD, T ( x, y) T (5, y) y 5y 5 on 3 y 3; T (5, y) 2y 5 0 y 5 and x 5;
2
T 5, 5 45 , T (5, 3) 11 and T (5, 3) 19
2 4
2
(iv) On AD, T ( x, y) T ( x, 3) x 9x 9 on 0 x 5; T ( x, 3) 2x 9 0 x 9 and y 3;
2
T 9 , 3 45 , T (0, 3) 9 and T (5, 3) 11
2 4
(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 (4, 2) is an interior critical point with T (4, 2) 12. Therefore the absolute maximum
is 19 at (5, 3) and the absolute minimum is 12 at (4, 2).
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35. (i) On OC, T ( x, y) T ( x, 0) x 6x 2 on
0 x 5; T ( x, 0) 2x 6 0 x 3 and y 0;
T (3, 0) 7, T (0, 0) 2, and T (5, 0) 3
2
(ii) On CB, T ( x, y) T (5, y) y 5y 3 on
(iii) On
y T y y y 5
and
2
5 37 T
2 4
3 0; (5, ) 2 5 0
x 5; T 5,
and (5, 3) 9
2
AB, T ( x, y) T ( x, 3) x 9x 11 on 0 x 5; T ( x, 3) 2x 9 0 x 9
and y 3;
9
, 3
37
2 4
T and T (0, 3) 11
2
(iv) On AO, T ( x, y) T (0, y) y 2 on 3 y 0; T (0, y) 2y 0 y 0 and x 0, but (0, 0) is not
an interior point of AO
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