James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

968 chapter 14 Partial Derivatives
4. f sx, yd − 3x 2 x 3 2 2y 2 1 y 4
_2.9
_2.7
_2.5
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_1.5
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y
1
0
0.5
1
1.5
1.5
1.7
1.9
1
x
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23–26 Use a graph or level curves or both to estimate the local
maximum and minimum values and saddle point(s) of the
function. Then use calculus to find these values precisely.
23. f sx, yd − x 2 1 y 2 1 x 22 y 22
24. f sx, yd − sx 2 yde 2x2 2y 2
25. f sx, yd − sin x 1 sin y 1 sinsx 1 yd,
0 < x < 2, 0 < y < 2
26. f sx, yd − sin x 1 sin y 1 cossx 1 yd,
0 < x < y4, 0 < y < y4
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5–20 Find the local maximum and minimum values and saddle
point(s) of the function. If you have three-dimensional graphing
software, graph the function with a domain and viewpoint that
reveal all the important aspects of the function.
5. f sx, yd − x 2 1 xy 1 y 2 1 y
6. f sx, yd − xy 2 2x 2 2y 2 x 2 2 y 2
7. f sx, yd − sx 2 yds1 2 xyd
8. f sx, yd − yse x 2 1d
9. f sx, yd − x 2 1 y 4 1 2xy
10. f sx, yd − 2 2 x 4 1 2x 2 2 y 2
11. f sx, yd − x 3 2 3x 1 3xy 2
12. f sx, yd − x 3 1 y 3 2 3x 2 2 3y 2 2 9x
13. f sx, yd − x 4 2 2x 2 1 y 3 2 3y
14. f sx, yd − y cos x
15. f sx, yd − e x cos y
16.
f sx, yd − xye 2sx2 1y 2 dy2
17. f sx, yd − xy 1 e 2xy
18. f sx, yd − sx 2 1 y 2 de 2x
19. f sx, yd − y 2 2 2y cos x, 21 < x < 7
20.
f sx, yd − sin x sin y, 2 , x , , 2 , y ,
21. Show that f sx, yd − x 2 1 4y 2 2 4xy 1 2 has an infinite
number of critical points and that D − 0 at each one. Then
show that f has a local (and absolute) minimum at each
critical point.
22. Show that f sx, yd − x 2 ye 2x2 2y 2 has maximum values at
(61, 1ys2 ) and minimum values at (61, 21ys2 ). Show
also that f has infinitely many other critical points and
D − 0 at each of them. Which of them give rise to maximum
values? Minimum values? Saddle points?
;
;
27–30 Use a graphing device as in Example 4 (or Newton’s
method or solve numerically using a calculator or computer) to
find the critical points of f correct to three decimal places. Then
classify the critical points and find the highest or lowest points on
the graph, if any.
27. f sx, yd − x 4 1 y 4 2 4x 2 y 1 2y
28. f sx, yd − y 6 2 2y 4 1 x 2 2 y 2 1 y
29. f sx, yd − x 4 1 y 3 2 3x 2 1 y 2 1 x 2 2y 1 1
30. f sx, yd − 20e 2x2 2y 2 sin 3x cos 3y, | x | < 1, | y | < 1
31–38 Find the absolute maximum and minimum values of f on
the set D.
31. f sx, yd − x 2 1 y 2 2 2x, D is the closed triangular region
with vertices s2, 0d, s0, 2d, and s0, 22d
32. f sx, yd − x 1 y 2 xy, D is the closed triangular region
with vertices s0, 0d, s0, 2d, and s4, 0d
33. f sx, yd − x 2 1 y 2 1 x 2 y 1 4,
D − hsx, yd | | x | < 1, | y | < 1j
34. f sx, yd − x 2 1 xy 1 y 2 2 6y,
D − hsx, yd | 23 < x < 3, 0 < y < 5j
35. f sx, yd − x 2 1 2y 2 2 2x 2 4y 1 1,
D − hsx, yd | 0 < x < 2, 0 < y < 3j
36. f sx, yd − xy 2 , D − hsx, yd | x > 0, y > 0, x 2 1 y 2 < 3j
37. f sx, yd − 2x 3 1 y 4 , D − hsx, yd | x 2 1 y 2 < 1j
38. f sx, yd − x 3 2 3x 2 y 3 1 12y, D is the quadrilateral
whose vertices are s22, 3d, s2, 3d, s2, 2d, and s22, 22d
39. For functions of one variable it is impossible for a con tinuous
function to have two local maxima and no local minimum.
But for functions of two variables such functions exist. Show
that the function
f sx, yd − 2sx 2 2 1d 2 2 sx 2 y 2 x 2 1d 2
has only two critical points, but has local maxima at both of
them. Then use a computer to produce a graph with a carefully
chosen domain and viewpoint to see how this is
possible.
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