195
CHAPTER FOURTEEN
Solutions for Section 14.1
1. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point
and is a local maximum; the point C is a saddle point.
2. (a) P is a local maximum.
(b) Q is a saddle point.
(c) R is a local minimum.
(d) S is none of these.
3. Figure 14.1 shows the gradient vectors around
direction of increasing values of the function.
P
and
Q pointing perpendicular to the contours and in the
y
6
5
4
4
2
Q
,3 , 1
1
0
,6
P
3
5
1
6
,5
,6 x
,4 ,2
,3
,1
,1
4
2
0
1
5
3
1 6i
,6
1
S
R
0
,4 ,2
0
R- P
,6 x
4
,6
5
3
2
,5
,3 ,5
,1
0
0
,5
y
,2 4
,
Q II iY
I
R
Rz RR )
W
,1:
0
,4
S
,3
,2
,6
1
0
2
1
,1
3
6
6
Figure 14.1
Figure 14.2
4. Figure 14.2 shows the direction of rf at the points where krf k is largest, since at those points the contours
are closest together.
5. The partial derivatives are fx (x; y) = 3x2 , 3 which vanishes for x = 1 and fy (x; y) = 3y2 , 3 which
vanishes for y = 1. The points (1; 1), (1; ,1), (,1; 1), (,1; ,1) where both partials vanish are the critical
points. To determine the nature of these critical points we calculate their discriminant and use the second
derivative test. The discriminant is
2
(x; y ) = (6x)(6y ) , 0 = 36xy .
D = fxx (x; y)fyy (x; y) , fxy
At (1; 1) and (,1; ,1) the discriminant is positive. Since fxx (1; 1) = 6 is positive, (1; 1) is a local minimum.
And since fxx (,1; ,1) = ,6 is negative, (,1; ,1) is a local maximum. The remaining two points, (1; ,1)
and (,1; 1) are saddle points since the discriminant is negative.
196
CHAPTER FOURTEEN /SOLUTIONS
6. First, we identify the critical points. The partials are fx (x; y) = 3x2 and fy (x; y) = ,2ye,y . These will
vanish simultaneously when x = 0 and y = 0, so our only critical point is (0; 0). The discriminant is
2
2
2 ,y
D = fxx (x; y)fyy (x; y) , fxy
, 2e,y ) , 0 = 12xe,y (2y2 , 1).
(x; y ) = (6x)(4y e
2
2
2
Unfortunately, the discriminant is zero at the origin so the second derivative test can tell us nothing about our
critical point. We can, however, see that we are at a saddle point by looking at the behavior of f (x; y) along
the line y = 0. Here we have f (x; 0) = x3 + 1, so for positive x, we have f (x; 0) > 1 = f (0; 0) and for
negative x, we have f (x; 0) < 1 = f (0; 0). So f (x; y) has neither a maximum nor a minimum at (0; 0).
7. Find the critical point(s) by setting
fx = (xy + 1) + (x + y) y = y2 + 2xy + 1 = 0;
fy = (xy + 1) + (x + y) x = x2 + 2xy + 1 = 0;
then we get x2 = y2 , and so x = y or x = ,y.
If x = y, then x2 + 2x2 + 1 = 0, that is, 3x2 = ,1, and there is no real solution. If x = ,y, then
2
x , 2x2 + 1 = 0, which gives x2 = 1. Solving it we get x = 1 or x = ,1, then y = ,1 or y = 1,
respectively. Hence, (1; ,1) and (,1; 1) are critical points.
Since
fxx (x; y) = 2y;
fxy (x; y) = 2y + 2x
fyy (x; y) = 2x;
and
the discriminant is
2
D(x; y) = fxx fyy , fxy
2
= 2y 2x , (2y + 2x)
2
2
= ,4(x + xy + y ):
thus
8.
D(1; ,1) = ,4(12 + 1 (,1) + (,1)2 ) = ,4 < 0;
D(,1; 1) = ,4((,1)2 + (,1) 1 + 12) = ,4 < 0:
Therefore (1; ,1) and (,1; 1) are saddle points.
At a critical point, fx = 0, fy = 0.
fx = 8y , (x + y)3 = 0; we know 8y = (x + y)3 :
fy = 8x , (x + y)3 = 0; we know 8x = (x + y)3 :
Therefore we must have x = y. Since (x + y)3 = (2y)3 = 8y3 , this tells us that 8y , 8y3 = 0. Solving gives
y = 0 ; 1 :
Thus the critical points are (0; 0); (1; 1); (,1; ,1).
fyy = fxx = ,3(x + y)2 , and fxy = 8 , 3(x + y)2 .
The discriminant is
2
D(x; y) = fxx fyy , fxy
,
4
2
4
= 9(x + y ) , 64 , 48(x + y ) + 9(x + y )
2
= ,64 + 48(x + y ) :
14.1 SOLUTIONS
D(0; 0) = ,64 < 0, so (0; 0) is a saddle point.
D(1; 1) = ,64 + 192 > 0 and fxx (1; 1) = ,12 < 0, so (1; 1) is a local maximum.
D(,1; ,1) = ,64 + 192 > 0 and fxx (,1; ,1) = ,12 < 0, so (,1; ,1) is a local maximum.
9. To find critical points, set partial derivatives equal to zero:
Ex = sin x = 0
x = 0; ; 2; when
Ey = y = 0
when
y = 0:
The critical points are
(,2; 0); (,; 0); (0; 0); (; 0); (2; 0); (3; 0) To classify, calculate D = ExxEyy , (Exy )2 = cos x.
At the points (0; 0); (2; 0); (4; 0); (6; 0); D = (1) > 0
and
Exx > 0
Exx(0; 2k) = cos(2k) = 1):
(Since
Therefore (0; 0); (2; 0); (4; 0); (6; 0); are local minima.
At the points (; 0); (3; 0); (5; 0); (7; 0); , we have cos(2k + 1) = ,1, so
D = (,1) < 0:
Therefore (; 0); (3; 0); (5; 0); (7; 0); are saddle points.
10. At a critical point,
fx = cos x sin y = 0
so
cos x = 0 or sin y
=
0;
fy = sin x cos y = 0
so
sin x = 0 or cos y
=
0:
and
Case 1: Assume cos x = 0. This gives
x = , 32 ; , 2 ; 2 ; 32 ; (This can be written more compactly as: x = k + =2, for k = 0; 1; 2; .)
If cos x = 0, then sin x = 1 6= 0. Thus in order to have fy = 0 we need cos y = 0, giving
(More compactly, y = l + =2, for l
Case 2: Assume sin y = 0. This gives
y = , 32 ; , 2 ; 2 ; 32 ; =
0; 1; 2; )
y = , 2; ,; 0; ; 2; (More compactly, y = l, for l = 0; 1; 2; )
If sin y = 0, then cos y = 1 6= 0, so to get fy = 0 we need sin x = 0, giving
x = ; ,2; ,; 0; ; 2; 197
198
CHAPTER FOURTEEN /SOLUTIONS
(More compactly, x = k for k = 0; 0; 1; 2; )
Hence we get all the critical points of f (x; y). Those from Case 1 are as follows:
, 2 ; , 2 ; , 2 ; 2 ; , 2 ; 32 2 ; , 2 ; 2 ; 2 ; 2 ; 32 32 ; , 2 ; 32 ; 2 ; 32 ; 32 Those from Case 2 are as follows:
(,; ,); (,; 0); (,; ); (,; 2) (0; ,); (0; 0); (0; ); (0; 2) (; ,); (; 0); (; ); (; 2) More compactly these points can be written as,
(k; l );
and (k + ; l + );
for
k = 0; 1; 2; ; l = 0; 1; 2; k = 0; 1; 2; ; l = 0; 1; 2; for
2
2
To classify the critical points, we find the discriminant. We have
fxx = , sin x sin y;
fyy = , sin x sin y;
and
fxy = cos x cos y:
Thus the discriminant is
2
D(x; y) = fxx fyy , fxy
2
= (, sin x sin y )(, sin x sin y ) , (cos x cos y )
2
2
2
2
= sin x sin y , cos x cos y
2
2
2
2
= sin y , cos x: (Use: sin x = 1 , cos x and factor.)
At points of the form (k; l) where k = 0; 1; 2; ; l = 0; 1; 2; , we have
D(x; y) = ,1 < 0 so (k; l) are saddle points.
At points of the form (k + 2 ; l + 2 ) where k = 0; 1; 2; ; l = 0; 1; 2; D(k + 2 ; l + 2 ) = 1 > 0; so we have two cases:
If k and l are both even or k and l are both odd, then
fxx = , sin x sin y = ,1 < 0, so (k + 2 ; l + 2 ) are local maximum points.
If k is even but l is odd or k is odd but l is even, then
fxx = 1 > 0 so (k + 2 ; l + 2 ) are local minimum points.
11. The partial derivatives are
Set Px
=
0 and Py
Px = ,6x , 4 + 2y
=
and
Py = 2x , 10y + 48:
0 to find the critical point, thus
2y , 6x = 4 and
10y , 2x = 48:
Now, solve these equations simultaneously to obtain x = 1 and y = 5.
Since Pxx = ,6, Pyy = ,10 and Pxy = 2 for all (x; y), at (1; 5) the discriminant D
(2)2 = 56 > 0 and Pxx < 0. Thus P (x; y ) has a local maximum value at (1; 5).
,6)(,10) ,
= (
14.1 SOLUTIONS
199
12. At a local maximum value of f ,
@ f = ,2 x , B = 0 :
@x
We are told that this is satisfied by x = ,2. So ,2(,2) , B = 0 and B = 4. In addition,
@ f = ,2y , C = 0
@y
and we know this holds for y = 1, so ,2(1) , C = 0, giving C = ,2. We are also told that the value of f
is 15 at the point (2; 1), so
15 = f (,2; 1) = A , ((,2)2 + 4(,2) + 12 , 2(1)) = A , (,5); so A = 10:
Now we check that these values of A, B , and C give f (x; y) a local maximum at the point (,2; 1). Since
fxx (,2; 1) = ,2;
fyy (,2; 1) = ,2
and
13.
14.
fxy (,2; 1) = 0;
we have that the discriminant D = (,2)(,2) , 0 > 0 and fxx (,2; 1) < 0. Thus, f has a local maximum
value 15 at (,2; 1).
At the origin f (0; 0) = 0. Since x6 0 and y6 0, the point (0; 0) is a local (and global) minimum. The
second derivative test does not tell you anything since D = 0.
At the origin g(0; 0) = 0. Since y3 0 for y > 0 and y3 < 0 for y < 0, the function g takes on both positive
and negative values near the origin, which must therefore be a saddle point. The second derivative test does
not tell you anything since D = 0.
15. At the origin h(0; 0) = 1. Since cos x and cos y are never above 1, the origin must be a local (and global)
maximum. The second derivative test
,
D = hxx hyy , (hxy )2 = (, cos x cos y)(, cos x cos y) , (sin x sin y)2 x=0;y=0
,
= cos2 x cos2 y , sin2 x sin2 y x=0;y=0
=1>0
and hxx (0; 0) < 0, so (0; 0) is a local maximum.
16.
(a) To find the critical points, we must solve the equations
@f = ex (1 , cos y) = 0
@x
@f = ex (sin y) = 0:
@y
The first equation has solution
y = 0; 2; 4; : : ::
The second equation has solution
y = 0; ; 2; 3; : : ::
Since x can be anything, the lines
are lines of critical points.
y = 0; 2; 4; : : :
200
CHAPTER FOURTEEN /SOLUTIONS
(b) We calculate
D = (fxx )(fyy )2 , (fxy )2 = ex (1 , cos y)ex cos y , (ex sin y)2
2x
2
2
= e (cos y , cos y , sin y )
2x
= e (cos y , 1)
At any critical point on one of the lines y
=
0, y
D=e
2
2, y = 4, : : :,
=
x (1 , 1) = 0:
Thus, D tells us nothing. However, all along these critical lines, the value of the function, f , is zero.
Since the function f is never negative, the critical points are all both local and global minima.
17.
(a)
; 3) is a critical point. Since fxx > 0 and the discriminant
2
2
= fxx fyy , 0 = fxx fyy > 0;
D = fxx fyy , fxy
the point (1; 3) is a minimum.
(1
y
0
1
4
32
16
64
12
0
(b)
3
x
1
Figure 14.3
2
2
a; b) is a critical point. Since the discriminant D = fxx fyy , fxy
= ,fxy
< 0, (a; b) is a saddle point.
(
y
9
7
53
,7 5
,
,3
,1
0
1
1
1
3
5
7
9
x
0
(a)
(b)
0
18.
1
0
,1
,3
,5 7
,
Figure 14.4
14.1 SOLUTIONS
201
19. The partial derivatives are
fx (x; y) = 3x2 , 3y2
fy (x; y) = ,6xy.
and
Now fx (x; y) will vanish if x = y and fy (x; y) will vanish if either x = 0 or y = 0. Since the
partial derivatives are defined everywhere, the only critical points are where fx (x; y) and fy (x; y) vanish
simultaneously. (0; 0) is the only critical point.
To find the contour for f (x; y) = 0, we solve the equation x3 , 3xy2 = 0. This can be factored into
p
f (x; y) = x(x ,
p
p
3 y)(x +
p
3 y)
=
0
whose roots are x = 0, x = 3 y and x = , 3 y. Each of these roots describes a line through the origin;
the three of them divide the plane into six regions. Crossing any one of these lines will change the sign of
only one of the three factors of f (x; y), which will change the sign of f (x; y).
y
f >0
f <0
,2
,,11
2
11
y=x
p
3
0
,2 ,1
f<0
0
0
0
0
1
2
f >0 x
,,11
,2
11
2
f >0
f <0
p
y = ,x
3
Figure 14.5
20. The first order partial derivatives are
fx (x; y) = 2kx , 2y
and
fy (x; y) = 2ky , 2x.
And the second order partial derivatives are
fxx (x; y) = 2k
fxy (x; y) = ,2
fyy (x; y) = 2k
Since fx (0; 0) = fy (0; 0) = 0, the point (0; 0) is a critical point. The discriminant is
D = (2k)(2k) , 4 = 4(k2 , 1).
For k = 2, the discriminant is positive, D = 12. When k = 2, fxx (0; 0) = 4 which is positive so we have
a local minimum at the origin. When k = ,2, fxx (0; 0) = ,4 so we have a local maximum at the origin. In
the case k = 0, D = ,4 so the origin is a saddle point.
202
CHAPTER FOURTEEN /SOLUTIONS
Lastly, when k = 1 the discriminant is zero, so the second derivative test can tell us nothing. Luckily,
we can factor f (x; y) when k = 1. When k = 1,
f (x; y) = x2 , 2xy + y2 = (x , y)2 .
This is always greater than or equal to zero. So f (0; 0) = 0 is a minimum and the surface is a trough-shaped
parabolic cylinder with its base along the line x = y.
When k = ,1,
f (x; y) = ,x2 , 2xy , y2 = ,(x + y)2 .
This is always less than or equal to zero. So f (0; 0)
with its top ridge along the line x = ,y.
y
k = ,2
,16
,12
,8
,4
=
0 is a maximum. The surface is a parabolic cylinder,
y
k = ,1
,30
,20
,1
x
,5
,1
12
8
,5
4
x
,10
,8
,20
y
,4
1
,1
,1
1
,4
,8
x
4
8
,12
,30
k=1
,16
,12
16
,10
,1
y
k=0
,16
k=2
12
16
y
16
12
30
20
8
10
4
5
1
1
x
x
1
5
10
20
30
Figure 14.6
Solutions for Section 14.2
1. Mississippi lies entirely within a region designated as “80s” so we expect both the high and low daily
temperatures within the state to be in the 80s. The South-Western most corner of the state is close to a region
designated as “90s”, so we would expect the temperature here to be in the high 80s, say 87-88. The northern
most portion of the state is relatively central to the “80s” region. We might expect the temperature there to
be between 83-87.