My answers to the exercises

14.7 1, 2, 5, 7, 9, 13, 17, 29, 33, 35, 39, 41.
1. Suppose (1; 1) is a critical point of a function f with continuous
second derivatives. In each case, what can you say about f ?
(a) fxx (1; 1) = 4, fxy (1; 1) = 1, fyy (1; 1) = 2
2
= 4 2 1 = 7 > 0, and fxx = 4 > 0, so f
D = fxx fyy fxy
has a local minimum at (1; 1).
(b) fxx (1; 1) = 4, fxy (1; 1) = 3, fyy (1; 1) = 2
2
= 4 2 32 = 1 < 0, so (1; 1) is a saddle
D = fxx fyy fxy
point of f .
2. Suppose (0; 2) is a critical point of a function g with continuous
second derivatives. In each case, what can you say about g?
(a) gxx (0; 2) = 1, gxy (0; 2) = 6, gyy (0; 2) = 1
2
= 1 1 36 = 37 < 0, so (0; 2) is a saddle
D = gxx gyy gxy
point of g.
(b) gxx (0; 2) = 1, gxy (0; 2) = 2, gyy (0; 2) = 8
2
= ( 1) ( 8) 4 = 4 < 0, and gxx < 0, so
D = gxx gyy gxy
g has a local minimum at (0; 2).
(c) gxx (0; 2) = 4, gxy (0; 2) = 6, gyy (0; 2) = 9
2
= 4 9 36 = 0, so you can’t say anything.
D = gxx gyy gxy
5–18 Find the local maximum and minimum values and saddle points
of the function.
5. f (x; y) = x2 + xy + y 2 + y
fx (x; y) = 2x + y and fy (x; y) = x + 2y + 1, so the critical
points satisfy 2x + y = 0 and x + 2y + 1 = 0. So there is
only one critical point: (1=3; 2=3). The second derivatives
are fxx (x; y) = 2, fxy (x; y) = 1, and fyy (x; y) = 1, so D =
2 1 12 = 1 > 0. As fxx (x; y) = 2 > 0, the function f has a
local minimum at (1=3; 2=3). The value of f at (1=3; 2=3)
is f (1=3; 2=3) = 1=9 2=9 + 4=9 2=3 = 1=3.
7. f (x; y) = (x y) (1 xy) = x y x2 y + xy 2
fx (x; y) = 1 2xy + y 2 and fy (x; y) = 1 + 2xy x2 , so the
critical points satisfy 1 2xy + y 2 = 0 and 1 + 2xy x2 = 0.
Adding these two equations gives y 2 = x2 , so y = x or y = x.
If y = x, then 1 2x2 + x2 = 0, so x = 1. If y = x,
1
then 1 + 2x2 + x2 = 0 which is impossible. So the critical
points are (1; 1) and ( 1; 1). The second derivatives are
fxx (x; y) = 2y, fxy (x; y) = 2x + 2y, and fyy (x; y) = 2x,
so
fxx fxy fyy D
(1; 1)
2 0
2
4
( 1; 1) 2
0
2
4
Both points are saddle points.
9. f (x; y) = y 3 + 3x2 y 6x2 6y 2 + 2
fx (x; y) = 6xy 12x and fy (x; y) = 3y 2 + 3x2 12y, so the
critical points satisfy 6xy 12x = 0 and 3y 2 + 3x2 12y = 0.
The …rst equation says that either x = 0 or y = 2. If x = 0,
then the second equation says that 3y 2 = 12y, so either y = 0
or y = 4. So (0; 0) and (0; 4) are critical points. If y = 2, then
the second equation says 12 + 3x2 24 = 0, so x2 = 4, that is
x = 2. So the other critical points are (2; 2) and ( 2; 2).
The second derivatives are fxx (x; y) = 6y 12, fxy (x; y) = 6x,
and fyy (x; y) = 6y 12, so
fxx
12
12
0
0
(0; 0)
(0; 4)
(2; 2)
( 2; 2)
fxy
0
0
12
12
fyy
12
12
0
0
D
144
144
144
144
The points ( 2; 2) are saddle points, f has a local maximum
at (0; 0), and a local minimum at (0; 4). Note that f (0; 0) =
2 is a local maximum value and f (0; 4) = 30 is a local
minimum value.
13. f (x; y) = ex cos y
fx (x; y) = ex cos y and fy (x; y) = ex sin y, so the critical
points satisfy ex cos y = 0 and ex sin y = 0, that is, cos y =
sin y = 0. But that never happens because cos2 y + sin2 y = 1,
so there are no critical points.
17. f (x; y) = y 2 2y cos x,
1 x 7
fx (x; y) = 2y sin x and fy (x; y) = 2y 2 cos x, so the critical
points satisfy 2y sin x = 0 and y = cos x. The …rst equation
says that y = 0 or sin x = 0. If y = 0, then x = =2 + n ,
2
whereas if sin x = 0, then x = n . So the critical points are
( =2 + n ; 0) and (n ; ( 1)n ). For 1
x
7, these are
( =2; 0), (3 =2; 0), (0; 1), ( ; 1), (2 ; 1).
The second derivatives are fxx (x; y) = 2y cos x, fxy (x; y) =
2 sin x, and fyy (x; y) = 2.
fxx fxy fyy
( =2; 0) 0
2
2
(3 =2; 4) 0
2 2
(0; 1)
2
0
2
( ; 1)
2
0
2
(2 ; 1)
2
0
2
D f
4 0
4 16
4
1
4
1
4
1
so ( =2; 0) and (3 =2; 4) are saddle points, and f has a local
minimum of 1 at the other three points.
29–36 Find the absolute maximum and minimum values of f on the set
D.
29. f (x; y) = x2 + y 2 2x,
D is the closed triangular region
with vertices (2; 0), (0; 2), and (0; 2).
fx = 2x 2 and fy = 2y, so the only critical point is (1; 0),
which is in the interior of the region, and f (1; 0) = 1. The
values of f at the vertices are f (0; 2) = f (0; 2) = 4 and
f (2; 0) = 2. The left side of the triangle is f(0; y) : 2 y 2g,
and f (0; y) = y 2 which is 0 at (0; 0), the unique critical point
on that side. The top side of the triangle is f(x; 2 x) : 0 x 2g,
and f (x; 2 x) = 2x2 6x + 4 which has the unique critical
point x = 3=2 and f (3=2; 1=2) = 9=4 + 1=4 3 = 1=2. The
bottom side of the triangle is f(x; 2 + x) : 0 x 2g, and
f (x; 2 + x) = 2x2 6x + 4 again, so it has a unique critical
point at x = 3=2 and f (3=2; 1=2) = 1=2. So the maximum
value of f on D is 4, achieved at (0; 2) and (0; 2), and the
minimum value is is 1, achieved at (1; 0).
33. f (x; y) = x4 +y 4 4xy+2,
D = f(x; y) : 0 x 3, 0 y 2g
fx = 4x3 4y and fy = 4y 3 4x, so if both are zero, y = x3
and x = y 3 , from which it follows that x = x9 , so x = 0 or
x = 1. Thus the only critical point in the interior of the
region is (1; 1), at which f has the value 0. The values of f on
the vertices are f (0; 0) = 2, f (3; 0) = 83, f (0; 2) = 18, and
3
f (3; 2) = 75. On the bottom, the values of f are x4 + 2 which
has a unique critical point at x = 0, the vertex (0; 0) again.
On the top, the values
of f are x4 8x+18 which has
a unique
p
p
3
3
critical point at x = 2 where f is equal to 18 6 2 10:44.
On the left side the values of f are y 4 + 2 which has a unique
critical point at y = 0, the vertex (0; 0). On the right side the
values ofpf are y 4 12y + 83 which has p
a unique critical point
70: 020. So the
at y = 3 3 where f is equal to 83 9 3 3
maximum value of f on D is 83, achieved at (3; 0), and the
minimum value of f on D is 0, achieved at (1; 1).
35. f (x; y) = 2x3 + y 4 ,
D = f(x; y) : x2 + y 2 1g
fx (x; y) = 6x2 and fy (x; y) = 4y 4 , so the only critical point
is (0; 0) at which the value of f is 0. The boundary of D
is the unit circle f(cos t; sin t) : 0 t 2 g. The value of f
at these points is g (t) = f (cos t; sin t) = 2 cos3 t + sin4 t. To
…nd the maximum and minimum values of g, set g 0 (t) = 0. So
6 cos2 t sin t+4 sin3 t cos t = 0, that is 6 cos t + 4 sin2 t cos t sin t =
0. The …rst factor can be written as 6 cos t + 4 (1 cos2 t) =
4 cos2 t 6 cos t + 4. So the places where g 0 (t) = 0 are where
cos t = 0 or sin t = 0 or 2 cos2 t + 3 cos t 2 = 0. Solving that
quadratic equation gives
p
25
3
cos t =
4
so cos t = 2 or cos t = 1=2. The …rst
p is impossible and the
3=2 . The points where
second occurs at the points 1=2;
cos t = 0 or sin t = 0 are (0; 1) and ( 1; 0). The values of f
on these 6 points, and on (0; 0), are
p
p
(0; 0) (0; 1) (0; 1) (1; 0) ( 1; 0) 1=2; 3=2
1=2;
3=2
f
0
1
1
2
2
7=16
7=16
so the maximum value of f is 2, which is achieved at (1; 0), and
the minimum value of f is 2, which is achieved at ( 1; 0).
39. Find the shortest distance from the point (2; 0; 3) to the plane
x + y + z = 1.
We want to minimize (x 2)2 +y 2 +(z + 3)2 where z = 1 x y, so
we look for critical points of f (x; y) = (x 2)2 +y 2 +(4 x y)2 .
4
We have fx = 2 (x 2) 2 (4 x y) and fy = 2y 2 (4 x y),
so if both are zero, then 2 (x 2) 2y, that is, y = x 2. But
fx (x; x 2) = 2 (x 2) 2 (6 2x) = 6x 16, so x = 8=3 and
y = 2=3 and z = 7=3. This has to be where the minimum is
achieved, so the shortest distance is
p
p
4=9 + 4=9 + 4=9 = 2= 3
41. Find the points on the cone z 2 = x2 + y 2 that are closest to the
point (4; 2; 0).
We want to minimize (x 4)2 + (y 2)2 + z 2 , where z 2 = x2 + y 2 .
So we want to minimize
f (x; y) = 2x2
8x + 2y 2
4y + 20
Setting fx = 4x 8 and fy = 4y 4 equalpto zero we get xp= 2
5 ,
and y = 1, so the closest points are 2; 1; 5 and 2; 1;
p
p
and the distance is 4 + 1 + 5 = 10.
5

Download Report