Ism et chapter_12

12. (a) R(150, 2000) = 333
(b) R(160, 2000) = 354
(c) R(170, 2000) = 375
(d) The extra distance gained appears to be
a constant 21 feet.
Chapter 12 13. H(80, 20) = 77.4, H(80, 40) = 80.4, and
H(80, 60) = 82.8. It appears that at 80◦ ,
increasing the humidity by 20% increases the
heat index by about 3.
Functions of 14. H(90, 20) = 86.5, H(90, 40) = 92.3, and
H(90, 60) = 100.5. At 90◦ , each extra 20%
Several Variables humidity adds about 7 to the heat index (but
this is also not constant).
and Partial 15. x2 + y 2 = 1 or 2 or 3

Diﬀerentiation 1.5

1.0
z=3

z=2

z=1
0.5

0.0

12.1 Functions of Several −1.5 −1.0
x
−0.5 0.0 0.5 1.0 1.5

−0.5

Variables y
−1.0

1. Domain = {(x, y)|y = −x}
−1.5

2
2. Domain = {(x, y)|y = x }
z = y2
3. Domain = (x, y)| x2 + y 2 ≥ 1
2.0

4. Domain = (x, y) | 1 < x2 + y 2 ≤ 4
1.5

5. Domain = {(x, y, z)|x2 + y 2 + z 2 < 4}
z 1.0
6. Domain = {(x, y, z)|x2 + y 2 = z}
0.5
7. (a) Range = {z|z ≥ 0}
(b) Range = { z | 0 ≤ z ≤ 2 } 0.0
−1.0 −0.5 0.0 0.5 1.0
8. (a) Range = {z| − 1 ≤ z ≤ 1} y

(b) Range = { z | − 1 ≤ z ≤ 1 }
9. (a) Range = {z|z ≥ −1} z = x2 + y 2
π π
(b) Range = z | − ≤ z <
4 2 8

10. (a) Range = {z|z > 0} 6

(b) Range = z | 0 < z ≤ e2
4

11. (a) R(150, 1000) = 312
2
(b) R(150, 2000) = 333
0
(c) R(150, 3000) = 350 2 1 1 00 −1 −2
−1 −2
2 y
x
(d) The distance gained varies from 17 feet to
21 feet.

660

12.1. FUNCTIONS OF SEVERAL VARIABLES 661

16. For x = 0, z = −y 2 3

2
y
−1.0 −0.5 0.0 0.5 1.0 z=3

0.0 1 z=2
z=1

−0.5 0

−3 −2 −1 0 1 2 3
x
−1
z −1.0
y

−2
−1.5

−3
−2.0

x = 0 ⇒ z = |y|
For y = 0, z = x2 5

2.0
4

1.5
3
z

z 1.0 2

1
0.5

0
0.0 −5.0 −2.5 0.0 2.5 5.0
−1.0 −0.5 0.0 0.5 1.0 y
x

z= x2 + y 2
z = x2 − y 2
10

8 5

6
4

4
3
2
2
−5.0 0
−2.5 −5.0
−2.5 1
y 0.0
0.0
2.5 −2
2.5 x
5.0
5.0 0 4
−4 −4 −2 2
00
−2 2 4
x
−4
−6 y

−8

−10

x2 − y 2 = 1 or − 1. For z = −1; z = 1. 18. 2x2 − y = 0 or 1 or 2
10
8

8
7
6
6

4
5

2
4

0
3
−10 −8 −6 −4 −2 0 2 4 6 8 10 2.7
−2
x 2
.7
−4
1
y
−6
0 1.7

−8 −2 −1 0 1 2
−1
x
−10
−2

17. x2 + y 2 = 1 or 2 or 3 z = −y

662 CHAPTER 12. FUNCTIONS OF SEVERAL VARS. AND PARTIAL DIFF.

5 2.0

4 1.6

3 1.2

2 0.8

1 0.4

0 0.0
−5 −4 −3 −2 −1 0 1 2 3 4 5 −2 −1 0 1 2
−1 −0.4
y x

−2 −0.8
z y
−3 −1.2

−4 −1.6

−5 −2.0

z = 2x2 − y z= 4 − x2 − y 2
10.0
2.0

7.5
1.5

5.0
1.0

2.5 4 0.5

2 4
−4
−4 0
0.0
−2 0.0 2
−2
y −2
x 0 x 00
2
−4 4 −2 2
−2.5
y 4
−4

19. z = 4 − y2 20. z = 0 or y 2 − x2 = 4
5 10

4
8

3
6

2 4

1
2

0
0
−5 −4 −3 −2 −1 0 1 2 3 4 5
−1 −10 −8 −6 −4 −2 0 2 4 6 8 10
y −2
x
−2
−4
z
y
−3
−6

−4
−8

−5
−10

√
z= 4 − x2 z= 5 − y 2 . For x =1 or -1
5.0

2.5
2

z
1
0.0
−5 −4 −3 −2 −1 0 1 2 3 4 5
x
0

z −2.5 −2 −1 0 1 2
y

−5.0

4 − x2 − y 2 = 0 or 1 z= 4 − y2


2 2

z 1 1

0 4 3 0
2 1 1 2 3
−2 −1 0
0
−2 −1 0 1 2 y −1 −2
−3 x
y
−1

z= 4 + x2 − y 2

2.0
2.2
4
1.5

2.1 1.0
2 3
y
0.5 2
2.0 1
0.0
0 x
0
−1
−0.5
1.9 −2
−1.0 −2

1.8

1.0 1.0
0.5 0.5
0.0
−0.5 −0.5
−1.0 −1.0
x y

22. (a)
21. (a)
2.0
3
x 2
1.6 −3 −2 −1 y 0 1
0 1 2
−1
0 3
1.2 −2
−3
0.8 −10

0.4
−20
3 0.0
2 −3
1 −1 −2
0
0 −30
2 1 −1
3 −0.4 −2 x
−3
y
−0.8 −40

−1.2
−50

−1.6

−2.0

2.0
x y
3 2
−3 −2 1 −1 0 1 −1 2 3
−2 −3
1.6
0

1.2

−10
0.8

−3
0.4
−2 −20
−1 3
0.0 2
x 1
0
0
−1 −30
−2 −0.4 1
−3 y 2
−0.8
3 −40
−1.2

−1.6 −50

−2.0

(b) (b)


0.75

0.5
−3
3
−2 x 12 2 0.25
8 3
−1 1
4 −3 2
00 0 −2 0.0 1
−1
−1 −4 x 00
1 1
y −8 −1
−2 −0.25 2
−12 2 −2 3
y
−3 −3
−16
3
−4 −20 −0.5
−24
−28
−0.75
0.75

0.5

−4
12
0.25
−3
8
y −2
x
−2 4
−10
−1 3 0.0
1 2 2 −3
3 1 −2
0
0 0
0 −1
2 1 −1
3 −2 y
1−4 −3
x
−0.25
2
−8
3
−12
−0.5
−16

−20

23. (a) 24. (a)
0.8

0.6 −1.0
10
0.4 z
−0.5 x −10
5 −5
0.2 0
0.0
0
5
3
0.0 2 10 −5 y
−2 1 0.5
−1 00
−1 1 2
x −0.2
−2 −10
−3 1.0
y
−0.4

−0.6

−0.8

1.0

0.8

0.6 −10
−10
0.4 x zy −1.0
−5 −5
−0.5
0.2 −3
0.0 00
−2
0.5
3 0.0
−1
2 1.0 5
1 0 y
0 5
1 −0.2 −1 −2
−3 10
2 x
10
3 −0.4

−0.6

−0.8

−1.0

(b) (b)


2.0
5.0
1.5
3.0
2.5
3 2.5 2.0
1.0 1.5 3
z y 1.0
2 2
0.5
3 0.0
0.0 1
0.5
1 2 0
−1
1
0
0.0 0 −2
−2.5
−3 x
−1 −1 z
−2
x y −5.0
−3 −2

−3

2.0

3.0
1.5
2.5 5.0
3 z 2.0
1.0
1.5 y
2 2.5
1.0
0.5
−3 1 0.5
−2 3
2 0.0
0.0
−1 0.00 1
x 0
0 −1
−1 1 −2
y 2 −2.5 x −3
−2 3 z
−3
−5.0

25. (a) 26. (a)
3
−3
−3 −3
2 −2 3
2
−2 x z −2 yx −1 1
0
1 −1
−1 −1 −2 0.0*100
0
−3
00 0 1
1 1 3 2
−5.0*10
−1
2 y 2 3

3 −2 3 4
−1.0*10

−3
4
−1.5*10

4
−2.0*10

4
−2.5*10

y x −3
3 −3 −2
−2 −1
−1
0
0
1 0
1
2
2 0.0*10 3
2 3

3
1 −5.0*10

0 4
−1.0*10
3 2 1 0 −1 −2 −3
x
−1 4
−1.5*10
z
−2
4
−2.0*10

−3
4
−2.5*10

(b) (b)


2
3,000

2,500
y 1
2,000

1,500

0
1,000 -2 -1 0 1 2
x
500
−3 3 -1
−2 0 2
−1 1
x −1 00 1
−2 2
−3 3
y
-2

3,000
31.
2,500
10
2,000

1,500
y 5
1,000

3
2 500
1 −3
−2 0
00 −1
−1 x -10 -5 0 5 10
0 y
1 −2 x
2 −3
3
-5

-10

27. Function a → Surface 1
Function b → Surface 4 32.
Function c → Surface 2
Function d → Surface 3
2

y
1

28. Function a → Surface 1 0
-3 -2 -1 0 1 2 3
Function b → Surface 4 x

Function c → Surface 2 -1

Function d → Surface 3 -2

29.
33.
4

1
y
2

y
0.5
0
-4 -2 0 2 4
x
0
-2 -1 -0.5 0 0.5
x

-4 -0.5

-1

30. 34.


1 2

y 0.5 y 1

0 0
-1 -0.5 0 0.5 1 -2 -1 0 1 2
x x

-0.5 -1

-1 -2

35.
2

39. Surface a −→ Contour Plot A
y 1 Surface b −→ Contour Plot D
Surface c −→ Contour Plot C
Surface d −→ Contour Plot B
0
-2 -1 0 1 2
x

-1

-2
40. Density Plot a −→ Contour Plot A
Density Plot b −→ Contour Plot D
Density Plot c −→ Contour Plot C
Density Plot d −→ Contour Plot B
36.
2

y 1
41.
f(x,y,z)=0

0
-2 -1 0 1 2
x

-1
3

2

-2 1

z 0

-1 3
2
1
-2 0
37. x
-1
-2
-2
-3
-3
-3 0 -1
2 1
3
y
4

f(x,y,z)=2

y 2

0
-4 -2 0 2 4
x
3

-2 2

1
3
2
z 0 1
-4
0
-1 -1
x -2
-2 -3
-3
-2
-1
0
-3 1 y
2
3
38.


f(x,y,z)=-2

3

2

1
3
2
z 0 1
0
-1 -1
x -2
-2 -3
-3
-2
-1
0
-3 1 y
2
3

42.

45. Viewed from positive x-axis: View B
Viewed from positive y-axis: View A
25

20

15

10

5
46. Viewed from positive x-axis: View A
-3 -2 -1
0
-1 -2
-3 Viewed from positive y-axis: View B
2 1 0 1 2
3 -5 3
x y
-10

47. Plot of z = x2 + y 2 :

43.
f=-2,f=2

8

6

6 4

4 2

-3 -3
2 -2 0 -1 -2
-1 00
2 1 1 2
3 3
0 x y

-3
-2
-1
0 y
1
3 2 1 0 -1 -2 -3
3
2
Plot of x = r cos t, y = r sin t, z = r2 :
x

44. For f (x, y, z) = 1
f (x, y, z) = 0 8

6

4

2
2
−2 −2
1y x
−1 −1 -3
-3 -2 0 -2
-1 -1
0
1 0 1 2
0 0 3 2 3
0

1 1
z −1
2 2

−2 The graphs are the same surface. The grid is
diﬀerent.

f (x, y, z) = −1 48. The graphs are the same surface.


plotted on a very large scale.
(2) When the level curves just get very close to
1 each other and appear to intersect at a point
0 P, then it means that the function has a limit
-1 at the point P though can not be continuous
-2
at P.
-3
For the discontinuity at P, there are two pos-
-4
sibilities for an existing limit at that point.
-5 -1
-0.5
1 0.5 0 -0.5
0.5
0 (i) The limit along with value of the function
-1 1
at that point P exists but there aren’t
equal to each other. This demonstrates
49. Parametric equations are: the case (1).
x = r cos t, y = r sin t, z = cos r2 (ii) The limit exists but the value of the func-
The graphs are the same surface, but the para- tion doesn’t exist at that point. This
metric equations make the graph look much demonstrates the case (2).
cleaner:
53. Point A is at height 480 and “straight up” is
to the northeast. Point B is at height 470 and
“straight up” is to the south. Point c is at
-4
1
height between 470 and 480 and “straight up”
-4
-2
0.5
-2 is to the northwest.
0 00
2
4 -0.5
2
54. The two peaks are located inside the inner cir-
4

-1
cles. The peak on the left has elevation be-
tween 500 and 510. The peak on the right has
elevation between 490 and 500.

55. The curves at the top of the figure seem to
have more effect on the temperature, so those
50. The graphs are the same surface.
are likely from the heat vent and the curves to
the left are likely from the window. The cir-
cular curves could be from a cold air return or
something as simple as a cup of coffee.
1

0.5

0

-0.5 -1
-0.5
-1 0
1
0.5 0.5
0
-0.5 1
-1

51. Let the function be f (x, y) and whose contour
plot includes several level curves f (x, y) = ci
for i = 1, 2, 3...... Now, any two of these con-
tours intersect at a point P (x1 , y1 ), if and only
if f (x1 , y1 ) = cm = cn for cm = cn , which is 56. The point of maximum power will be inside all
not possible, as f can not be a function in that the contours, slightly toward the handle from
case. Hence different contours can not inter- the center. This is maximum because power
sect. increases away from the rim of the racket.

52. (1) When the level curves are very close to each 57. It is not possible to have a PGA of 4.0. If
other they appear to be intersecting at a point a student earned a 4.0 grade point average in
that is the point P, especially when they are high school, and 1600 on the SAT’s, their PGA


would be 3.942. It is possible to have a neg- 2
whenever (x − a) + (y − b) < δ.
2
ative PGA, if the high school grade point av- Hence lim (x + y) = a + b is verified.
erage is close to 0, and the SAT score is the (x,y)→(a,b)
lowest possible. It seems the high school grade
point average is the most important. The max- 2. We have lim x = 1 and lim y= 2
(x,y)→(1,2) (x,y)→(1,2)
imum possible contribution from it is 2.832. Therefore, by the definition 2.1, there exist
The maximum possible contribution from SAT ε1 , ε2 > 0, such that
verbal is 1.44, and from SAT math is 0.80. | x − 1 | < ε1 and | y − 2 | < ε2 ,
2 2
58. p(2, 10, 40) ≈ 0.8653 whenever (x − 1) + (y − 2) < δ.
p(3, 10, 40) ≈ 0.9350 Now consider
p(3, 10, 80) ≈ 0.9148 | (2x + 3y) − 8 | = |(2x − 2) + (3y − 6)|
p(3, 20, 40) ≈ 0.9231 = | 2 (x − 1) + 3 (y − 2)|
Thus we see that being ahead by 3 rather than ≤ 2|x − 1| + 3|y − 2|
by 2 increase the probability of winning. We < 2ε1 + 3ε2 = ε (say)
also see that having the 80 yards to the goal Thus, we have | (2x + 3y) − 8 | < ε,
instead of 40 yards decreases the probability of 2 2
whenever (x − 1) + (y − 2) < δ.
winning. Less time remaining in the game also Hence lim (2x + 3y) = 8 is verified.
increases the probability of winning. (x,y)→(1,2)

59. If you drive d miles at x mph, it will take you 3. We have lim f (x, y) = L and
(x,y)→(a,b)
d
hours. Similarly, driving d miles at y miles lim g (x, y) = M
x (x,y)→(a,b)
d Therefore, by the definition 2.1, there exist
per hour takes hours. The total distance
y ε1 , ε2 > 0, such that
d d | f (x, y) − L | < ε1 and | g (x, y) − M | < ε2 ,
traveled is 2d, and the time taken is + =
x y 2 2
d(x + y) whenever (x − a) + (y − b) < δ.
. The average speed is total distance Now consider
xy
2xy | (f (x, y) + g (x, y)) − (L + M ) |
divided by total time, so S(x, y) = . = |(f (x, y) − L) + (g (x, y) − M )|
x+y
60y ≤ | f (x, y) − L | + | g (x, y) − M |
If x = 30, then S(30, y) = = 40. < ε1 + ε2 = ε (say)
30 + y
We solve to get 60y = 1200 + 40y Thus, we have
20y = 1200 and y = 60 mph. If we replace 40 | (f (x, y) + g (x, y)) − (L + M ) | < ε,
with 60 in the above solution, we see that there 2 2
whenever (x − a) + (y − b) < δ.
is no solution. It is not possible to average 60 Hence lim (f (x, y) + g (x, y)) = L + M
mph in this situation. (x,y)→(a,b)
is verified.
60. We have P = RE. Substituting gives
d d 4. We have lim f (x, y) = L.
Y = = (x,y)→(a,b)
P RE Therefore, by the definition 2.1, there exist
ε1 > 0, such that
12.2 Limits and Continuity | f (x, y) − L | < ε1 ,
2 2
1. We have lim x = a and lim y= b whenever (x − a) + (y − b) < δ.
(x,y)→(a,b) (x,y)→(a,b)
Therefore, by the definition 2.1, there exist Now consider
ε1 , ε2 > 0, such that | (cf (x, y)) − cL | = |c| |(f (x, y) − L)|
|x − a| < ε1 and |y − b| < ε2 , < |c| ε1 = ε (say)
2 2
Thus, we have | (cf (x, y)) − cL | < ε,
whenever (x − a) + (y − b) < δ. 2 2
Now consider whenever (x − a) + (y − b) < δ.
|(x + y) − (a + b)| = |(x − a) + (y − b)| Hence lim cf (x, y) = cL is verified.
(x,y)→(a,b)
≤ |x − a| + |y − b|
< ε1 + ε2 = ε (say) x2 y
Thus, we have |(x + y) − (a + b)| < ε, 5. lim =3
(x,y)→(1,3) 4x2 − y

12.2. LIMITS AND CONTINUITY 671
√
x+y 1 3x3 x2 3
6. lim 2 − 2xy
= lim =
(x,y)→(2,−1) x 8 (x,x2 )→(0,0) x4 + x4 2
cos xy −1 Since the limits along these two paths do not
7. lim 2+1
= agree, the limit does not exist.
(x,y)→(π,1) y 2
exy 1 15. Along the path x = 0
8. lim 2 + y2
= 0
(x,y)→(−3,0) x 9 lim =0
(0,y)→(0,0) y 3
9. Along the path x = 0 Along the path x = y 3
0 y3 1
lim =0 lim =
(0,y)→(0,0) y 2 (y 3 ,y)→(0,0) 2y 3 2
Along the path y = 0 Since the limits along these two paths do not
3x2 agree, the limit does not exist.
lim =3
(x,0)→(0,0) x2
Since the limits along these two paths do not 16. Along the path x = 0
agree, the limit does not exist. 0
lim =0
(x,y)→(0,0) 8y 6
10. Along the path x = 0 Along the path x = y 3
2y 2 2y 6 2
lim = −2 lim =
(0,y)→(0,0) −y 2 (y 3 ,y)→(0,0) y 6 + 8y 6 9
Along the path y = 0 Since the limits along these two paths do not
0
lim =0 agree, the limit does not exist.
(x,0)→(0,0) 2x2
lim =0
(x,y)→(0,0) y 2
11. Along the path x = 0 Along the path y = x
0
lim =0 x sin x 1
(0,y)→(0,0) 3y 2 lim =
(x,x)→(0,0) 2x2 2
Along the path y = x
Since the limits along these two paths do not
4x2
lim =2 agree, the limit does not exist.
(x,x)→(0,0) 2x2
lim =0
(0,y)→(0,0) x3 + y 3
12. Along the path x = 0
0 Along the path y = x
lim =0
(0,y)→(0,0) 2y 2
x(cos x − 1)
Along the path x = y lim
2x2 2 (x,x)→(0,0) 2x3
lim 2 + 2x2
= (cos x − 1) − sin x
(x,x)→(0,0) x 3 = lim 2
= lim
Since the limits along these two paths do not x→0 2x x→0 4x
agree, the limit does not exist. − cos x 1
= lim =−
x→0 4 4
13. Along the path x = 0
0 where the last equalities are by L’Hopital’s
lim =0
(0,y)→(0,0) y 2 rule. Since the limits along these two paths
Along the path y = x3/2 do not agree, the limit does not exist.
2x4
lim 4 + x3
= 2. 19. Along the path x = 1
(x,x3/2 )→(0,0) x
Since the limits along these two paths do not 0
lim =0
agree, the limit does not exist. (1,y)→(1,2) y 2 − 4y + 4
Along the path y = x + 1
14. Along the path x = 0 x2 − 2x + 1 1
0 lim =
(x,x+1)→(1,2) 2x2 − 4x + 2 2
lim =0
(0,y)→(0,0) y 2 Since the limits along these two paths do not
Along the path y = x2 agree, the limit does not exist.


20. Along the path y = 0 x2 y x2 y
0 |f (x, y) − L| = ≤ = |y|
lim =0 x2 +y 2 x2
(x,0)→(2,0) (x − 2)2 Since lim |y| = 0,
(x,y)→(0,0)
Along the path x = 2
2y 2 Theorem 2.1 gives us that
lim =2 x2 y
(2,y)→(2,0) y 2 lim =0
(x,y)→(0,0) x2 + y 2
agree, the limit does not exist.
27. If the limit exists, it must be equal to 0 (to see
21. Along the path x = 0, y = 0 this use the path x = 0). To show that L = 0,
0 2x2 sin y 2x2 sin y
lim =0 |f (x, y) − L| = ≤
(0,0,z)→(0,0,0) z 2 2x2 + y 2 2x2
Along the path x2 = y 2 + z 2 = | sin y|
3(y 2 + z 2 ) 3 Since lim | sin y| = 0,
lim
2 +z 2 ,y,z)→(0,0,0) 2(y 2 + z 2 )
= (x,y)→(0,0)
(y 2
Theorem 2.1 gives us that
2x2 sin y
agree, the limit does not exist. lim =0
(x,y)→(0,0) 2x2 + y 2
22. Along the path x = 0, y = 0
z2 28. If the limit exists, it must be equal to 0 (to see
lim =1 this use the path x = 0). To show that L = 0,
(0,0,z)→(0,0,0) z 2
Along the path x = 0, z = 0
y2 x3 y + x2 y 3
lim = −1 |f (x, y) − L| =
(0,y,0)→(0,0,0) −y 2 x2 + y 2
Since the limits along these two paths do not x y + x2 y 3
3
agree, the limit does not exist. ≤
x2
23. Along the path x = 0, y = 0 = xy + y 3
0
lim =0
(0,0,z)→(0,0,0) z 3 Since lim |xy + y 3 | = 0,
Along the path x = y = z (x,y)→(0,0)
x4 1 Theorem 2.1 gives us that
lim = x3 y + x2 y 3
(x,x,x)→(0,0,0) 3x4 3 lim =0
Since the limits along these two paths do not (x,y)→(0,0) x2 + y 2
agree, the limit does not exist.
24. Along the path x = 0, y = 0 this use the path x = 0). To show that L = 2,
0
lim =0
(0,0,z)→(0,0,0) z 4
x3 + 4x2 + 2y 2
Along the path x = y = z |f (x, y) − L| = −2
x4 1 2x2 + y 2
lim 4
= x3
(x,x,x)→(0,0,0) 3x 3 =
Since the limits along these two paths do not 2x2 + y 2
agree, the limit does not exist. x3
≤
2x2
x
this use the path x = 0). To show that L = 0, =
xy 2 xy 2 2
|f (x, y) − L| = 2 2
≤ = |x|
x +y y2 x
Since lim |x| = 0, Since lim = 0,
(x,y)→(0,0) 2
(x,y)→(0,0)
Theorem 2.1 gives us that Theorem 2.1 gives us that
xy 2 x3 + 4x2 + 2y 2
lim =0 lim =0
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0) 2x2 + y 2

26. If the limit exists, it must be equal to 0 (to see 30. If the limit exists, it must be equal to −1 (to
this use the path x = 0). To show that L = 0, see this use the path x = 0). To show that


L = −1, L = 0,
3x3
x2 y − x2 − y 2 |f (x, y) − L| =
|f (x, y) − L| = +1 x2 + y 2 + z 2
x2 + y 2
3x3
x2 y ≤ = |3x|
= 2 x2
x + y2
Since lim |3x| = 0,
x2 y (x,y,z)→(0,0,0)
≤ = |y|
x2 Theorem 2.1 gives us that
3x3
lim =0
Since lim |y| = 0, (x,y,z)→(0,0,0) x2 + y 2 + z 2
(x,y)→(0,0)
Theorem 2.1 gives us that 34. If the limit exists, it must be equal to 0 (to see
x2 y − x2 − y 2 this use the path x = 0, y = 0). To show that
lim = −1
(x,y)→(0,0) x2 + y 2 L = 0,
x2 y 2 z 2
31. If the limit exists, it must be equal to 0 (To see |f (x, y) − L| =
x2 + y 2 + z 2
this use the path x = 0 or y = 1).
x2 y 2 z 2
To show that L = 0 , ≤ = y2 z2
x2 y − x2 x2
| f (x, y) − L | = 2
x + y 2 − 2y + 1 Since lim |y 2 z 2 | = 0,
(x,y,z)→(0,0,0)
x2 (y − 1) Theorem 2.1 gives us that
= 2
x2 + (y − 1) x2 y 2 z 2
2 lim =0
x (y − 1) (x,y,z)→(0,0,z) x2 + y 2 + z 2
≤ ≤ |y − 1| . √
x2
35. Since t is continuous for all t ≥ 0 and because
Since lim |y − 1| = 0 , 9 − x2 − y 2 is a polynomial, f is continuous
(x,y)→(0,1)
where x2 + y 2 ≤ 9.
Theorem 2.1 gives us that
x2 y − x2 36. Since e3x−4y , x2 and −y are continuous for all
lim 2 + y 2 − 2y + 1
= 0.
(x,y)→(0,1) x x and y, the sum is continuous for all x and y.
37. Since ln t is continuous for all t > 0, and
32. If the limit exists, it must be equal to 0 (To see 3−x2 +y is a polynomial, f is continuous where
this use the path x = −1 or y = 2). x2 − y < 3.
To show that L = 0,
38. Since tan t is continuous for all
2 (2n + 1)π
xy 2 − 4xy + 4x + (y − 2)
|f (x, y) − L| = t= and x + y is a polynomial, f is
2x2 + 4x + y 2 − 4y + 6 2
continuous oﬀ the collection of lines
2 2 (2n + 1)π
x(y − 2) + (y − 2) x+y =
= 2
2x2 + 4x + 2 + y 2 − 4y + 4 √
2
39. Since t is continuous for all t ≥ 0 and because
(x + 1) (y − 2) x2 +y 2 +z 2 −4 is a polynomial, f is continuous
= 2 2
2(x + 1) + (y − 2) where x2 + y 2 + z 2 ≥ 4.
2
√
(x + 1) (y − 2) 40. Since t is continuous for all t ≥ 0 and because
≤ 2 = |x + 1| . z − x2 − y 2 is a polynomial, f is continuous
(y − 2)
where z − x2 − y ≥ 0.
Since lim |x + 1| = 0 ,
(x,y) → (−1,2) 41. Along any line y = y0 , for y0 = 2, the limit
Theorem 2.1 gives us that 1
2 lim (y0 − 2) cos 2 does not exist.
xy 2 − 4xy + 4x + (y − 2) (x,y0 )→(0,y0 ) x
lim = 0. If y0 = 2, then
(x,y)→(−1,2) 2x2 + 4x + y 2 − 4y + 6
lim f (x, y) = 0 = f (0, 2).
(x,2)→(0,2)
33. If the limit exists, it must be equal to 0 (to see f is continuous for x = 0 and at the point
this use the path x = 0, y = 0). To show that (0, 2).


42. As a composition of continuous functions f is 1 − cos xy
lim
continuous for x2 + y 2 < 1. If (x0 , y0 ) satisﬁes (x,y)→(0,0) x2 y 2 + x2 y 3
x2 + y0 = 1, the limit
0
2
1 2 1 4
lim f (x) = 1 = f (x0 , y0 ) 1 − 1 − (xy) + (xy) ...
2 4!
(x,y)→(x0 ,y0 ) = lim
so the function is continuous for all (x, y) with (x,y)→(0,0) x2 y 2 (1 + y)
x2 + y 2 ≤ 1. 1 2 1 4 1 6
(xy) − (xy) + (xy) − ...
= lim 2 4! 6!
43. Since the limit (x,y)→(0,0) x2 y 2 (1 + y)
x2 − y 2 1 1 2 1 4
lim − (xy) + (xy) − ...
(x,y)→(x0 ,x0 ) x − y 2 4! 6!
= lim
(x − y)(x + y) (x,y)→(0,0) 1+y
= lim
(x,y)→(x0 ,x0 ) x−y 1
= = 0.5
= lim x + y = 2x0 = f (x0 , x0 ) 2
(x,y)→(x0 ,x0 )
for any x0 , the function f (x, y) is continuous
for all (x, y). 50.
x y f (x, y)
0.1 0.1 2.7273
1 −0.1 −0.1 3.3333
44. Since lim cos does not exist
(x,y)→(0,0) x2
+ y2 0.01 0.01 2.9703
along any path, the function is not continu- −0.01 −0.01 3.0303
ous at (0, 0). The function is continuous for all 0.001 0.001 2.9970
(x, y) = (0, 0). −0.001 −0.001 3.0030

45. Along the path x = 0, Thus, we estimate that the limit is 3.
y
lim , which is indetermi- 3 sin xy 2
(0,y)→(0,0) y 2+4−2
lim
nate. (x,y)→(0,0) x2 y 2 + xy 2
Therefore the limit does not exist. 1 3 1 5
3 xy 2 − xy 2 + xy 2 ...
3! 5!
= lim
46. Along the path y = x, (x,y)→(0,0) xy 2 (x + 1)
x2 − x2 1 1
lim √ √ , which is indeterminate. 3 1−
2
xy 2 +
4
xy 2 ...
(x,x)→(0,0) 3
x− 3x 3! 5!
Therefore the limit does not exist. = lim
(x,y)→(0,0) x+1
=3
47. Along the path y = −x,
x e−1 − 1
lim , which does not exist. 51. True. The limit is L, then the limit computed
(x,−x)→(0,0) x−x
Therefore the limit does not exist. along the line y = b must also be L.

48. Along the path x = 0, 52. False. The limit along a particular path be-
0 ing L does not imply that the limit is L (see
lim , which is indeterminate.
(0,y)→(0,0) 0 Example 2.3).
Therefore the limit does not exist.
53. False. The limit along two paths being L does
49. not imply that the limit is L. The limit must
x y f (x, y)
be the same along any path.
0.1 0.1 0.4545
−0.1 −0.1 0.5555
0.01 0.01 0.4950 54. True. As (x, y) → (0, 0), (cx, y) → (0, 0). Re-
−0.01 −0.01 0.5050 placing x by cx does not change the limit.
0.001 0.001 0.4995
−0.001 −0.001 .5005 55. The density plot shows sharp color changes
Thus, we estimate that the limit is 0.5. near the origin.


59. (a) The limit along the line y = kx:
1 xy 2 k 2 x3
lim = lim 2
(x,kx)→(0,0) x2 + y 4 x→0 x + k 4 x4
y 0.5
k2 x
= lim = 0.
x→0 1 + k 4 x2

-1 -0.5
0
0 0.5 1 (b) The limit along the line y = kx:
x
2xy 3
lim .
(x,kx)→(0,0) x2 + 8y 6
-0.5

2k 3 x4
-1 = lim 2
x→0 x + 8k 6 x6
2k 3 x2
= lim =0
56. The density plot shows sharp color changes x→0 1 + 8k 6 x4

near the origin. 60. (a) As shown in example 2.5, the limit as
(x, y) → (0, 0) does not exist, therefore
1
the function cannot be continuous there.
y 0.5
(b) The limit along the line y = kx
xy 2 k 2 x3
lim 2 + y4
= lim 2
0
(x,kx)→(0,0) x x→0 x (1 + k 4 x2 )
-1 -0.5 0 0.5 1 2
k x
x
= lim = 0.
-0.5
x→0 1 + k 4 x2
Therefore, along any straight line y = kx,
the function ”acts” continuous.
-1

61. As the several level curves of the function f
meet at (a, b). With the reference to the exer-
57. The density plot shows sharp color changes cise 52 of the section12.1, the lim f (x, y)
near the origin. (x,y)→(a,b)
exists, but f is not continuous at (a, b).
1
62. Given that f and g are continuous at (a, b) then
y 0.5
by the deﬁnition of continuity we have
lim f (x, y) = f (a, b) and
(x,y)→(a,b)
0 lim g (x, y) = g (a, b).
-1 -0.5 0 0.5 1 (x,y)→(a,b)
x
To prove f + g is continuous.
-0.5
lim (f + g) (x, y)
(x,y)→(a,b)

-1 = lim (f (x, y) + g (x, y))
(x,y)→(a,b)

= lim f (x, y) + lim g (x, y)
(x,y)→(a,b) (x,y)→(a,b)
58. The density plot shows sharp color changes
near the origin. = f (a, b) + g (a, b)
= (f + g) (a, b)
0.2 Thus f + g is continuous.
Similarly we can prove f − g is continuous.
y
0.1 lim (f − g) (x, y)
(x,y)→(a,b)

0
= lim (f (x, y) − g (x, y))
-0.2 -0.1 0 0.1 0.2 (x,y)→(a,b)
x

= lim f (x, y) − lim g (x, y)
-0.1 (x,y)→(a,b) (x,y)→(a,b)

= f (a, b) − g (a, b)
-0.2
= (f − g) (a, b)
Thus f − g is continuous.


2 2 2 2
63. Converting to polar coordinates, fx = ey −(x+y) − ey −x ,
2 2
x2 + y 2 r fy = ey −(x+y) + 2yf (x, y).
lim = lim =1
(x,y)→(0,0) sin x 2 + y2 r→0 sin r
y
by L’Hopital’s Rule. 2 y −1
9. fx = 3 ln(x yz) + 6 + x z ,
z
64. Converting to polar coordinates, y
2 2
ex +y − 1
2
er − 1 3x ln x
lim = lim =1 fy = + xz ,
(x,y)→(0,0) x2 + y 2 r→0 r2 y z
y
by L’Hopital’s Rule. 3x y ln x
fz = − xz .
z z2
65. Converting to polar coordinates,
xy 2 r3 cos θ sin2 θ xy xy
lim = lim −2x x2 y
(x,y)→(0,0) x2 + y 2 r→0 r2 10. fx = − 2xe z − e z ,
2
3 z
= lim r cos θ sin θ = 0
r→0 (x2 + y 2 + z 2 ) 2
xy
66. Converting to polar coordinates, −2y x3
fy = − e z ,
3 z
x2 y r3 cos2 θ sin θ (x 2 + y2 + z2 ) 2
lim = lim xy
(x,y)→(0,0) x2 + y 2 r→0 r2
−2z x3 y
= lim r cos2 θ sin θ = 0 fz = + 2 e z .
r→0
3 z
(x2 + y 2 + z 2 ) 2
12.3 Partial Derivatives ∂2f ∂2f ∂2f
11. = 6x, = −8x, = −8y
1. fx = 3x2 − 4y 2 , fy = −8xy + 4y 3 ∂x2 ∂y 2 ∂y∂x

2. fx = 2xy 3 − 3, fy = 3x2 y 2 ∂2f ∂2f ∂2f
12. = 2y, = −3 sin y, = 2x
∂x2 ∂y 2 ∂y∂x
3. fx = 2x sin xy + x2 y cos xy,
fy = x3 cos xy − 9y 2 4
13. fxx = − − 6y 3 ,
x2
2 1 2 5
4. fx = 6xyex y
− √ , fy = 3x2 ex y
fxy = −18xy 2 + ,
2 x−1 1 + y2
10y
x fxyy = −36xy − 2.
(1 + y 2 )
4e y y
5. fx = − 2 √
y x + y2 y
x 14. fxx = 16e4x + sin(x + y 2 ) + 3/2 ,
4x
4xe y x fxy = 2y sin x + y 2 − √ ,
1
fy = − 2 + 2 4 xy
y x + y2
fyyx = 2 sin x + y 2 + 4y 2 cos x + y 2
cos(x − y) 1
6. fx = + 2x tan y, + √ .
y 8y xy
cos(x − y) sin(x − y)
fy = − − + x2 sec2 y.
y y2 xy 3
15. fxx = 3/2
,
y
2
y
2
x
2
(1 − x2 y 2 )
7. f (x, y) = sin t dt = sin t dt − sin t dt fyz = yz sin(yz) − cos(yz),
x 0 0
fx = − sin x2 , fy = sin y 2 . fxyz = 0.

x+y 2z 2
y 2 −t2
16. fxx = 4y (xy + 1) e2xy − 3,
8. f (x y) = e dt (x + y)
2z 2
x
x+y x
fyy = 4x3 e2xy − 3 − xz sin(y + z),
2 2 2 2
(x + y)
−t −t
= ey dt − ey dt fyyzz = −
4
3 −2x cos(y+z)+xz sin(y+z).
0 0 (x + y)

12.3. PARTIAL DERIVATIVES 677

17. fww = 2tan−1 (xy) − z 2 ewz , nRV 3
·
2w 1 − x2 y 2 P V − n2 aV + 2n3 ab
3
fwxy = 2 , = −1.
(1 + x2 y 2 )
fwwxyz = 0. If we misunderstand the chain rule and con-
sider each of ∂P, ∂V and ∂T as separate quan-
y 1/2 tities and not the operators then, we get
18. fxx = − 3/2
− 6x sin w2 + z 2 , ∂ (T (P, V )) ∂ (P (V, T )) ∂ (V (P, T ))
4(x + z)
1/2 ∂P ∂V ∂T
(x + z) ∂T ∂P ∂V
fyy = − , = · · =1
4y 3/2 ∂P ∂V ∂T
fwxyz = 0.
∂P (V, T )
19. Taking partial derivative implicitly: 21. Find implicitly in
∂T
n2 a ∂ (V (P, T )) 14
P+ 2 P + 2 (V − 0.004) = 12T
V ∂T V
2n2 a ∂ (V (P, T )) ∂P (V, T )
+ − 3 . (V − nb) (V − 0.004) = 12 and
V ∂T ∂T
= nR. ∂P (V, T ) 12
= .
Multiply through by V 3 and ∂T V − 0.004
The increase in pressure due to an increase in
∂V 12
solve for to get: one degree will be (assuming V is much
∂T V
∂V larger than 0.004).
(P V 3 − n2 aV + 2n3 ab) = nRV 3
∂T
∂ (V (P, T )) nRV 3 22. In this case, we have
⇒ = 3 − n2 aV + 2n3 ab
. ∂T 7 7 125P
∂T PV =− 2 + +
∂V 6V 750V 3 12
20. We have (from Example 3.3) Thus, the increase in T due to an increase in
nRT n2 a V depends on V and P .
P = − 2
V − nb V
n2 a cL4
⇒ P + 2 (V − nb) = nRT . 23. S =
V wh3
∂S cL4 1 cL4 1
Therefore, on taking partial derivative implic- =− 2 3 =− =− S
∂w w h w wh3 w
itly:
∂P (T, V ) 2n2 a cL4
− (V − nb) 24. S =
∂V V3 wh3
n2 a ∂S cL4 3 cL4 3
+ P + 2 (1) = 0 or = −3 4 = − =− S
V ∂h wh h wh3 h
∂P (T, V ) 2n2 a P V 2 + n2 a
= − 2 25. The variable with the largest exponent has the
∂V V3 V (V − nb) largest proportional eﬀect. In this case h has
∂P (T, V ) 2n2 aV − 2n3 ab − P V 3 − n2 aV the greatest proportional eﬀect.
=
∂V V 3 (V − nb)
∂P (T, V ) − P V − n2 aV + 2n3 ab
3 26. The variable with the largest exponent (in ab-
= ... (i) solute value) has the greatest proportional ef-
∂V V 3 (V − nb)
fect.
∂ (T (P, V )) V − nb
Also, = ... (ii) ∂C ∂C
∂P nR 27. (10, 10) ≈ 1.4, (10, 10) ≈ −2.4
and from exercise 19, we have ∂t ∂s
∂ (V (P, T )) nRV 3 When the temperature is 10◦ and the wind
= 3 − n2 aV + 2n3 ab
.... (iii) speed is 10mph, an increase in temperature of
∂T PV
1◦ will increase the wind chill by approximately
Therefore, from (i) , (ii) and (iii)
1.4 degrees, whereas an increase in wind speed
∂ (T (P, V )) ∂ (P (V, T )) ∂ (V (P, T )) of 1mph will decrease the wind chill by 2.4 de-
∂P ∂V ∂T grees.
∂C
P V 3 − n2 aV + 2n3 ab V − nb If (10, 10) = 1, then a 1◦ increase in tem-
=− · ∂t
V 3 (V − nb) nR perature would correspond to a 1◦ increase in


the wind chill temperature. It is perhaps sur- 32. (a) We consider the y = 0 trace.
prising that a 1◦ increase in temperature leads ∂f
(1, 0) = 1 is the slope of this trace
to a greater increase in the “felt” temperature ∂x
(when the wind speed is 10mph). at (1, 0, 1).

∂C ∂C
28. (10, 20) ≈ 1.5, (10, 20) ≈ −1.1
∂t ∂s
For a ﬁxed temperature of 10◦ , the wind chill
is decreasing faster at a wind speed of 10mph 2.5

than at a wind speed of 20mph. 2

1.5
∂f
29. (170, 3000) ≈ 2.2 feet per ft/sec. 1

∂v 0.5
-2

An increase of 1 foot per second of velocity -1

0 0 x
increases the range by approximately 2.2 feet. -2
-1 1

∂f y
0
1 2
(170, 3000) ≈ 0.0195 feet per rpm. 2

∂w
An increase in 1 rpm increases the range by (b) We consider the x = 0 trace.
0.0195 feet. ∂f
(0, 2) = 1 is the slope of this trace
30. We need an additional 5ft of ﬂight. Since ∂y
∂f at (0, 2, 2).
(170, 3000) ≈ 0.02, there is an additional
∂w
5
= 250 rpm of backspin needed. 4
0.02
31. (a) We consider the y = 1 trace. 3

∂f
(1, 1) = −2 is the slope of this trace 2
∂x
at (1, 1, 2). 1
3
2
1
0 0 y
−1
−2
−3 −2 −1 −3
0 1 2 3
x

4

2 ∂f ∂f
33. = 2x, = 2y
0 ∂x ∂y
-2
∂f
-2
= 0 at x = 0.
-4
-1 ∂x
-2
0
∂f
-1 x = 0 at y = 0.
y
0
1
1 ∂y
2 2 This means there are horizontal tangent lines
to the trace in the y = 0 plane and the x = 0
(b) We consider the x = 2 trace.
∂f plane at (0, 0). This corresponds to the mini-
(2, 0) = 0 is the slope of this trace mum value of the function.
∂y
at (2, 0, 0). ∂f ∂f
34. = 2x − 4x3 , = 2y
∂x ∂y
∂f 1
= 0 at x = 0, ± √ .
2.5 ∂x 2
0.0 ∂f
= 0 at y = 0.
−2.5
∂y
−5.0
This means there are horizontal tangent lines
−7.5
2
3
to the trace in the y = 0 plane and the x = 0
−10.0

−12.5
1 plane at (0, 0) and in the plane perpendic-
0 y
1
−3
−2 −2
−1
ular to the x- and y-axes at √ , 0 and
−1
0
1 −3
2
2
x 3 1
−√ , 0 .
2


∂f ∂f f (0, 0) = 6, f (0, 1) = 8, f (0, −1) = 4
35. = cos x sin y, = sin x cos y
∂x ∂y
∂f π ∂f f (0, ±1) − f (0, 0)
= 0 when either x = + nπ, or y = mπ. ⇒ (0, 0) ≈ = 2.
∂x 2 ∂y ±1
∂f π ∂f f (−1, 1) − f (0, 1)
= 0 when either x = nπ, or y = + mπ. (b) (0, 1) ≈
∂y 2 ∂x −1
π π
When x = + nπ and y = + mπ, 6−8
2 2 = = 2.
f (x, y) = 1 if m and n are both even and if m −1
∂f f (0, 1) − f (0, 0)
and n are both odd, and f (x, y) = −1 if one is (0, 1) ≈
odd and the other is even. These are maximum ∂y 1
8−6
and minimum points. If x = nπ and y = mπ, = = 2.
f (x, y) = 0 and these points are neither min- 1
ima nor maxima. ∂f f (2 + 0.5, 0) − f (2, 0)
(c) (2, 0) ≈ ... (1)
∂f 2 2 ∂f 2 2
∂x 0.5
36. = −2xe−x −y , = −2ye−x −y . ∂f f (2 − 1.5, 0) − f (2, 0)
∂x ∂y (2, 0) ≈ ... (2)
∂f ∂x −1.5
= 0 at x = 0.
∂x (1) - (2) ⇒
∂f ∂f f (2.5, 0) − (0.5, 0)
= 0 at y = 0. 4 (2, 0) ≈
∂y ∂x 0.5
This means that there are horizontal tangent ∂f f (2.5, 0) − f (0.5, 0)
lines to the trace in the x = 0 plane and the (2, 0) ≈
∂x 2
y = 0 plane at (0, 0). 10 − 8
= =1
2
c
37. fx = − cos (x + y) , Now using (1),
y−b f (2, 0) ≈ f (2.5, 0) − 0.5
−cx
fy = 2 − cos (x + y) ,
∂f
(y − b) ⇒ (2, 0) = 9.5
c ∂x
fyx = − ∂f f (2, 0.5) − f (2, 0)
2 + sin (x + y) , ⇒ (2, 0) ≈ = 1.
(y − b) ∂y 0.5
c
fxy = − 2 + sin (x + y)
(y − b) ∂f f (−1, 0) − f (0, 0)
40. (a) (0, 0) ≈ = −2.
Thus, fyx = fxy . ∂x −1
∂f f (0, 1) − f (0, 0)
(0, 0) ≈ = −2.
b e(b/(x−y) ) ∂y 1
38. fx = cy xcy−1 − 2 ,
(x − y) ∂f f (−1, 1) − f (0, 1)
2b e(b/(x−y) ) (b) (0, 1) ≈ = −2
fxy = cx cy−1 2
+ c yx cy−1
ln x − ∂x −1
(x − y)
3 ∂f f (0, 0) − f (0, 1)
(0, 1) ≈ = −2
2 (b/(x−y) )
b e ∂y −1
+ 4 ,
(x − y) (c) f (2.5, 0) = 4, f (0, 0) = 6
b e(b/(x−y) ) ∂f f (2.5, 0) − f (2, 0)
fy = cxcy ln x − , (2, 0) ≈
(x − y)
2 ∂x 0.5
∂f
2b e(b/(x−y) ) ⇒ (0.5) · (2, 0) ≈ 4 − f (2, 0) ... (1)
fyx = cxcy−1 + c2 yxcy−1 ln x − 3
∂x
(x − y) ∂f f (0, 0) − f (2, 0)
2 (b/(x−y) ) (2, 0) ≈
b e ∂x −2
+ 4 ∂f
(x − y) ⇒ 2· (2, 0) ≈ f (2, 0) − 6... (2)
∂x
Thus, fyx = fxy . ∂f
( 1 ) + ( 2) ⇒ (2.5) · (2, 0) ≈ −2 or
∂x
39. (a) f (0, 0) = 6, f (0.5, 0) = 8, ∂f
(2, 0) ≈ −0.8.
f (−0.5, 0) = 4, ∂x
∂f f (±0.5, 0) − f (0, 0) Therefore, from (2) we get f (2, 0) = 4.4.
⇒ (0, 0) ≈ Hence,
∂x ±0.5
±2 ∂f f (2, −1) − f (2, 0)
= = 4. (2, 0) ≈ = −1.6
±0.5 ∂y −1


∂f f (x + h, y, z) − f (x, y, z) ∂A 1
41. (x, y, z) = lim Then = P − 2L.
∂x h→0 h ∂L 2
∂f f (x, y + h, z) − f (x, y, z) Using the formula A = LW ,
(x, y, z) = lim ∂A 1
∂y h→0 h we get = W = P − L.
∂f f (x, y, z + h) − f (x, y, z) ∂L 2
(x, y, z) = lim The difference between the two derivatives can
∂z h→0 h be explained by noting that in the first case,
42. There are 9 second order partial derivatives of the perimeter is being held constant while the
f (x, y, z). Assuming the equality of mixed par- length is changing, and therefore the width
tial derivatives fxy = fyx , fxz = fzx , is changing as well. In the second case, the
fyz = fzy , there are 6 different second order width is being held constant while the length
partial derivatives. is changing, and therefore the perimeter must
be changing too. We should not expect these
y(x4 + 4x2 y 2 − y 4 )
43. fx (x, y) = , and to be the same!
(x2 + y 2 )2
∂f
x(x4 − 4x2 y 2 − y 4 ) 47. (x0 , y0 ) is equal to the slope at x0 of the
fy (x, y) = ∂x
(x2 + y 2 )2 curve obtained by intersecting the surface
for (x, y) = (0, 0). ∂2f
For (x, y) = (0, 0), the limit definition gives z = f (x, y) with the plane y = y0 . (x0 , y0 )
∂x2
fx = fy = 0. We compute is the concavity of this curve at the point
fx (0, 0 + h) − fx (0, 0) x = x0 .
fxy (0, 0) = lim
h→0 h ∂2f
−h5 48. (x0 , y0 ) is the concavity of the curve
lim = −1. ∂y 2
h→0 h5
fy (0 + h, 0) − fy (0, 0) formed by intersecting z = f (x, y) with the
fyx (0, 0) = lim plane x = x0 ,at the point y = y0 .
h→0 h
h5 f (0, 1) − f (1, 1) 0−1
lim = 1. 49. (a) fx (1, 1) ≈ = =1
h→0 h5 −1 −1
The mixed partial derivatives are not continu- f (−1, 1) − f (0, 1) 1−0
ous on an open set containing (0, 0). (b) fx (0, 1) ≈ = =
−1 −1
44. For x = 0 the function f is the constant func- −1
tion 0. Similarly if y = 0. Therefore the two f (1, 2) − f (1, 0) 0−2
(c) fy (1, 0) ≈ = =
first order partial derivatives are 0 at the point 2 2
−1
(0, 0). f (1, 2) − f (1, 1) 0−1
(d) fy (1, 1) ≈ = =
∂f −1 y2 1 1
45. = 2 sin xy 2 + cos xy 2 −1
∂x x x
∂2f −2xy 2y f (h, 0) − f (0, 0)
= 2
cos xy 2 + cos xy 2 50. fx (0, 0) ≈ ≈ 0.
∂x∂y x x h
y2 f (0, h) − f (0, 0)
+ (−2xy) sin xy 2 fy (0, 0) ≈ ≈ ∞.
x h
= −2y 3 sin xy 2 . 51. Consider fx = 2x sin y + 3x2 y 2 , therefore

∂f 2yx f= fx dx = x2 sin y + x3 y 2 + g (y)
= cos xy 2 = 2y cos xy 2
∂y x ⇒ fy = x2 cos y + 2x3 y + g (y),
∂2f √
= −2y 3 sin xy 2 . but fy = x2 cos y + 2x3 y + y.
∂y∂x √ 2 √
Therefore, g (y) = y or g(y) = y y + c,
∂2f ∂2f 3
= , but differentiating with respect 2 √
∂x∂y ∂y∂x ⇒ f (x, y) = x2 sin y + x3 y 2 + y y + c.
to y first was much easier! 3
x
46. The perimeter, length and width are related by 52. Consider fx = yexy + 2 , therefore
x +1
P − 2L 1
P = 2L + 2W , so that W = . f = fx dx = exy + ln x2 + 1 + g (y)
2 2
Substitute into the area formula to get
P − 2L 1 ⇒ fy = xexy + g (y),
A = LW = L = LP − L2 . but fy = xexy + y cos y
2 2


Therefore, g (y) = y cos y or −0.5
= V.
g (y) = y sin y + cos y + c. Thus 1 + 0.1(1 − T )
1/2 The inflation rate has a greater influence
f (x, y) = exy +ln x2 + 1 +y sin y+cos y+c
on V .
2x 2 ∂V 5
53. Consider fx = + 2 (b) =− V
x2 + y 2 x −1 ∂I 1+I
∂V 3.6
f= fx dx = V
∂r 1 + 0.72r
The inflation rate has the greater influ-
x−1
= ln x2 + y 2 + ln + g (y) ence on V .
x+1
2y x−1 ∂p
⇒ fy = 2 + ln + g (y), 57. = cos x cos t. This describes the change in
x + y2 x+1 ∂x
3 2y the position of the string at a fixed time as the
but fy = 2 + . distance along the string changes.
y + 1 x2 + y 2
3 ∂p
Therefore, g (y) = 2 or = − sin x sin t. This describes the change in
y +1 ∂t
position of the string at a fixed distance from
g (y) = 3tan−1 y + c the end as time changes.
x−1
⇒ f = ln x2 + y 2 + ln
x+1 ∂p
58. = p0 (x − ct) −µe−µt − cp0 e−µt
+ 3tan−1 y + c ∂t
= −µp − cp0 eµt
y
54. Consider fx = + 2 cos (2x + y) ∂p
√ x = p0 e−µt
f = 2 xy + sin(2x + y) + g(y) ∂x
∂p ∂p
x Therefore = −c − µp
fy = + cos (2x + y) + g (y), ∂t ∂x
y ∂p
x gives the rate of change of the concentra-
but fy = + cos (2x + y) . ∂t
y tion of pollutant at fixed position.
Therefore, g (y) = 0 or g(y) = c ∂p
√ gives the rate of change of the concentra-
⇒ f (x, y) = 2 xy + sin (2x + y) + c ∂x
tion of pollutant at a particular time as the
∂fn location in the stream varies.
55. (a) = nπ cos nπx cos nπct ∂p ∂p
∂x = −c − µp says that the time rate of
2
∂ fn ∂t ∂x
= −n2 π 2 sin nπx cos nπct change of concentration at a particular loca-
∂x2 tion is related to the rate of change with dis-
∂fn
= −nπc sin nπx sin nπct tance along the stream and also to the current
∂t
2
∂ fn concentration and the decay rate.
= −n2 π 2 c2 sin nπx cos nπct
∂t2 H
∂ 2 fn ∂ 2 fn 59. (a) G/T = − S, so
So c2 2
= . T
∂x ∂t2 ∂(G/T ) −H
= 2 .
∂2f ∂T T
(b) (x − ct) = f (x − ct) = c2 f (x − ct)
∂x2 1
(b) Let U = .
∂2f T
= c2 2 (x − ct) G
∂x = UH − S
dx df /dt f (x − ct)(−c) T
= = = −c ∂(G/T ) ∂(G/T )
dt df /dx f (x − ct) = =H
so c gives the velocity. ∂(1/T ) ∂U

∂V (1 + 0.1(1 − T ))5 ∂R (R1 R2 + R1 R3 + R2 R3 )(R2 R3 )
56. (a) = −5000 60. =
∂I (1 + I)6 ∂R1 (R1 R2 + R1 R3 + R2 R3 )2
−5 (R1 R2 R3 )(R2 + R3 )
= V. −
1+I (R1 R2 + R1 R3 + R2 R3 )2
2
∂V (1 + 0.1(1 − T ))4 2 2
R2 R3 R
= (5)(−0.1)1000 = = .
∂T (1 + I)5 (R1 R2 + R1 R3 + R2 R3 )2 R1


Due to the symmetry we can easily write: (a) fx (2, 1) = 4, fy (2, 1) = 2.
2 2
∂R R ∂R R The tangent plane at (2, 1, 4) is
= , and = 4(x − 2) + 2(y − 1) − (z − 4) = 0
∂R2 R2 ∂R3 R3
The normal line is
(100)(60) x = 2 + 4t, y = 1 + 2t, z = 4 − t
61. P (100, 60, 15) = = 400.
15 (b) fx (0, 2) = 0, fy (0, 2) = 4.
Since 100 animals were tagged, we estimate
1 The tangent plane at (0, 2, 3) is
that we tagged of the population. 0(x − 0) + 4(y − 2) − (z − 3) = 0
4
∂P TS ∂P The normal line is
= − 2 , so (100, 60, 15) ≈ −27. x = 0, y = 2 + 4t, z = 3 − t
∂t t ∂t
If one more recaptured animal were tagged, our 2
−y 2 2
−y 2
estimate of the total population would decrease 2. fx = −2xe−x , fy = −2ye−x
by 27 animals.
(a) fx (0, 0) = 0, fy (0, 0) = 0.
62. Temperature is colder to the north (greater lat- The tangent plane at (0, 0, 1) is
itude and therefore greater y). Therefore we 0(x − 0) + 0(y − 0) − (z − 1) = 0
∂T The normal line is
expect < 0. x = 0, y = 0, z = 1 − t
∂y
If a cold front is moving from east to west, (b) fx (1, 1) = −2e−2 , fy (1, 1) = −2e−2 .
then we expect the temperature to be colder The tangent plane at (1, 1, e−2 ) is
in the east (which has smaller longitude, x) −2e−2 (x−1)−2e−2 (y −1)−(z −e−2 ) = 0
∂T
and therefore > 0. −2e−2 (x + y) − 5e−2 − z = 0
∂x The normal line is
∂P x = 1 − 2e−2 t, y = 1 − 2e−2 t, z = e−2 − t
63. (a) = 0.75L−0.25 K 0.25
∂L 3. fx = cos x cos y, fy = − sin x sin y
∂P
= 0.25L0.75 K −0.75
∂K (a) fx (0, π) = −1, fy (0, π) = 0.
(b) From Part (a), The tangent plane at (0, π, 0) is
∂P −1(x − 0) + 0(y − π) − (z − 0) = 0
= 0.75L−0.25 K 0.25 The normal line is
∂L
∂P x = −t, y = π, z = −t
= 0.25L0.75 K −0.75 and therefore, π π
∂K (b) fx ( , π) = 0, fy ( , π) = 0.
2 2
∂2P π
= −0.1875L−1.25 K 0.25 < 0 The tangent plane at ( , π, −1) is
∂L2 2
π
∂2P 0(x − ) + 0(y − π) − (z + 1) = 0
= −0.1875L0.75 K −1.75 < 0 2
∂K 2 The normal line is
π
x = , y = π, z = −1 − t
∂2P 2
(c) = 0.1875L−0.25 K −.75 > 0
∂L∂K
4. fx = 3x2 − 2y, fy = −2x
∂D1 ∂D2
64. = −5, and = −6. They are com- (a) fx (−2, 3) = 6, fy (−2, 3) = 4.
∂p2 ∂p1
plementary because an increase in the price of The tangent plane at (−2, 3, 4) is
one decreases the demand for the other. 6(x + 2) + 4(y − 3) − (z − 4) = 0
The normal line is
65. They are called substitute commodities because x = −2 + 6t, y = 3 + 4t, z = 4 − t
they behave similarly. (b) fx (1, −1) = 5, fy (1, −1) = −2.
The tangent plane at (1, −1, 3) is
5(x − 1) − 2(y + 1) − (z − 3) = 0
12.4 Tangent Planes and Lin- The normal line is
x = 1 + 5t, y = −1 − 2t, z = 3 − t
ear Approximations
x y
5. fx = , fy =
1. fx = 2x, fy = 2y x2 + y2 x2 + y2

12.4. TANGENT PLANES AND LINEAR APPROXIMATIONS 683

−3 4 1
(a) fx (−3, 4) = , fy (−3, 4) = . 9. fx = √ , fy = zsec2 (yz) ,
5 5 1 − x2
The tangent plane at (−3, 4, 5) is fz = ysec2 (yz)
−3 4
(x + 3) + (y − 4) − (z − 5) = 0 1 1 1
5 5 (a) fx 0, π, = 1, fy 0, π, = ,
The normal line is 4 4 2
3 4
x = −3 − t, y = 4 + t, z = 5 − t 1
5 5 fz 0, π, = 2π
4 3 4
(b) fx (8, −6) = , fy (8, −6) = − . 1 1
5 5 L 0, π, = 1 + (x − 0) + (y − π)
The tangent plane at (8, −6, 10) is 4 2
4 3 1
(x − 8) − (y + 6) − (z − 10) = 0 + 2π z −
5 5 4
The normal line is y
4 3 = x + + 2πz + 1 − π
x = 8 + t, y = −6 − t, z = 10 − t 2
5 5
1 √ 1
4 4x (b) fx √ , 2, 0 = 2, fy √ , 2, 0 = 0,
6. fx = , fy = − 2 2
y y2
1
fz √ , 2, 0 = 2
(a) fx (1, 2) = 2, fy (1, 2) = −1. 2
The tangent plane at (1, 2, 2) is π √ 1
L (x, y, z) = + 2 x − √
2(x − 1) − 1(y − 2) − (z − 2) = 0 4 2
The normal line is + 0 (y − 2) + 2 (z − 0)
√ π
x = 1 + 2t, y = 2 − t, z = 2 − t = 2x + 2z + − 1
1 4
(b) fx (−1, 4) = 1, fy (−1, 4) = . 1
4 10. fx = eyz − ,
The tangent plane at (−1, 4, −1) is 2 x − y2
1 y
(x + 1) + (y − 4) − (z + 1) = 0 fy = xzeyz + ,
4 x − y2
The normal line is yz
1 fz = xye
x = −1 + t, y = 4 + t, z = −1 − t
4 1
(a) fx (4, 1, 0) = 1 − √ ,
x y 2 3
7. fx = , fy = 1
x 2 + y2 x 2 + y2
fy (4, 1, 0) = √ ,
3
(a) fx (3, 0) = 1, fy (3, 0) = 0 fz (4, 1, 0) = 4
L(x, y) = 3 + 1(x − 3) + 0(y − 0) = x √
L(x, y, z) = 3 − 3
(b) fx (0, −3) = 0, fy (0, −3) = −1
L(x, y) = 3 + 0(x − 0) − 1(y + 3) = −y 1
+ 1− √ (x − 4)
2 3
8. fx = cos x cos y, fy = − sin x sin y 1
+ √ (y − 1) + 4(z − 0)
π π 1 π π −1 3
(a) fx , = , fy , =
4 4 2 4 4 2 1 1 2
1 1 π = 1− √ x + √ y + 4z − 1 − √
L (x, y) = + x− 2 3 3 3
2 2 4
1 π 1
− y− (b) fx (1, 0, 2) = ,
2 4 2
x y 1 fy (1, 0, 2) = 2,
= − +
2 2 2 fz (1, 0, 2) = 0 L(x, y, z)
√ √
π π 3 π π 3 1
(b) fx , = , fy , =− = 0 + (x − 1) + 2(y − 0) + 0(z − 2)
3 6 4√ 3 6 4 2
1 1
3 3 π = x + 2y −
L (x, y) = + x− 2 2
4 4√ 3
3 π 11. fw = 2wxy − yzewyz ,
− y− fx = w2 y,
√ 4
√ 6
3 3 3 π fy = w2 x − wzewyz ,
= x− y+ − √ fz = −wyewyz
4 4 4 8 3


(a) fx (−2, 3, 1, 0) = −12 17. As in example 4.5,
fy (−2, 3, 1, 0) = 4 ∂S
(36, 2, 6) = 0.1728
fz (−2, 3, 1, 0) = 12 ∂L
fw (−2, 3, 1, 0) = 2 ∂S
(36, 2, 6) = −0.7776
L(x, y, z, w) = −11 − 12(w + 2) ∂h
∂S
+ 4(x − 3) + 12(y − 1) + 2(z − 0) (36, 2, 6) = −0.7776
∂w
= −12w + 4x + 12y + 2z − 37 S(36, 2, 6) = 1.5552
(b) fw (0, 1, −1, 2) = 2 The maximum sag occurs if (L − 36) = 0.5,
fx (0, 1, −1, 2) = 0 (w − 2) = −0.2 and (h − 6) = −0.5. The linear
fy (0, 1, −1, 2) = 0 approximation predicts the change in sag will
fz (0, 1, −1, 2) = 0 be
L(w, x, y, z) = −1 + 2(w − 0) = 2w − 1 0.5(0.1728) + 0.2(0.7776) + 0.5(0.7776)
= 0.6307.
12. fx = −yz sin(xyz) − 2w3 x, The range of sags will be 1.5552 ± 0.6307.
fy = −zx sin(xyz),
fz = −xy sin(xyz), ∂S
18. (32, 2, 8) = 0.0512
fw = −3w2 x2 ∂L
∂S
(a) fx (2, −1, 4, 0) = 16 (32, 2, 8) = −0.1536
∂h
fy (2, −1, 4, 0) = 0
∂S
fz (2, −1, 4, 0) = 0 (32, 2, 8) = −0.2048
fw (2, −1, 4, 0) = −12 ∂w
L(w, x, y, z) = −7 − 12(w − 2) + 16(x + 1) S(32, 2, 8) = 0.4096
= −12w + 16x + 33 The maximum sag occurs if (L − 32) = 0.4,
(w − 2) = −0.3 and (h − 8) = −0.4. The linear
(b) fx (2, 1, 0, 1) = −16 approximation predicts the change in sag will
fy (2, 1, 0, 1) = 0 be
fz (2, 1, 0, 1) = 0 0.4(0.0512) + 0.3(0.2048) + 0.4(0.1536)
fw (2, 1, 0, 1) = −12 = 0.1434.
L(w, x, y, z) = −7 − 12(w − 2) − 16(x − 1) The range of sags will be 0.4096 ± 0.1434.
= −12w − 16x + 33
19. g(9.9, 930) ≈ 4 + 0.3(−0.1) − 0.004(30)
13. L(x, y) = x = 3.85
x y L(x, y) f (x, y)
3 −0.1 3 3.00167 20. g(10.2, 910) ≈ 4 + 0.3(0.2) − 0.004(10)
3.1 0 3.1 3.1 = 4.02
3.1 −0.1 3.1 3.10161 21. The linear approximation will be
g(s, t) ≈ 4 + 0.1(s − 10) − 0.001(t − 900)
14. L(x, y) = −y
g(10.2, 890) ≈ 4.03
x y L(x, y) f (x, y)
0.1 −3 3 3.00167 ∂g −0.04
22. ≈ = 0.8
0 −3.1 3.1 3.1 ∂G −0.05
0.1 −3.1 3.1 3.10161 g(10.15, 905, 3.98)
≈ 4 + 0.3(0.15) − 0.004(5) + 0.8(−0.02)
y = 4.009
15. L (x, y, z) = x + + 2πz + 1 − π
2
x y z L(x, y, z) f (x, y, z) 23. fx = 2y, fy = 2x + 2y
0 3 0.25 0.929204 0.931596 ∆z = f (x + ∆x, y + ∆y) − f (x, y)
0.1 π 0.25 1.1 1.100167 = 2(x + ∆x)(y + ∆y) + (y + ∆y)2
0.1 π 0.2 0.685841 0.726543 − (2xy + y 2 )
√ = (2y)∆x + (2x + 2y)∆y
π
16. L (x, y, z) = 2x + 2z +−1 + (2∆y)∆x + (∆y)∆y
4
x y z L(x, y, z) f (x, y, z) Here
0.7 2 0 0.775348 0.775397 ∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y,
0.7 1.9 0 0.775348 0.775397 where ε1 = 2∆y and ε2 = ∆y.
0.7 2 0.1 0.975347 0.978108 Therefore, f is diﬀerentiable, if fx and fy


2 3
are continuous on some open region contain- (∆x + 2∆y) (∆x + 2∆y)
ing (a, b) of the domain of f , as ε1 → 0 and + ex+2y +
2! 3!
ε2 → 0 for (∆x, ∆y) → (0, 0) +.... )
= ex+2y (∆x) + 2ex+2y (∆y)
24. fx = 2(x + y), fy = 2(x + y) 2 2
∆z = f (x + ∆x, y + ∆y) − f (x, y) (∆x) + 2 (∆x) (2∆y) + (2∆y)
+ex+2y
= [(x + ∆x) + (y + ∆y)]2 − (x + y)2 2!
= (2x + 2y)∆x + (2x + 2y)∆y 3 2 2 3
(∆x) + 3(...) (...) + 3 (..) (...) + (...)
+ (∆x + 2∆y)∆x + (∆y)∆y +
3!
+...]
Here
= ex+2y (∆x) + 2ex+2y (∆y)
∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y, (∆x) + (2∆y)
where ε1 = ∆x + 2∆y and ε2 = ∆y. + ex+2y
2!
Therefore, f is differentiable, if fx and fy 2
(∆x) + 3 (∆x) (2∆y)
are continuous on some open region contain- + + ... (∆x)
3!
ing (a, b) of the domain of f , as ε1 → 0 and
ε2 → 0 for (∆x, ∆y) → (0, 0). (∆x) + (2∆y)
+ ex+2y
2!
25. fx = 2x, fy = 2y 2
3 (∆x) (2∆y) + (2∆y)
∆z = f (x + ∆x, y + ∆y) − f (x, y) + + ... (2∆y)
3!
= (x + ∆x)2 + (y + ∆y)2 − (x2 + y 2 )
= (2x)∆x + (2y)∆y + (∆x)∆x + (∆y)∆y Here
Here ∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y,

∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y, where
where ε1 = ∆x and ε2 = ∆y. ε1 = ex+2y
Therefore, f is differentiable, if fx and fy (∆x) + (2∆y)
are continuous on some open region contain- 2!
2
ing (a, b) of the domain of f , as ε1 → 0 and (∆x) + 3 (∆x) (2∆y)
ε2 → 0 for (∆x, ∆y) → (0, 0) + + ...
3!
and
26. fx = 3x2 − 3y, fy = −3x
ε2 = ex+2y
∆z = f (x + ∆x, y + ∆y) − f (x, y) (∆x) + (2∆y)
= (x + ∆x)3 − 3(x + ∆x)(y + ∆y)
2!
− (x3 − 3xy) 2
= (3x2 − 3y)∆x − 3x∆y 3 (∆x) (2∆y.) + (2∆y)
+ + ...
+ [3x∆x + (∆x)2 ]∆x − 3∆x∆y 3!
Here Therefore, f is differentiable, if fx and fy
are continuous on some open region contain-
∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y,
2 ing (a, b) of the domain of f , as ε1 → 0 and
where ε1 = 3x∆x + (∆x) and ε2 = −3∆x. ε2 → 0 for (∆x, ∆y) → (0, 0)
Therefore, f is differentiable, if fx and fy
28. fx = 2x sin y and fy = x2 cos y
ing (a, b) of the domain of f , as ε1 → 0 and ∆z = f (x + ∆x, y + ∆y) − f (x, y)
ε2 → 0 for (∆x, ∆y) → (0, 0) 2
∆z = (x + ∆x) sin (y + ∆y) − x2 sin y
= x2 sin (y + ∆y) − x2 sin y
27. fx = ex+2y and fy = 2ex+2y
+ 2x (∆x) sin (y + ∆y)
∆z = f (x + ∆x, y + ∆y) − f (x, y) 2
+ (∆x) sin (y + ∆y)
= ex · e∆x · e2y · e2∆y − ex · e2y 2
= x cos y (sin ∆y − ∆y + ∆y)
= ex+2y e∆x+2∆y − 1
+ x2 sin y (cos ∆y − 1)
= ex+2y (1 + (∆x + 2∆y)
2 + (2x sin y) (cos ∆y − 1 + 1) ∆x
(∆x + 2∆y) + (2x cos y sin ∆y) ∆x
+ + ... − 1
2! 2
+ (∆x) sin (y + ∆y)
= ex+2y (∆x) + 2ex+2y (∆y) = x2 cos y ∆y + (2x sin y) ∆x


+ [2x sin y (cos ∆y − 1) 2 2
30. fx = − 2 and fy = − 2
+2x cos y sin ∆y (x + y) (x + y)
+∆x sin (y + ∆y)] (∆x) ∆z = f (x + ∆x, y + ∆y) − f (x, y)
sin ∆y 2 2
+ x2 cos y −1 = −
∆y (x + ∆x) + (y + ∆y) x + y
(cos ∆y − 1) −2∆x − 2∆y
+ sin y (∆y) = ,
(∆y) (x + y + ∆x + ∆y) (x + y)
Here Writing −2∆x − 2∆y = −2∆x (1) − 2∆y (1)
∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y, and replacing 1 by
Where (x + y + ∆x + ∆y)
ε1 = [2x sin y (cos ∆y − 1) + 2x cos y sin ∆y (x + y)
+∆x sin (y + ∆y)] (x + y + ∆x + ∆y)
− +1 ,
and (x + y)
sin ∆y we get
ε2 = x2 cos y −1
∆y 2 2
(cos ∆y − 1) ∆z = − 2 ∆x − 2 ∆y
+ sin y (x + y) (x + y)
(∆y) 2 (∆x + ∆y)
+ ∆x
Therefore, f is differentiable, if fx and fy (x + y + ∆x + ∆y) (x + y)
are continuous on some open region contain- 2 (∆x + ∆y)
+ ∆y
ing (a, b) of the domain of f , as (x + y + ∆x + ∆y) (x + y)
ε1 → 0 and ε2 → 0 for (∆x, ∆y) → (0, 0) Here
2x x2 ∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y,
29. fx = and fy = − 2
y y Where
∆z = f (x + ∆x, y + ∆y) − f (x, y) 2 (∆x + ∆y)
2
ε1 = and
(x + ∆x) x2 (x + y + ∆x + ∆y) (x + y)
= −
(y + ∆y) y 2 (∆x + ∆y)
2 ε2 =
x2 2x∆x (∆x) x2 (x + y + ∆x + ∆y) (x + y)
= + + −
y + ∆y y + ∆y y + ∆y y Therefore, f is differentiable, if fx and fy
2
x2 ∆y 2x∆x (∆x) are continuous on some open region contain-
=− + + ing (a, b) of the domain of f , as ε1 → 0 and
y (y + ∆y) y + ∆y y + ∆y
x2 y ε2 → 0 for (∆x, ∆y) → (0, 0)
=− 2 + 1 − 1 ∆y
y y + ∆y 31. fx = yex + cos x, fy = ex
2x y
+ − 1 + 1 ∆x dz = (yex + cos x)dx + ey dy
y y + ∆y
(∆x)
2 1 1
+ 32. fx = √ , fy = √
y + ∆y 2 x+y 2 x+y
2 1 1
2x x2 (∆x) dz = √ dx + √ dy
= ∆x − 2 ∆y + 2 x+y 2 x+y
y y y + ∆y
2
x2 (∆y) 2x (∆x) (∆y) 1 1
+ 2 − 33. fx = −
y (y + ∆y) y (y + ∆y) x 1 + (x − y − z)2
Here 1 1
fy = +
∆z = fx (a, b) ∆x+fy (a, b) ∆y +ε1 ∆x+ε2 ∆y, y 1 + (x − y − z)2
Where 1 1
2
fz = + 2
∆x x ∆y − 2xy∆x z 1 + (x − y − z)
ε1 = and ε2 = .
(y + ∆y) y 2 (y + ∆y) 1 1
Therefore, f is differentiable, if fx and fy dw = − dx
x 1 + (x − y − z)2
ing (a, b) of the domain of f , as ε1 → 0 and 1 1
+ + dy
ε2 → 0 for (∆x, ∆y) → (0, 0) y 1 + (x − y − z)2


1 1 The function is differentiable if
+ + 2 dz ∆z = fx ∆x + fy ∆y + ε1 ∆x + ε2 ∆y
z 1 + (x − y − z)
where ε1 and ε2 both go to zero as
x2 y (∆x, ∆y) → (0, 0). If the function is differen-
2x2 y z tiable we must be able to write
34. fx = 1+ e z −
z 2 (x + y) ln (x + y) ∆x∆y 2
= ε1 ∆x + ε2 ∆y.
2
x y ∆x2 + ∆y 2
x3 z To see that this is impossible, assume that we
fy = e z − have such an expression, solve for ε1 , and ex-
z 2 (x + y) ln (x + y)
amine the limit:
x2 y ∆y 2 ∆y
x3 y lim ε1 = − ε2
fz = − 2 e z − ln (x + y) (∆x,∆y)→(0,0) ∆x 2 + ∆y 2 ∆x
z 2 1
x y Along the line ∆y = ∆x, this gives 0 = + 0.
z + 2x2 y 2
dw = e z
 Therefore the function f is not differentiable.
z
37. (a) f (0, 0) = 6. We can get from the z = 6
z level curve to the z = 8 level curve by
− dx moving 1 in the y direction, or by moving
2 (x + y) ln (x + y)
  0.5 in the x direction.
3
x2 y ∂z 2 ∂z 2
x z ≈ =4 ≈ =2
+ e z −  dy ∂x 0.5 ∂y 1

z 2 (x + y) ln (x + y) L(x, y) = 6 + 4x + 2y

x2 y
 (b) f (1, 0) ≈ 8
3
 x y z We can get from z = 8 level curve to the
+ − e − ln (x + y) dz

z z = 10 level curve by moving 1.75 in the
x-direction, 2.25 in y-direction
f (0 + h, 0) − f (0, 0) ∂z 2 8 ∂z 2 8
35. fx (0, 0) = lim =0 ≈ = , ≈ =
h→0 h ∂x 1.75 7 ∂y 2.25 9
f (0, 0 + h) − f (0, 0) 8 8
fy (0, 0) = lim =0 L (x, y) ≈ 8 + (x − 1) + y
h→0 h 7 9
Using Definition 4.1, at the origin we have (c) f (0, 2) ≈ 8.9 using
2(0 + ∆x)(0 + ∆y) f (0, 2) − f (0, 2 − h1 ) f (0, 2 + h2 ) − f (0, 2)
∆z = ≈
(0 + ∆x)2 + (0 + ∆y)2 h1 h2
2∆x∆y We can get the z = 10 level curve by mov-
= .
∆x2 + ∆y 2 ing 1 in the x-direction, 1.25 in y-direction
The function is differentiable if ∂z ∂z 1.1
≈ 1.1, ≈ = 0.88
∆z = fx ∆x + fy ∆y + ε1 ∆x + ε2 ∆y ∂x ∂y 1.25
where ε1 and ε2 both go to zero as L (x, y) ≈ 8.9 + 1.1x + 0.88 (y − 2)
(∆x, ∆y) → (0, 0). If the function is differen-
38. (a) f (0, 0) = 0
tiable we must be able to write ∂z 2
2∆x∆y (0, 0) ≈ = −2
= ε1 ∆x + ε2 ∆y, ∂x −1
∆x2 + ∆y 2
but the function on the left does not have a ∂z −2
limit as (∆x, ∆y) → (0, 0). (The limit is dif- (0, 0) ≈ =1
∂y −2
ferent along the lines ∆y = ∆x and along L(x, y) = −2x + y
∆y = −∆x.) (b) f (1, 0) ≈ −1.5 using
f (0 + h, 0) − f (0, 0) f (1 + h1 , 0) − f (1, 0)
36. fx (0, 0) = lim =0 h1
h→0 h
f (0, 0 + h) − f (0, 0) f (1, 0) − f (1 − h2 , 0)
fy (0, 0) = lim =0 ≈
h→0 h h2
Using Definition 4.1, at the origin we have We can get the z = 0 level curve by mov-
(0 + ∆x)(0 + ∆y)2 ing -1 in the x-direction, 0.5 in y-direction
∆z = ∂z −1.5 ∂z −1.5
(0 + ∆x)2 + (0 + ∆y)2 ≈ = 1.5, ≈ = −3
∆x∆y 2 ∂x −1 ∂y 0.5
= . L (x, y) ≈ −1.5 + 1.5 (x − 1) − 3y
∆x2 + ∆y 2


(c) f (0, 2) ≈ 3 using
f (0, 2) − f (0, 2 − h1 ) 1

h1
f (0, 2 + h2 ) − f (0, 2) y 0.5
≈
h2
We can get the z = 4 level curve by -1 -0.5
0
0 0.5 1
moving -1 in the x-direction, 0.75 in y- x

direction -0.5

∂z −1 ∂z −1 4
≈ = 1, = =−
∂x −1 ∂y (3/4 ) 3 -1
4
L (x, y) ≈ 3 + x − (y − 2)
3
Moving from the z = 1 contour to the z = 2
contour moves 0.7 in the x direction or −0.7 in
39. (See exercise 27 from Section 12.3)
∂w ∂w the y direction. This makes our approximation
≈ 1.4, ≈ −2.4 ∂z ∂z
∂t ∂x of ≈ 1.43 and ≈ −1.43.
L(t, s) = −9 + 1.4(t − 10) − 2.4(s − 10) ∂x ∂y
∂z ∂z
L(12, 13) = −13.4 The exact values are = 1 and = −1.
∂x ∂y
Zoomed in so the level curves are equally
∂w ∂w
40. ≈ 1.35, ≈ −1.6 spaced , we get
∂t ∂x (with −0.1 ≤ x ≤ 0.1 and −0.1 ≤ y ≤ 0.1):
L(t, s) = −18 + 1.35(t − 10) − 1.6(s − 15)
L(12, 13) = −12.1 0.1
The diﬀerence is due to the fact that the change
in wind chill is not as rapid at (10, 15) as it is
y 0.05
at (10, 10).

0
41. Use level curves for z-values between 0.9 and -0.1 -0.05 0 0.05 0.1
x
1.1 with a graphing window of
−0.1 ≤ x ≤ 0.1 and −0.1 ≤ y ≤ 0.1. -0.05

To move from the z = 1.00 level curve to the
z = 1.05 level curve you move 0.025 to the -0.1

∂f 0.05
right, so ≈ = 2.
∂x 0.025 Moving from the z = 1 contour to the z = 1.05
To move from the z = 1.00 level curve to the
contour moves 0.05 in the x direction or −0.05
z = 1.05 level curve you move 0.05 down, so
∂f 0.05 in the y direction. This makes our approxima-
≈ ≈ −1. ∂z ∂z
∂y −0.05 tion of ≈ 1.0 and ≈ −1.0. In the ﬁrst
We also have f (0, 0) = 1. Therefore ∂x ∂y
estimate, the function values were changing
L(x, y) ≈ 1 + 2(x − 0) − 1(y − 0)
much more rapidly away from (0, 0) than they
0.1
were at (0, 0). In the second estimate, the spac-
ing between contours was even, so the function
y 0.05
values were changing roughly the same amount
throughout the window.

0
-0.1 -0.05 0 0.05 0.1
x

-0.05

∂f ∂f
43. 0, 1, (a, b) × 1, 0, (a, b)
-0.1 ∂y ∂x
i j k
∂f
= 0 1 ∂y (a, b)
42. With window −1 ≤ x ≤ 1 and −1 ≤ y ≤ 1, ∂f
the contour plot is: 1 0 (a, b)
∂x

12.5. THE CHAIN RULE 689

∂f ∂f ru = u, 1, 0
1 (a, b) 0 (a, b)
=i ∂y −j ∂y ru (2, 1) = 2, 1, 0
∂f ∂f rv = 0, 0, 1
0 (a, b) 1 (a, b)
∂x ∂x rv (2, 1) = 0, 0, 1
0 1
+k ru (2, 1) × rv (2, 1) = 1, −2, 0
1 0
Therefore the tangent plane is
∂f ∂f
= (a, b)i + (a, b)j − k (x − 2) − 2(y − 2) + 0(z − 1) = 0
∂x ∂y
∂f ∂f 49. As f (x) is a differentiable at x = a,
= (a, b), (a, b), −1
∂x ∂y
f (a + ∆x) − f (a)
f (a) = lim . Therefore
44. The main point here is that if we fix v, then the ∆x→0 ∆x
equation r(u, v) defines a curve on the para- f (a + ∆x) − f (a)
metric surface (if v is fixed then there is only f (a) + ε = , where ε → 0
∆x
one parameter—u). The tangent vector to this or f (a + ∆x) − f (a) = f (a) ∆x + ε ∆x,
curve is ru and this tangent vector lies in the ∆y = f (a) ∆x + ε ∆x, where
tangent plane. Similarly for rv .
Therefore, ru × rv must be normal to the vec- lim ε ∆x = 0, since ε → 0 and ∆x → 0
∆x→0
tors ru and rv and therefore ru × rv is normal
Now, if y = f (x, z) differentiable at (a, b) then
to the tangent plane.
we can write:
45. r = 2u, v, 4uv and ∆y = fx (a, b) ∆x + fz (a, b) ∆z + ε1 ∆x + ε2 ∆z,
r(1, 2) = 2, 2, 8 and
where ε1 → 0 and ε2 → 0 for
ru = 2, 0, 4v
ru (1, 2) = 2, 0, 8 (∆x, ∆z) → (0, 0).
rv = 0, 1, 4u Consider z to be a constant, so ∆z = 0 and
rv (1, 2) = 0, 1, 4 ∂f
ru (1, 2) × rv (1, 2) = −8, −8, 2 fx = = f (a) and ε = ε1 gives
∂x
∆y = f (a) ∆x + ε ∆x, where lim ε ∆x = 0,
−8(x − 2) − 8(y − 2) + 2(z − 8) = 0 ∆x→0
since ε → 0 and ∆x → 0
46. r = 2u2 , uv, 4uv 2 and
r(−1, 1) = 2, −1, −4 and 50. Consider the Taylor series about x = a, which
ru = 4u, v, 4v 2 converges to f (x).
ru (−1, 1) = −4, 1, 4 Therefore
rv = 0, u, 8uv ∞
f (k) (a) k
rv (−1, 1) = 0, −1, −8 f (x) = (x − a) ... (1),
ru (−1, 1) × rv (−1, 1) = −4, −32, 4 k!
k=0
Therefore the tangent plane is ∞
f (k) (a) k−1
−4(x − 2) − 32(y + 1) + 4(z + 4) = 0 ⇒ f (x) = k(x − a) ... (2)
k!
k=0
47. r = cos u, sin u, v for 0 ≤ u ≤ 2π and
Therefore from (1) and (2)f (a),f (a) = 0
0 ≤ v ≤ 2.
The point (1,0,1) corresponds to (u, v) = (0, 1). ∆y − dy
Also, we have ε = ,⇒ ε ∆x = ∆y − dy
r(0, 1) = 1, 0, 1 and ∆x
ru = − sin u, cos u, 0 ε ∆x = f (a + ∆x) − f (a) − f (a) ∆x ... (3)
ru (0, 1) = 0, 1, 0 Therefore, from (1) , (2) and (3)
rv = 0, 0, 1
∞
rv (0, 1) = 0, 0, 1 f (k) (a) k
ε ∆x = f (a + ∆x) = (∆x)
ru (0, 1) × rv (0, 1) = 1, 0, 0 k!
k=0
(x − 1) = 0
12.5 The Chain Rule
u2
48. r = , u, v and
2 1. g(t) = (t2 − 1)2 esin t
r(2, 1) = 2, 2, 1 and g (t) = 2(t2 − 1)(2t)esin t + (t2 − 1)2 cos tesin t


2 2 2
2. g(u, v) = e(3u sin v)(4v u) = e12u v sin v 2 v sin u
= 3eu (v 2 + 1) √
∂g 2 2 v2 + 1
= 24uv 2 sin v e12u v sin v u2
√
∂u = 3e v sin u v 2 + 1
∂g 2 2
= 12u2 (v 2 cos v + 2v sin v)e12u v sin v ∂f dx ∂f dy ∂f dz
∂v 7. g (t) = + +
∂x dt ∂y dt ∂z dt
∂f dx ∂f dy
3. g (t) = + ∂g ∂f ∂x ∂f ∂y ∂f ∂z
∂x dt ∂y dt 8. = + +
∂f ∂f ∂u ∂x ∂u ∂y ∂u ∂z ∂u
= 2xy, = x2 − cos y ∂g ∂f ∂x ∂f ∂y ∂f ∂z
∂x ∂y = + +
dx t dy ∂v ∂x ∂v ∂y ∂v ∂z ∂v
=√ , = et
dt t2 + 1 dt ∂g ∂f ∂x ∂f ∂y
t 9. = +
g (t) = 2xy √ + (x2 − cos y)et ∂u ∂x ∂u ∂y ∂u
t 2+1
t ∂g ∂f ∂x ∂f ∂y
= 2 t2 + 1et √ +[(t2 +1)−cos et ]et = +
2+1 ∂v ∂x ∂v ∂y ∂v
t ∂g ∂f ∂x ∂f ∂y
= (2t + t2 + 1 − cos et )et = +
∂w ∂x ∂w ∂y ∂w
∂f dx ∂f dy ∂g ∂f ∂x ∂f ∂y ∂f ∂z
4. g (t) = + 10. = + +
∂x dt ∂y dt ∂u ∂x ∂u ∂y ∂u ∂z ∂u
∂f x ∂f y ∂g ∂f ∂x ∂f ∂y ∂f ∂z
= , = = + +
∂x x 2 + y2 ∂y x2 + y2 ∂v ∂x ∂v ∂y ∂v ∂z ∂v
dx dy ∂g ∂f ∂x ∂f ∂y ∂f ∂z
= cos t = 2t = + +
dt dt ∂w ∂x ∂w ∂y ∂w ∂z ∂w
x y 11. Use the fact involved in the exercise 8, here
g (t) = cos t + 2t
x2 + y2 x2 + y2 x (u, v) = u + v
sin t cos t y (u, v) = u − v
=
sin t2 + (t2 + 2)2 z (u, v) = u2 + v 2 .
(t2 + 2)2t Therefore,
+ ∂x ∂x
sin t2 + (t2 + 2)2 = 1; =1
∂u ∂v
sin t cos t + 2t(t2 + 2)
= ∂y ∂y
sin t2 + (t2 + 2)2 = 1; = −1
∂u ∂v
∂z ∂z
∂g ∂f ∂x ∂f ∂y = 2u; = 2v
5. = + ∂u ∂v
∂u ∂x ∂u ∂y ∂u Thus
= 8xy 3 (3u2 − v cos u) + 12x2 y 2 (8u) ∂g ∂g ∂g ∂g
= 8(u3 − v sin u)(4u2 )3 (3u2 − v cos u) = + + 2u and
∂u ∂x ∂y ∂z
+ 12(u3 − v sin u)2 (4u2 )2 (8u)
∂g ∂f ∂x ∂f ∂y ∂g ∂g ∂g ∂g
= + = − + 2v
∂v ∂x ∂v ∂y ∂v ∂v ∂x ∂y ∂z
= 8xy 3 (− sin u) + 12x2 y 2 (0) 12. Use the fact involved in the exercise 8, here
= 8(u3 − v sin u)(4u2 )3 (− sin u)
x (u, v) = u2 v
∂g ∂f ∂x ∂f ∂y y (u, v) = v
6. = + z (u, v) = v cos u
∂u ∂x ∂u ∂y ∂u √
2
= (y 3 − 8x)(2ueu ) + 3xy 2 ( v 2 + 1 cos u) Therefore,
2 2
= [(v 2 + 1)3/2 − 8eu ](2ueu ) ∂x ∂x
2 √ = 2uv, = u2
+ 3eu (v 2 + 1)( v 2 + 1 cos u) ∂u ∂v
2
= eu −16u + (v 2 + 1)3/2 (2u + 3 cos u) ∂y ∂y
= 0, =1
∂g ∂f ∂x ∂f ∂y ∂u ∂v
= + ∂z ∂z
∂v ∂x ∂v ∂y ∂v = −v sin u, = cos u
v sin u ∂u ∂v
= (y − 8x)(0) + 3xy 2 √
3
Thus
v2 + 1


∂g ∂g ∂g ∂P
= 2uv − v sin u and 16. (4, 3) ≈ 4.0296
∂u ∂x ∂z ∂k
∂g ∂g ∂g ∂g ∂P
= u2 + + cos u (4, 3) ≈ 16.1185
∂v ∂x ∂y ∂z ∂l
k (t) = −0.2, l (t) = 0.08
13. Use the fact involved in the exercise 10, here ∂P ∂P
g (t) = k (t) + l (t)
x (u, v, w) = uv, ∂k ∂l
u ≈ (4.0296)(−0.2) + (16.1185)(0.08)
y (u, v, w) =
v
z (u, v, w) = w2 = 0.4835
Therefore,
∂P 16 −2/3 2/3 ∂P 32 1/3 −1/3
∂x ∂x ∂x 17. = k l , = k l
=v =u =0 ∂k 3 ∂l 3
∂u ∂v ∂w ∂P ∂P
∂y 1 ∂y u ∂y (4, 3) ≈ 4.4026, (4, 3) ≈ 11.7402
= =− 2 =0 ∂k ∂l
∂u v ∂v v ∂w k (t) = −0.2, l (t) = 0.08
∂z ∂z ∂z
=0 =0 = 2w ∂P ∂P
∂u ∂v ∂w g (t) = k (t) + l (t)
Thus, ∂k ∂l
= (4.4026)(−0.2) + (11.7402)(0.08)
∂g ∂g 1 ∂g
=v + , = 0.0587
∂u ∂x v ∂y
∂g ∂g u ∂g
=u − 2 and ∂P 16 −2/3 2/3 ∂P 32 1/3 −1/3
∂v ∂x v ∂y 18. = k l , = k l
∂g ∂g ∂k 3 ∂l 3
= 2w ∂P ∂P
∂w ∂z (5, 2) ≈ 2.8953, (5, 2) ≈ 14.4768
∂k ∂l
14. Use the fact involved in the exercise 10, here k (t) = −0.1, l (t) = 0.04
x (u, v, w) = u2 + w2 ∂P ∂P
g (t) = k (t) + l (t)
y (u, v, w) = u + v + w ∂k ∂l
z (u, v, w) = u cos v. = (2.8953)(−0.1) + (14.4768)(0.04)
Therefore, = 0.2895
∂x ∂x ∂x
= 2u =0 = 2w 19. I(t) = q(t)p(t)
∂u ∂v ∂w dq dp
∂y ∂y ∂y = 0.05q(t), = 0.03p(t)
=1 =1 =1 dt dt
∂u ∂v ∂w
∂z ∂z ∂z dI ∂I dq ∂I dp
= cos v = −u sin v =0 = +
∂u ∂v ∂w dt ∂q dt ∂p dt
Thus dq dp
∂g ∂g ∂g ∂g = p(t) + q(t)
= 2u + + cos v , dt dt
∂u ∂x ∂y ∂z
= p(t)[0.05q(t)] + q(t)[0.03p(t)]
∂g ∂g ∂g
= − u sin v and = 0.08p(t)q(t)
∂v ∂y ∂z
= 0.08I(t)
∂g ∂g ∂g
= 2w +
∂w ∂x ∂y Income increases at a rate of 8% as claimed.
∂P 20. I(t) = q(t)p(t)
15. (4, 6) ≈ 3.6889 dq dp
∂k = −0.03q(t), = 0.05p(t)
∂P dt dt
(4, 6) ≈ 16.6002
∂l dI ∂I dq ∂I dp
k (t) = 0.1, l (t) = −0.06 = +
dt ∂q dt ∂p dt
∂P ∂P dq dp
g (t) = k (t) + l (t) = p(t) + q(t)
∂k ∂l dt dt
≈ (3.6889)(0.1) + (16.6002)(−0.06) = p(t)[−0.03q(t)] + q(t)[0.05p(t)]
= −0.6271 = 0.02p(t)q(t) = 0.02I(t)


Income increases at a rate of 2% in this situa- 26. F (x, y, z) = ln(x2 + y 2 ) − z − tan−1 (x + z)
tion. 2x 1
2 3
Fx = − 2
21. F (x, y, z) = 3x z + 2z − 3yz x2 + y 2 1 + (x + z)
2
Fx = 6xz 2x 1 + (x + z) − x2 + y 2
= ,
Fy = −3z (x2 + y 2 ) 1 + (x + z)
2

Fz = 3x2 + 6z 2 − 3y 2y
Fy = 2
∂z Fx −6xz x + y2
=− = 2
∂x Fz 3x + 6z 2 − 3y 1 2 + (x + z)
2
∂z Fy 3z Fz = −1 − 2 =− 2
=− = 2 1 + (x + z) 1 + (x + z)
∂y Fz 3x + 6z 2 − 3y
2
∂z Fx 2x 1 + (x + z) − x2 + y 2
22. F (x, y, z) = xyz − 4y 2 z 2 + cos xy =− =
∂x Fz (x2 + y 2 ) 2 + (x + z)
2
Fx = yz − y sin xy
Fy = zx − 8yz 2 − x sin xy 2y 1 + (x + z)
2
∂z Fy
Fz = xy − 8y 2 z =− ==
∂y Fz (x2 + y 2 ) 2 + (x + z)
2
∂z Fx −yz + y sin xy
=− =
∂x Fz xy − 8y 2 z
27. The chain rule gives
∂z Fy −zx + 8yz 2 + x sin xy ∂x ∂y
=− = fθ = fx + fy
∂y Fz xy − 8y 2 z ∂θ ∂θ
∂x ∂y
23. F (x, y, z) = 3exyz − 4xz 2 + x cos y = −r sin θ and = r cos θ
∂θ ∂θ
Fx = 3yzexyz − 4z 2 + cos y So, fθ = −fx r sin θ + fy r cos θ.
Fy = 3xzexyz − x sin y
28. From Exercise 27,
Fz = 3xyexyz − 8xz fθ = −fx r sin θ + fy r cos θ
∂z Fx −3yzexyz + 4z 2 − cos y
=− = fθθ =(−fxx r sin θ + fxy r cos θ)(−r sin θ)
∂x Fz 3xyexyz − 8xz
∂z Fy −3xzexyz + x sin y + fx (−r cos θ)
=− =
∂y Fz 3xyexyz − 8xz + (−fyx r sin θ + fyy r cos θ)(r cos θ)
+ fy (−r sin θ)
24. F (x, y, z) = 3yz 2 − e4x cos 4z − 3y 2
=fxx r2 sin2 θ − 2fxy r2 sin θ cos θ
Fx = −4e4x cos 4z
+ fyy r2 cos2 θ − fx r cos θ − fy r sin θ
Fy = 3z 2 − 6y
Fz = 6yz + 4e4x sin 4z
29. From exercises 27 and 28, and example 5.4, we
∂z Fx 4e4x cos 4z have:
=− =
∂x Fz 6yz + 4e4x sin 4z fr = fx cos θ + fy sin θ
∂z Fy −3z 2 + 6y frr = fxx cos2 θ + 2fxy cos θ sin θ + fyy sin2 θ
=− = fθ = −fx r sin θ + fy r cos θ
∂y Fz 6yz + 4e4x sin 4z
fθθ = fxx r2 sin2 θ − 2fxy r2 sin θ cos θ
25. F (x, y, z) = xyz − cos(x + y + z) + fyy r2 cos2 θ − fx r cos θ
1 1
Fx = yz + sin(x + y + z), − fy r sin θfrr + fr + 2 fθθ
r r
Fy = xz + sin(x + y + z) and = (fxx cos2 θ + 2fxy cos θ sin θ + fyy sin2 θ)
Fz = xy + sin(x + y + z) 1
+ (fx cos θ + fy sin θ)
∂z Fx yz + sin(x + y + z) r
=− =− , 1
∂x Fz xy + sin(x + y + z) + 2 (fxx r2 sin2 θ − 2fxy r2 sin θ cos θ
r
∂z Fy xz + sin(x + y + z) + fyy r2 cos2 θ − fx r cos θ − ry sin θ)
=− =− = fxx + fyy
∂y Fz xy + sin(x + y + z)


30. The mistake is treating θ as a constant and not p2 v0
V (t) = .
applying the chain rule. p + v0 ct
Taking the derivative of x = r cos θ, with re- But V (0) = 1
spect to x and using the chain rule, gives, v
⇒ pv0 = 1 ⇒ V =
∂r ∂θ v0
1 = cos θ − r sin θ v0
∂x ∂x ⇒ v (t) = 2
∂r 1 sin θ ∂θ (1 + v0 ct)
= +r 1
∂x cos θ cos θ ∂x ⇒ V (t) = 2
(1 + v0 ct)
∂θ
Of course, if θ is constant then = 0 and ft ft
∂x Here the units of v = , those of v0 =
∂r 1 sec sec
= , but θ is not constant. ⇒ Units of V = 1. It may also be observed
∂x cos θ
sec
x that units of c = 2.
31. Make the change of variables X = and (f t)
L
α2 ∂X ∂T ⇒ V is dimensionless.
T = t. Then, ut = uX + uT T dt 1
L2 ∂t ∂t Now, T = qt ⇒ t = ⇒ = .
α2 ∂X ∂T q dT q
= 2 uT ux = uX + uT
L ∂x ∂x dV
1 1 ∂X ∂T Also,⇒ = −V is the simpliﬁed initial value
= uX uxx = uXX + uXT dT
L L ∂x ∂x problem.
1
= 2 uXX dV dt V (t)
L ⇒ · = −V or q = = cv0 v (t).
The heat equation then becomes dt dT −V (t)
1 α2 1
α2 2 uXX = 2 uT , or simply uXX = uT . Therefore units of q = , thus the units of
L L sec
ft 1
The dimensions of X are = 1, and the di- T = · sec = 1 ⇒ T is dimensionless.
ft sec
2
ft /sec
mensions of T are sec = 1. 34. Let the variables V and T be such that
ft2
Both X and T are dimensionless. V = pv and T = qt.
Therefore,
32. Making this change of variables, dV dv
=p ⇒ V (t) = pv (t)
∂X ∂T 1 dt dt
ux = uX + uT = uX dv
∂x ∂x L ⇒ −g + cv 2 = p
1 ∂X ∂T 1 dt
uxx = uXX + uXT = 2 uXX dv dv dv
L ∂x ∂x L using = −g we get, − + cv 2 = p
dt dt dt
a2 dv
Similarly, utt = 2 uT T . ⇒ cv 2 = (p + 1)
L dt
Putting these into the wave equation: dv cv 2 c dv
1 a2 ⇒ = dt = 2
a2 uXX = 2 uT T or uXX = uT T dt (p + 1) (p + 1) v
L2 L t v
c dv
33. Let the variables V and T be such that V = pv ⇒ dt =
(p + 1) v2
dV dv 0 v0
and T = qt. Therefore =p
dt dt ct 1 1
⇒ = −
⇒ V (t) = pv (t) (p + 1) v0 v
1 (p + 1) − v0 ct
dv dv ⇒ =
⇒ −cv 2 = p or − cdt = p 2 v (p + 1) v0
dt v (p + 1) v0
t v ⇒v=
dv (p + 1 − v0 ct)
⇒ −c dt = p p (p + 1) v0
v2 ⇒V =
0 v0 (p + 1 − v0 ct)
pv0 (p + 1) v0
⇒ v (t) = or ⇒ v (t) = and
(p + v0 ct) (p + 1 − v0 ct)


p (p + 1) v0 ∂g ∂f ∂x ∂f ∂y
V (t) = 38. = +
(p + 1 − v0 ct) ∂u ∂x ∂u ∂y ∂u
v ∂2g ∂f ∂ 2 x ∂ 2 f ∂x ∂ 2 f ∂y ∂x
But V (0) = 1 ⇒ pv0 = 1 ⇒ V = = + +
v0 ∂u∂v ∂x ∂u∂v ∂x 2 ∂v ∂x∂y ∂v ∂u
ft
Here the units of v = , and those of v0 = ∂f ∂ 2 y ∂ 2 f ∂x ∂ 2 f ∂y ∂y
sec + + + 2
ft ∂y ∂u∂v ∂y∂x ∂v ∂y ∂v ∂u
sec 39. For g (u, v) = f (x , y, z),
⇒ Units of V = 1 ⇒ V is dimensionless.
T dt 1 where x = u + v; y = u − v and z = u − v
Now, T = qt ⇒ t = ⇒ = .
q dT q ∂g ∂f ∂x ∂f ∂y ∂f ∂z
dV 2 = · + · + ·
= −a + V is the simpliﬁed initial value ∂u ∂x ∂u ∂y ∂u ∂z ∂u
dT
problem. ∂2g ∂f ∂ 2 x
dV dt dV =
· = −a + V 2 ⇒ = q −a + V 2 . ∂u∂v ∂x ∂u∂v
dt dT dt 2
Now as V is dimensionless, ∂ f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂x
+ + +
−a + V 2 has to be dimensionless, ∂x 2 ∂v ∂x∂y ∂v ∂x∂z ∂v ∂u
dV 1 ∂f ∂ 2 y
also the unit of = , +
dt sec ∂y ∂u∂v
dV
1 ∂ 2 f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂y
Therefore the units of q = dt = . + + 2 +
(−a + V 2) sec ∂y∂x ∂v ∂y ∂v ∂y∂z ∂v ∂u
1 ∂f ∂ 2 z
Thus the units of T = · sec = 1 +
sec ∂z ∂u∂v
⇒ T is dimensionless. ∂ 2 f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂z
+ + + 2
∂z∂x ∂v ∂z∂y ∂v ∂z ∂v ∂u
35. g (t) = fx x (t) + fy y (t)
g (t) = fx x (t) + (fxx x (t) + fxy y (t))x (t) x (u, v) = u + v
+ fy y (t) + (fyx x (t) + fyy y (t))y (t) ∂x ∂x ∂2x
⇒ = 1; = 1; =0
= fxx (x (t))2 + 2fxy x (t)y (t) ∂u ∂v ∂u∂v
+ fyy (y (t))2 + fx x (t) + fy y (t)
y (u, v) = u − v
∂y ∂y ∂2y
36. g (t) = fx x (t) + fy y (t) + fz z (t) ⇒ =1; = −1 ; =0
∂u ∂v ∂u∂v
g (t) = fx x (t)
+ x (t) [fxx x (t) + fxy y (t) + fxz z (t)] z (u, v) = u2 + v 2
+ fy y (t) ∂z ∂z
+ y (t) [fyx x (t) + fyy y (t) + fyz z (t)] ⇒ = 2u ; = 2v ;
∂u ∂v
+ fz z (t)
∂2z
+ z (t) [fzx x (t) + fzy y (t) + fzz z (t)] =0
∂u∂v
= fx x (t) + fy y (t) + fz z (t)
+ (x (t))2 fxx + (y (t))2 fyy + (z (t))2 fzz ∂2g ∂2f ∂2f ∂2f
= − + 2v
+ 2x (t)y (t)fxy ∂u∂v ∂x2 ∂x∂y ∂x∂z
+ 2x (t)z (t)fxz ∂2f ∂2f ∂2f
+ − 2 + 2v
+ 2y (t)z (t)fyz ∂y∂x ∂y ∂y∂z
∂2f ∂2f ∂2f
+ − + 2v 2 2u
∂g ∂f ∂x ∂f ∂y ∂z∂x ∂z∂y ∂z
37. = +
∂u ∂x ∂u ∂y ∂u ∂2f ∂2f ∂2f ∂2f ∂2f
2
∂ g 2
∂f ∂ x ∂ 2 f ∂x ∂ 2 f ∂y ∂x = − 2 + 4uv 2 − + 2v
= + + ∂x2 ∂y ∂z ∂x∂y ∂x∂z
∂u2 ∂x ∂u2 ∂x2 ∂u ∂x∂y ∂u ∂u ∂2f ∂2f ∂2f ∂2f
∂f ∂ 2 y ∂ 2 f ∂x ∂ 2 f ∂y ∂y + + 2v + 2u − 2u
+ + + ∂y∂x ∂y∂z ∂z∂x ∂z∂y
∂y ∂u2 ∂y∂x ∂u ∂y 2 ∂u ∂u
2 40. For g (u, v) = f (x , y, z),
∂ 2 f ∂x ∂ 2 f ∂x ∂y
= +2
∂x2 ∂u ∂x∂y ∂u ∂u where x = u2 v; y = v and z = v cos u
2 2
∂ f ∂y ∂f ∂ 2 x ∂f ∂ 2 y ∂g ∂f ∂x ∂f ∂y ∂f ∂z
+ 2 + + = + +
∂y ∂u ∂x ∂u2 ∂y ∂u2 ∂v ∂x ∂v ∂y ∂v ∂z ∂v


∂2g ∂f ∂ 2 x w
= h = v w w ln v + v . Therefore
∂v 2 ∂x ∂v 2 v
2
∂ f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂x w w vw
+ + + g = uv v w w ln v + v ln u + u
∂x2 ∂v ∂x∂y ∂v ∂x∂z ∂v ∂v v u
t2 +2)(
3−t3 )
∂f ∂ 2 y Here g (t) = (sin t)( ;
+
∂y ∂v 2
where u (t) = (sin t); v (t) = t2 + 2 and
∂ 2 f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂y
+ + 2 + w (t) = 3 − t3 . Therefore
∂y∂x ∂v ∂y ∂v ∂y∂z ∂v ∂v
∂f ∂ 2 z u (t) = sin t
+
∂z ∂v 2 ⇒ u (t) = cos t;
∂ 2 f ∂x ∂ 2 f ∂y ∂ 2 f ∂z ∂z v (t) = t2 + 2 ⇒ v (t) = 2t and
+ + + 2
∂z∂x ∂v ∂z∂y ∂v ∂z ∂v ∂v
w (t) = 3 − t3 ⇒ w (t) = −3t2 thus,
2 ∂x ∂2x
x (u, v) = u v ⇒ = u2 ; =0 t2 +2)(
3−t3 )
∂v ∂v 2 g (t) = (sin t)(
∂y ∂2y (3−t3 )
y (u, v) = v ⇒ =1; =0 · t2 + 2 (ln sin t)
∂v ∂v 2
∂z ∂2z 2t 3 − t3
z (u, v) = v cos u ⇒ = cos u ; =0 · −3t2 ln t2 + 2 +
∂v ∂v 2 (t2 + 2)
2 2 2 2
∂ g ∂ f 2 ∂ f ∂ f
2
= 2
u + + cos u u2 + t2 + 2
(3−t3 )
cot t
∂v ∂x ∂x∂y ∂x∂z
∂2f 2 ∂2f ∂2f
+ u + 2 + cos u
∂y∂x ∂y ∂y∂z 43. Since g(h) = f (x + hu1 , y + hu2 ), it is clear
2
∂ f 2 ∂ f 2
∂ f 2 that g(0) = f (x, y).
+ u + + 2 cos u cos u ∂(x + hu1 ) ∂(y + hu2 )
∂z∂x ∂z∂y ∂z g (h) = fx + fy
2
∂h ∂h
∂2f 4 ∂2f ∂2f ∂2f = fx (x + hu1 , y + hu2 )u1
= 2
u + 2 + 2 cos2 u + u2 + fy (x + hu1 , y + hu2 )u2
∂x ∂y ∂z ∂x∂y
2 2 g (0) = fx (x, y)u1 + fy (x, y)u2 .
∂ f ∂ f ∂2f
+u2 cos u + u2 + cos u g (h) = fxx u2 + fxy u1 u2 + fyx u2 u1 + fyy u2
1 2
∂x∂z ∂y∂x ∂y∂z = fxx u2 + 2fxy u1 u2 + fyy u2
1 2
2 2
∂ f ∂ f where each second partial of f is evaluated at
+u2 cos u + cos u
∂z∂x ∂z∂y (x + hu1 , y + hu2 ). Therefore,
g (0) = fxx u2 + 2fxy u1 u2 + fyy u2
1 2
41. Apply the natural log to write:
where each second partial of f is evaluated at
ln g(t) = v(t) ln u(t) and diﬀerentiate to get
1 1 (x, y). Continuing in this vein, we see that
g (t) = v (t) ln u(t) + v(t) u (t) g (0) = fxxx u3 + 3fxxy u2 u2
1 1
g(t) u(t)
Now solve for g (t) (using g(t) = u(t)v(t) ). + 3fxyy u1 u2 + fyyy u3
2 2
v g (4) (0) = fxxxx u4 + 4fxxxy u3 u2 + 6fxxyy u2 u2
1 1 1 2
g = uv (v ln u + u ) + 4fxyyy u1 u3 + fyyyy u4
u 2
2 2
Applying this to g(t) = (2t + 1)3t yields The coeﬃcients are from the binomial expan-
2 3t2 sion (Pascal’s triangle), the number of partial
g (t) = (2t + 1)3t (6t2 ln(2t + 1) + (2)) derivatives with respect to x match the pow-
2t + 1
2 ers of u1 , and the number of partial derivatives
2 6t
= (2t + 1)3t (6t2 ln(2t + 1) + ) with respect to y match the powers of u2
2t + 1
v(t)w(t) 44. Using exercise 43, and computing the Taylor
42. g (t) = u(t) .
h(t)
Series: f (x + ∆x, y + ∆y)
Let g (t) = (u (t)) , = f (x + hu1 , y + hu2 ) = g(h)
where h (t) = (v (t))
w(t)
. g (0) 2
= g(0) + g (0)h + h
2!
Now using the exercise 41, we have g (0) 3
h + h + ···
g = uh h ln u + u and 3!
u = f (x, y) + [fx u1 + fy u2 ] h


1 fy (x, y) = x cos xy fy (0, 0) = 0
+ fxx u2 + 2fxy u1 u2 + fyy u2 h2
1 2
2! fxx (x, y) = −y 2 sin xy fxx (0, 0) = 0
1
+ fxxx u3 + 3fxxy u2 u2
1 1 fxy (x, y) = cos xy − xy sin xy
3! fxy (0, 0) = 1
+3fxyy u1 u2 + fyyy u2 h3 + · · ·
2 2
fyy (x, y) = −x2 sin xy fyy (0, 0) = 0
∆x ∆y
= f (x, y) + fx + fy h fxxx (0, 0) = fxxy (0, 0) = fxyy (0, 0)
h h
2
= fyyy (0, 0) = 0
1 ∆x 1
+ fxx f (∆x, ∆y) ≈ 2(1)∆x∆y = ∆x∆y
2! h 2!
∆x ∆y (b) Substituting u = xy, sin u = sin(xy)
+2fxy (xy)3 (xy)5
h h = (xy) − + + ···
∆y
2 3! 5!
+fyy h2 x3 y 3 x5 y 5
h = xy − + + ···
3 3! 5!
1 ∆x
+ fxxx
3! h 47. f (x, y) = e2x+y f (0, 0) = 1
∆x
2
∆y fx (x, y) = 2e2x+y fx (0, 0) = 2
+3fxxy fy (x, y) = e2x+y fy (0, 0) = 1
h h
∆x ∆y
2 fxx (x, y) = 4e2x+y fxx (0, 0) = 4
+3fxyy fxy (x, y) = 2e2x+y fxy (0, 0) = 2
h h
3
fyy (x, y) = e2x+y fyy (0, 0) = 1
∆y fxxx (x, y) = 8e2x+y fxxx (0, 0) = 8
+fyyy h3 + · · ·
h fxxy (x, y) = 4e2x+y fxxy (0, 0) = 4
= f (x, y) + [fx ∆x + fy ∆y] fxyy (x, y) = 2e2x+y fxyy (0, 0) = 2
1 fyyy (x, y) = e2x+y fyyy (0, 0) = 1
+ fxx ∆x2 + 2fxy ∆x∆y + fyy ∆y 2
2! f (∆x, ∆y) ≈ 1 + 2∆x + ∆y
1 1
+ fxxx ∆x3 + 3fxxy ∆x2 ∆y + [4∆x2 + 2(2)∆x∆y + ∆y 2 ]
3! 2!
+3fxyy ∆x∆y 2 + fyyy ∆y 3 + · · · 1
+ [8∆x3 + 3(4)∆x2 ∆y
3!
45. (a) f (x, y) = sin x cos y f (0, 0) = 0
+ 3(2)∆x∆y 2 + ∆y 3 ]
fx (x, y) = cos x cos y fx (0, 0) = 1
= 1 + 2∆x + ∆y + 2∆x2 + 2∆x∆y
fy (x, y) = − sin x sin y fy (0, 0) = 0 1 4
fxx (x, y) = − sin x cos y fxx (0, 0) = 0 + ∆y 2 + ∆x3 + 2∆x2 ∆y
2 3
fxy (x, y) = − cos x sin y fxy (0, 0) = 0 1
fyy (x, y) = − sin x cos y fyy (0, 0) = 0 + ∆x∆y 2 + ∆y 3
6
fxxx (x, y) = − cos x cos y fxxx (0, 0) = −1 48. Substituting u = 2x + y,
fxxy (x, y) = sin x sin y fxxy (0, 0) = 0 eu = e2x+y
fxyy (x, y) = − cos x cos y fxyy (0, 0) = −1 (2x + y)2
fyyy (x, y) = sin x sin y fyyy (0, 0) = 0 = 1 + (2x + y) +
2!
f (∆x, ∆y) ≈ 1∆x (2x + y)3 (2x + y)4
1 + + + ···
+ [1(−1)∆x3 + 3(−1)∆x∆y 2 ] 3! 4!
3! y2
1 1 = 1 + 2x + y + 2x2 + 2xy +
= ∆x − ∆x3 − ∆x∆y 2 2
6 2 4x3 2 2 y3
(b) (sin x)(cos y) + + 2x y + xy +
3 6
4
x3 x5 2x 4x3 y 2 2
= x− + + ··· + + +x y
3! 5! 3 3
3 4
y2 y4 xy y
· 1− + + ··· + + + ···
2! 4! 3 24
x3 xy 2 x5 x3 y 2 xy 4 x
= x− − + + + +· · · 49. f (x, y, z) = + yez ,
6 2 120 12 24 y
where x = t2 ; y = t + 4 and z = ln t2 + 1 .
46. (a) f (x, y) = sin xy f (0, 0) = 0 dx dy dz 2t
fx (x, y) = y cos xy fx (0, 0) = 0 Therefore, = 2t; = 1 and = 2 .
dt dt dt t +1


Now, using 1
53. (a) R(c, h) = R(1, 1) = 1
0.55 0.45
dg ∂f dx ∂f dy ∂f dz +
= + + c h
dt ∂x dt ∂y dt ∂z dt Rc (1, 1) = 0.55 Rh (1, 1) = 0.45
we get, R(c, h) ≈ 1 + 0.55∆c + 0.45∆h
dg 2t x 2tyez (b) The graph of R and the ﬁrst degree Taylor
= + − 2 + ez + polynomial with h = 40 is plotted below.
dt y y t2 + 1 h=40

y z
50. f (x, y, z) = tan−1 + tan−1 , 40
x y
t
where x = 2 cos t; y = 2 sin t and z = . 30
8
Therefore, R
20
dx dy dz 1
= −2 sin t; = 2 cos t and = .
dt dt dt 8 10

dg ∂f dx ∂f dy ∂f dz
Now, using = + +
dt ∂x dt ∂y dt ∂z dt 0
0 10 20 30 40 50
c
we get
If h = c, then R is equal to the ﬁrst de-
dg 2y sin t 2x cos t 2z cos t gree Taylor polynomial.
= 2 + − 2
dt x + y2 x 2 + y2 y + z2 If c = 5 and h = 40, then one would ex-
y 1 pect the rating to be low, especially con-
+ sidering that most driving is done in the
y2 + z2 8
city.
51. V = πr2 h, 54. If E = f (P, T ) and P = g(T, V ),
dr dh
= 0.2 and = −0.2 then substituting we get
dt dt E = f (g(T, V ), T ) = h(T, V ).
Therefore Using the chain rule
dV ∂V dr ∂V dh ∂E ∂f ∂T ∂f ∂g
= + = +
dt ∂r dt ∂h dt ∂T V ∂T ∂T ∂P ∂T
dV ∂E ∂E ∂P
⇒ = 0.2 (2πrh) − 0.2 πr2 = +
dt ∂T P ∂P T ∂T V
dV b−h
⇒ = (0.2) (πr) (2h − r). 55. (a) a = 55racbh − hb b2 =
dt b2
dV
Now the volume increases if > 0, for which (b) At h = 50 and b = 200, a = 0.250, and
dt
r a = 0.00375 ≈ 4 points.
(0.2) (πr) (2h − r) > 0 or h > .
2 If h = 100 and b = 400, a = 0.250 still,
and a = 0.01875 ≈ 2 points.
dr dh In general if the number of hits and at
52. Here, = 0.02r0 ; = −0.02h0 where
dt dt bats are doubled, then the rate of change
r0 , h0 are the initial radius and height of the
cylinder respectively. From the exercise 51, of the average is halved.
(c) At h = 50 and b = 200, an out results in
dV
= 0.02r0 (2πrh) − 0.02h0 πr2 or a=
50
≈ 0.249, a decrease in one point.
dt 201
dV At h = 100 and b = 400, an out results in
= (0.02) πr (2hr0 − h0 r) 100
dt a= ≈ 0.249, a decrease of one point.
401
dV So, the rounded values each change by a
Now the volume increases if > 0, for which
dt point.
(2hr0 − h0 r) > 0
56. This is the chain rule of Theorem 5.1.
h 1 r d
That is > . Clearly (kp(x, y)) = p(x, y).
h0 2 r0 dk


For the other side of the equation, 11. f = 2xy, x2 + 8y
d f (2, 1) = 4, 12
[p(kx, ky)] √
dk 1 3 √
d d Du f (2, 1) = 4, 12 · , =2+6 3
= px (kx, ky) (kx) + py (kx, ky) (ky) 2 2
dk dk
= xpx (kx, ky) + ypy (kx, ky)
Which gives us the equations 12. f = 3x2 y, x3 − 8y
p(x, y) = xpx (kx, ky) + ypy (kx, ky) f (2, −1) = −12, 16
1 1 √
This must be true for all values of k, in partic- Du f (2, −1) = −12, 16 · √ ,√ =2 2
ular for k = 1 and therefore, 2 2
p(x, y) = xpx (x, y) + ypy (x, y)
The total production is the term p(x, y). The x y
13. f= , ,
cost of capital at its level of marginal product x2 + y2 x2 + y2
is the term xpx (x, y). The cost of labor at its 3 4
level of marginal product is the term ypy (x, y). f (3, −4) = ,−
5 5
3 4 3 2
Du f (3, −4) = ,− · √ , −√
12.6 The Gradient and 5 5 13 13
17
Directional Derivatives = √
5 13
1. f = 2x + 4y 2 , 8xy − 5y 4 2
−y 2
−y
14. f = 8xe4x , −e4x
2 3y 3 3y 3
2. f = 3x e , 3x e − 4y f (1, 4) = 8, −1
2 1
3.
2 2 2
f = exy + xy 2 exy , 2x2 yexy − 2y sin y 2 Du f (3, −4) = 8, −1 · − √ , − √
√ 5 5
= −3 5
3y 3y/x 3
4. f= − e − 2xy 3 , e3y/x − 3x2 y 2
x2 x 15. f = −2 sin(2x − y), sin(2x − y)
f (π, 0) = 0, 0
8 4x/y −8x
5. f= e − 2, 2 e4x/y 1 1
y y u= √ ,√
2 2
f (2, −1) = −8e−8 − 2, −16e−8 1 1
Du f (π, 0) = 0, 0 · √ , √ =0
6. f = 3y cos 3xy, 3x cos 3xy + 2y 2 2
f (π, 1) = −3, −3π + 2
2 sin 4y sin 4y
16. f= , + 4 cos 4y · ln x2 y
7. f = 6xy + z sin x, 3x2 , − cos x x y
f (0, 2, −1) = 0, 0, −1 π 8
f −2, = −1, ,
8. f = 2z 2 e2x−y − 4z 2 , 8 π
16 −π
−z 2 e2x−y , 2ze2x−y − 8xz u= √ ,√
f (1, 2, 2) = −8, −4, −12 256 + π 2 256 + π 2
π
9. f = 2w cos x, −w2 sinx + 3zexz ln y, Du f −2,
8
(3exz /y ) , 3xexz ln y 8 16 π
= −1, · √ , −√
π 256 + π 2 256 + π 2
f (2, π, 1, 4) = −4, 0, 3e4π , 0
16 8 24
1 x1 x3 = −√ −√ = −√
256 + π 2 256 + π 2 256 + π 2
10. fx1 = cos −√
x2 x2 x1 x3
x1 x1 y
fx2 = − 2 cos 17. f= 3x2 yz 2 − , x3 z 2
x2 x2 x2 + y 2
x1 x
fx3 = −6x3 x4 x5 − √ + 2 , 2x3 yz
x1 x3 x + y2
fx4 = −3x2 x53 23 9
f (1, −1, 2) = − , , −4 ,
fx5 = −3x2 x43 2 2
f (2, 1, 2, −1, 4) 2 1
= cos 2 − 1, −2 cos 2, 47, −48, 12 u = √ , 0, − √
5 5

12.6. THE GRADIENT AND DIRECTIONAL DERIVATIVES 699

Du f (1, −1, 2) = −4 sin 4, 4 sin 4, −4 sin 4, 1
23 9 2 1 2 1 4
= − , , −4 · √ , 0, − √ u = − √ , 0, √ , √
2 2 5 5 21 21 21
23 4 19 Du f (2, −1, 1, 0)
= −√ + √ = −√
5 5 5 = −4 sin 4, 4 sin 4, −4 sin 4, 1
·
x 2 1 4
18. f= , − √ , 0, √ , √
x2 + y2 + z2 21 21 21
4 sin 4 + 4
y z = √
, 21
x2 + y 2 + z 2 x2 + y 2 + z 2 √ √
1 4 8 2x1 −x2 1 −2 3 x5 3 x4
f (1, −4, 8) = ,− , 23. f = , , , √ , √
9 9 9 x2 x2 2 1 − 4x2 2 x4 2 x5
3
1 1 −2 3
u= √ ,√ ,√ f (2, 1, 0, 1, 4) = 4, −4, −2, 3,
6 6 6 4
Du f (1, −4, 8) 1 −2 4 −2
1 4 8 1 1 −2 19 u= , 0, , ,
= ,− , · √ ,√ ,√ =− √ 5 5 5 5
9 9 9 6 6 6 9 6 Du f (2, 1, 0, 1, 4)
3 1 −2 4 −2 37
19. f = yexy+z , xexy+z , exy+z = 4, −4, −2, 3, , 0, , , =
4 5 5 5 5 10
f (1, −1, 1)= −1, 1, 1 ,
24. f = 3x3 x3 , 9x1 x2 x3 , 3x1 x3 − 4e4x3 ,
2 2 2
4 −2 3 1 1
u= √ , √ , √ ,
29 29 29 2x4 2x5
Du f (1, −1, 1) 1 1
f (−1, 2, 0, 4, 1) = 0, 0, −28, ,
4 −2 3 8 2
= −1, 1, 1 · √ ,√ ,√ 2 1 1 2
29 29 29 u = √ , − √ , 0, √ , − √
−3 10 10 10 10
=√ Du f (−1, 2, 0, 4, 1)
29 1 1
= 0, 0, −28, , ·
20. f = −y sin xy, −x sin xy, 1 8 2
2 1 1 2
f (0, −2, 4) = 0, 0, 1 √ , − √ , 0, √ , − √
10 10 10 10
3 4 −7
u= 0, , − = √
5 5 8 10
Du f (0, −2, 4) ∂z 4x − 4xy − z 2
3 4 4 25. = and
= 0, 0, 1 · 0, , − =− ∂x 3yz 2 − 2xz
5 5 5 ∂z 2x2 − z 3
=
w2 x ∂y 3yz 2 − 2xz
21. f= 2w x2 + 1, √ + 3z 2 exz , 4x − 4xy − z 2 2x2 − z 3
x2 + 1 f= , .
0, 3exz + 3xzexz 3yz 2 − 2xz 3yz 2 − 2xz
f (2, 0, 1, 0) = 4, 0, 0, 3 Further (x, y) = (1, 1) , ⇒ 2 − z 3 + z 2 − 2 = 0
1 3 4 −2 ⇒ z = 1, as z > 0.
u= √ ,√ ,√ ,√
30 30 30 30
Du f (2, 0, 1, 0) f (1, 1) = 1, 1 ;
1 3 4 −2 3 1
= 4, 0, 0, 3 · √ , √ , √ , √ u = √ , −√
30 30 30 30 10 10
−2 Therefore,
= √
30 3 1 2
Du f (1, 1) = 1, 1 · √ , −√ =√
22. 2 2
f = −2wxy sin(w xy), −w y sin(w xy), 2 10 10 10
−w2 x sin(w2 xy), 3 − 2 sec2 2z ∂z −ez
f (2, −1, 1, 0) 26. = z and
∂x xe − y 2 + 2y


∂z 2 (y − 1) z in the direction 4, 0
= z
∂y xe − y 2 + 2y and the minimum change is −4;
−ez 2 (y − 1) z in the direction −4, 0 .
f= ,
z − y 2 + 2y xez − y 2 + 2y
.
xe
(b) From part (a),
Further (x, y) = (1, 0) ⇒ 1 · ez − 0 + 0 = 2 f (−2, π) = −4, 0 and
⇒ ez = 2 or z = ln2. f (−2, π) = 4.
f (1, 0) = −1, − ln 2 ; Thus, the maximum change is 4;
1 2 in the direction −4, 0
u= −√ , √
5 5 and the minimum change is −4;
1 2 in the direction 4, 0 .
Du f (1, 0) = −1, − ln 2 · − √ , √
5 5
1 − ln 4 2x −1
= √ 30. (a) f= ,
5 2x2−y 2 2x2 − y
3 1
27. (a) f = 2x, −3y 2 f (3, 2) = ,−
2 8
√
f (2, 1) = 4, −3
145
f (2, 1) = 5 f (3, 2) =
The maximum change is 5; 8 √
145
in the direction 4, −3 The maximum change is ;
The minimum change is −5; 8
3 1
in the direction −4, 3 in the direction ,−
2 8 √
(b) From part (a), 145
The minimum change is − ;
f (−1, −2) = −2, −12 and 8
√ √ 3 1
f (−1, − 2) = 148 = 2 37. in the direction − ,
√ 2 8
Thus the maximum change is 2 37;
in the direction −2, −12 (b) From part (a),
√
And the minimum change is −2 37; 4 1
f (2, −1) = ,− and
in the direction 2, 12 . 3 6
√
65
28. (a) f = 4y 2 e4x , 2ye4x f (2, −1) = .
6 √
f (0, −2) = 16, −4
√ 65
f (0, −2) = 272 √ Thus, the maximum change is ;
6
The maximum change is 272; 4 1
in the direction 16, −4 √ in the direction ,−
3 6
The minimum change is − 272; √
in the direction −16, 4 65
and the minimum change is − ;
6
(b) From part (a),
4 1
f (3, −1) = 4e12 , −2e12 and in the direction − , .
√ 3 6
f (2, 1) = 2e12 5.
√
Thus, the maximum change is 2e12 5; x y
31. (a) f= ,
in the direction 4e12 , −2e12 and x2 + y2 + y2x2
√
the minimum change is −2e12 5; 3 4
f (3, −4) = ,−
in the direction −4e12 , 2e12 . 5 5
f (3, −4) = 1
29. (a) f = 2x cos (3xy) − 3x2 y sin (3xy) , The maximum change is 1;
−3x3 sin (3xy) 3 4
in the direction ,−
Therefore, f (2, 0) = 4, 0 and 5 5
The minimum change is −1;
f (2, 0) = 4. 3 4
Thus, the maximum change is 4; in the direction − ,
5 5


(b) From part (a), 33. (a) f = 8xyz 3 , 4x2 z 3 , 12x2 yz 2
4 5 f (1, 2, 1) = 16, √ 24
4,
f (−4, 5) = −√ , √ and
41 41 f (1, 2, −2) = 848 √
f (−4, 5) = 1. The maximum change is 848;
Thus, the maximum change is 1; in the direction 16, 4, 24 √
4 5 The minimum change is − 848;
in the direction − √ , √ in the direction −16, −4, −24
41 41
and the minimum change is −1; (b) From part (a), f (2, 0, 1) = 0, 16, 0 and
4 5 f (2, 0, 1) = 16.
in the direction √ , − √ . Thus, the maximum change is 16;
41 41
in the direction 0, 16, 0
32. (a) f (x, y) = xtan−1 x
y
and the minimum change is −16;
in the direction 0, −16, 0 .
x xy −x2
⇒ f= tan−1 + , 2 .
y x2 + y 2 x + y 2 x y
π 1 1 34. (a) f= , ,
Therefore, f (1, 1) = + ,− and x2 + y2 + z2 x2 + y2 + z2
4 2 2
z
π2 π 1
f (1, 1) = + + . x2 + y2 + z2
16 4 2 1 2 2
π2 π 1 f (1, 2, −2) = , ,−
Thus, maximum change is + + ; 3 3 3
16 4 2 f (1, 2, −2) = 1
π 1 1 The maximum change is 1;
in the direction + ,−
4 2 2 1 2 2
in the direction , ,−
and the minimum change is 3 3 3
π2 π 1 The minimum change is −1;
− + + ; 1 2 2
16 4 2 in the direction − , − ,
π 1 1 3 3 3
in the direction − + , . (b) From part (a),
4 2 2
3 1 1
(b) From part (a), f (3, 1, −1) = √ , √ , −√
√ 11 11 11
√ 1 2 1
f 1, 2 = tan−1 √ + ,− and f (3, 1, −1) = 1.
2 3 3 Thus, the maximum change is 1;
and √ 3 1 1
in the direction √ , √ , − √
f 1, 2 11 11 11
√ 2 and the minimum change is −1;
−1 1 2 1 3 1 1
= tan √ + + . in the direction − √ , − √ , √ .
2 3 9 11 11 11
Thus, the maximum change is 35. (a) The direction u = 1, 0 is sketched in
√ 2 with its initial point located the point
−1 1 2 1 (1, 0). The level curves to be considered
tan √ + + ;
2 3 9 are z = 1 and z = 2. From the graph we
in the direction can approximate the directional deriva-
√ ∆z
−1 1 2 1 tive by estimating , where ∆u = the
= tan √ + ,− ∆u
2 3 3 distance along the unit vector u, which
√
and the minimum change is −
is 2 √ 1, as z = 2 cuts√ the x-axis at
√ 2 x = 2. Thus, ∆u = 2 − 1. Fur-
−1 1 2 1 ther, the vector appears to extend from
− tan √ + + ;
2 3 9 the z = 1 level curve to the z = 2
in the direction level curve. Therefore, ∆z = 1 and our
√ estimate of the directional derivative is
1 2 1 ∆z 1
− tan−1 √ + , . =√ = 2.4142.
2 3 3 ∆u 2−1


(b) The direction u = 0, 1 is sketched in cos(x+y) 1, 1 . Further, it is stated that each
with its initial point located the point crest extends parallel to the shore line, hence
(1, 0). The level curves to be considered the gradient is perpendicular to the shore line.
are z = 1 and z = 0. Proceeding as The sketch of z = sin (2x − y)
done in part (a), we have ∆u = 1 and
∆z
∆z = 0 − 1 = −1, therefore = −1 1.0
∆u
36. (a) The direction u = 1, 0 is sketched in 0.5

with its initial point located the point
−3 0.0
(0, 1). The level curves to be considered −2
−1
0 −1
−2
−3

y1 0
are z = −1 and z = 0. Proceeding as 3
2 1
2
x
3
−0.5
done in Exercise 35 (a), we have ∆u = 1
∆z
and ∆z = 0−(−1) = 1, therefore =1 −1.0
∆u
(b) The direction u = 0, −1 is sketched in
with its initial point located the point
(0, 1). The level curves to be considered
are z = −1 and z = 0. Proceeding as
done in Part (a), we have ∆u = 1 and
∆z sin(2x − y) = 2 cos(2x − y), − cos(2x − y)
∆z = 0 − (−1) = 1, therefore =1 = cos(2x − y) 2, −1
∆u
37. The level curves (surfaces) are circles (spheres)
centered at the origin. The gradients will be
orthogonal to these, and therefore be parallel As the gradient cos(2x − y) 2, −1 is perpen-
to the position vectors. dicular to the shore line, as derived earlier the
vector perpendicular to the shore line (parallel
38. f (a, b) = 2ag (a2 + b2 ), 2bg (a2 + b2 ) to the gradient) is 2a, −a a = 0
= 2g (a2 + b2 ) a, b
Notice that the level curves are all concentric
circles, centered at the origin.

39. The sketch of z = sin (x + y)
40. sin(x + y) = cos(x + y), cos(x + y)
1.0
= cos(x + y) 1, 1
1, 1 · 100, −100 = 0
0.5 and therefore these vectors are perpendicular.
−3 5.0
The directional derivative
−2

x
−10.0
00.0
1
2.5
in the direction of 100, −100 must be zero
−2.5 2
−5.0
y −0.5
3 because this is the direction of a level curve–a
curve where the function in constant.
−1.0 sin(x + y) viewed from (100, −100, 0)

1

0.5
Here sin(x + y) = cos(x + y), cos(x + y)
= cos(x + y) 1, 1 -4 -2
0
0 2 4
x
y
Consider the level curves in the xy plane. They -0.5

are f (x, y) = c or sin (x + y) = c. For z = 0,
-1
sin (x + y) = 0 that is x + y = nπ for any in-
teger n or y = −x + nπ. Therefore, the crests
are in the directions parallel to u = 1, −1, 0 ,
which is perpendicular to the gradient that is sin(2x − y) viewed from (100, 200, 0)


f = 0, 0 when x = y and −x + x3 = 0.
3
x − x factors as x(x − 1)(x + 1).
1 So the places where the tangent plane is par-
allel to the xy-plane are at (0, 0), (1, 1), and
0.5
(−1, −1). These are possible local extrema.
0
4 -4 -2 0
2 2 -2 -4
x y
4
46. f = cos x cos y, − sin x sin y
-0.5 mπ
f = 0, 0 when (x, y) = , nπ
2
-1 mπ
or when (x, y) = nπ,
2
where n, m are integers. These are possible lo-
cal extrema.
41. f (x, y, z) = x2 + y 3 − z 47. f (x, y) = x2 y − 2y 2
f = 2x, 3y 2 , −1 ⇒ f = 2xy, x2 − 4y ,
f (1, −1, 0) = 2, 3, −1
The tangent plane is: Therefore
2(x − 1) + 3(y + 1) − z = 0 f (a, b) = 2ab, a2 − 4b = 4, 0
The normal line has parametric equations
gives 2ab = 4 and a2 − 4b = 0 ⇒ a = 2; b = 1
x = 1 + 2t, y = −1 + t, z = −t
48. f (x, y) = x2 y 2 − 2xy
42. f (x, y, z) = x2 + y 2 − z
⇒ f = 2xy 2 − 2y, 2x2 y − 2x ,
x y
f= , , −1 Therefore
x2 + y 2 x2 + y 2
3 4 f (a, b) = 2ab2 − 2b, 2a2 b − 2a = 4, 12
f (3, −4, 5) = , − , −1
5 5 gives 2ab2 − 2b = 4 and 2a2 b − 2a = 12
The tangent plane is:
3 4 ⇒ a = 3; b = 1
(x − 3) − (y + 4) − (z − 5) = 0
5 5
The normal line has parametric equations 49.
3
x=3+ t
5
4
y = −4 − t
5
z =5−t

43. f (x, y, z) = x2 + y 2 + z 2 − 6
f = 2x, 2y, 2z
f (−1, 2, 1) = −2, 4, 2
The tangent plane is:
−2(x + 1) + 4(y − 2) + 2(z − 1) = 0
x = −1 − 2t
y = 2 + 4t 50.
z = 1 + 2t

44. f (x, y, z) = x2 − y 2 − z 2
f = 2x, −2y, −2z
f (5, −3, −4) = 10, 6, 8
10(x − 5) + 6(y + 3) + 8(z + 4) = 0
x = 5 + 10t
y = −3 + 6t
z = −4 + 8t

45. f = 4x − 4y, −4x + 4y 3


2.2 − 1.8 But, lim f (x, y)
51. fx (0, 0) ≈ =2 (x,y)→(0,0)
0.1 − (−0.1)
1.6 − 2.4 2 (0) (y)
fy (0, 0) ≈ = −2 = lim =0
0.2 − (−0.2) (0,y)→(0,0)
2
(0) + y 2
f (0, 0) ≈ 2, −2
(along the path x = 0)
1.6 − 2.4 2 (x) (0)
52. fx (0, 0) ≈ = −2 = lim
0.2 − (−0.2) 2 =0
(x,0)→(0,0) x2 + (0)
2.1 − 1.9 1
fy (0, 0) ≈ = (along the path y = 0)
0.3 − (−0.3) 3
1 2x2
f (0, 0) ≈ −2, = lim = 1 (along the path y = x)
3 (x,x)→(0,0) 2x2

53. (a) True. Thus, lim f (x, y) does not exist
(x,y)→(0,0)
∂f ∂g ∂f ∂g
(f + g) = + , + ⇒ f is not continuous at (0, 0)
∂x ∂x ∂y ∂y
∂f ∂f ∂g ∂g 57. The gradient is
= , + ,
∂x ∂y ∂x ∂y f = − tan 10◦ , tan 6◦ ≈ −0.176, 0.105 ,
= f+ g and this gives the direction of maximum as-
(b) True. cent. The rate of change in this direction is
∂f ∂g ∂f ∂g the magnitude of the gradient, f ≈ 0.205,
(f g) = g+ f, g+ f and the rise in degrees is tan−1 0.205 ≈ 11.6◦ .
∂x ∂x ∂y ∂y
∂f ∂f ∂g ∂g
= , g+ , f 58. If the point on the mountain is (x0 , y0 , z0 ) with
∂x ∂y ∂x ∂y the positive x-direction as east, then the direc-
= ( f )g + f ( g) tion of steepest ascent will be
54. 2
f (x, y) = fxx (x, y) + fyy (x, y) = 6x + 2. f = − tan 4◦ , tan 3◦ ≈ −0.06993, 0.05241
f ≈ 0.08739
55. The function is not continuous because the The rise in the direction of f would be
limit along any line is 0, but the limit along tan−1 (0.08739) ≈ 4.99◦
1
the curve y = x2 is . Therefore the limit as 59. (a) f = −8x, −2y
2
(x, y) → (0, 0) does not exist and the function f (1, 2) = −8, −4
is not continuous. The rain will run in direction 8, 4 .
f (hu1 , hu2 ) − f (0, 0)
Du f (0, 0) = lim (b) The level curve is
h→0 h
h3 u2 u2 100 = 200 − y 2 − 4x2
1
= 6 6 =0 or y 2 + 4x2 = 100, which is an ellipse.
h u1 + 2h2 u2 2
for any vector u. All directional derivatives 60. This means that
may exist, even if the function is not continu- 2 units of stock 1 should be bought,
ous. 1 unit of stock 2 should be sold,
2xy 6 units of stock 3 should be bought,
56. f (x, y) = 2 , for (x, y) = (0, 0)
(x + y 2 ) nothing should be done with stock 4, and
2 units of stock 5 should be sold.
2y y 2 − x2 2x x2 − y 2
⇒ f (x, y) = 2 , 2
(x2 + y2 ) (x2 + y 2 ) 61. We have the function g(w, s, t), with
g(4, 10, 900) = 4. Then
Consider a vector u = u1 , u2 , therefore by ∂g 0.04
the deﬁnition of directional derivative (4, 10, 900) = = 0.8,
∂w 0.05
Du f (x, y) ∂g 0.06
(4, 10, 900) = = 0.3, and
f (hu1 , hu2 ) − f (0, 0) ∂s 0.2
= lim
h→0 h ∂g −0.04
(4, 10, 900) = = −0.004. Then
2 (hu1 ) (hu2 ) ∂t 10
= lim
h→0 (hu1 )2 + (hu2 )2 g(4, 10, 900) = 0.8, 0.3, −0.004
2u1 u2 and this gives the direction of maximum in-
= 2 , which exists. crease of gauge.
u1 + u2 2

12.7. EXTREMA OF FUNCTIONS OF SEVERAL VARIABLES 705

−10e−z −5e−z 1 1 x = (x2 )2 = x4
62. T = , , −5e−z +
x3 y2 x2 y x4 − x = 0
−5e−8 −25 −8 x(x3 − 1) = 0
T (1, 4, 8) = −10e−8 , , e
16 4 so x = 0 or x = 1, and the critical points are
Heat decreases most rapidly in the opposite di- (0, 0) and (1,1).
5 25 D(0, 0) < 0 so (0, 0) is a saddle point.
rection: 10, , .
16 4 D(1, 1) > 0 and fxx > 0, so (1, 1) is a local
Heat increases most rapidly in the same direc-
−5 −25 minimum.
tion as T : −10, , .
16 4 4. fx = 4y − 4x3
63. If the shark moves toward a higher electrical fy = 4x − 4y 3
charge, it should move in direction 12, −20, 5 . fxx = −12x2
fyy = −12y 2
64. An increase in v will greatly increase S, fxy = 4
An increase in t will decrease S, D = 144x2 y 2 − 16
An increase in e will increase S, and Solving f = 0, 0 gives equations
An increase in θ will decrease S. y = x3 and x = y 3 .
Substituting gives
12.7 Extrema of Functions of x = (x3 )3 = x9
x9 − x = 0
Several Variables x(x8 − 1) = 0
2 so x = 0 or x = ±1
1. fx = −2xe−x (y 2 + 1) This gives critical points at
2
fy = 2ye−x (0, 0), (1, 1), (−1, −1).
2
fxx = (4x2 − 2)e−x (y 2 + 1) D(0, 0) = −16 < 0, so saddle point at (0, 0).
2
fyy = 2e−x D(1, 1) = 128 > 0
fxy = −4xye−x
2
fxx (1, 1) = −12 < 0 and so there is a local
Solving f = 0, 0 gives critical point maximum at (1, 1).
(x, y) = (0, 0). D(−1, −1) = 128 > 0
D(0, 0) = (−2)2 − 02 = −4, so f has a saddle fxx (−1, −1) = −12 < 0 and so there is a local
point at (0, 0). maximum at (1, 1).

2. fx = −2 cos x sin x = − sin 2x 5. fx = 2xy + 2x
fy = 2y fy = 2y + x2 − 2
fxx = −2 cos 2x fxx = 2y + 2
fyy = 2 fyy = 2
fxy = 0 fxy = 2x
D = −4 cos 2x D = 2y + 4 − 4x2
Solving f = 0, 0 gives critical points at Solving f = 0, 0 gives equations
nπ 2x(y + 1)= 0 and x2 + 2y − 2 = 0.
(x, y) = ,0 .
2 Either x = 0 or y = −1.
π
If x is an odd multiple of , then D > 0 This gives critical points at
2
and fxx > 0 and these critical points are local (0, 1) and (±2, −1).
minima. The other critical points are saddle D(0, 1) = 6 > 0 and fxx = 4 > 0, so (0, 1) is a
points. local minimum.
D(±2, −1) = −14 < 0, so (±2, −1) are both
3. fx = 3x2 − 3y saddle points.
fy = −3x + 3y 2
fxx = 6x 6. fx = 4x − 2xy
fyy = 6y fy = 3y 2 − x2 − 3
fxy = −3 fxx = 4 − 2y
D = 36xy − 9 fyy = 6y
Solving f = 0, 0 gives equations fxy = −2x
y = x2 and x = y 2 . D = (4 − 2y)(6y) − 4x2 = 24y − 12y 2 − 4x2
Substituting gives Solving f = 0, 0 gives equations


2x(2 − y) = 0 and 3y 2 − x2 − 3 = 0. 2xey = 0, x2 − y 2 − 2y = 0
Solving gives critical points at ey = 0 gives x = 0 , y = 0 or y= − 2
(0, 1), (0, −1), (3, 2), (−3, 2). So critical point are (0, 0) and (0,-2)
D(0, 1) = 12, fxx (0, 1) = 2 D (0, 0) = −4 < 0
and so there is a local minimum at (0, 1). so f has saddle point at (0, 0).
D(0, −1) = −26 D (0, −2) = 4e−4 > 0 and fxx > 0
and so there is a saddle at (0, −1). so there is local minimum at (0,-2).
D(3, 2) = −36
2 2
and so there is a saddle at (3, 2). 11. fx = 1 − 2x2 e−x −y
D(−3, 2) = −36 2
fy = −2xye−x −y
2

and so there is a saddle at (−3, 2). 2 2
fxx = 2xe−x −y (2x2 − 3)
2 2

7. fx = −2xe−x −y
2 2 fyy = 2xe−x −y (2y 2 − 1)
2 2
fy = −2ye−x −y
2 2
fxy = −2xye−x −y
2 2
D = e−2x −2y (12x2 − 30x2 y 2 − 8x4 − 4y 2 )
2 2
fxx = (4x2 − 2)e−x −y
2 2
fyy = (4y 2 − 2)e−x −y
2 2
fxy = 4xye−x −y Solving f = 0, 0 gives equation gives
2 2
D = (4x2 − 2)(4y 2 − 2)e−2x −2y 1 1
2 2 x = ± √ . So critical points are ± √ , 0 .
− 16x2 y 2 e−2x −2y 2 2
Solving f = 0, 0 gives critical point (0, 0) 1 1
D(0, 0) = 4, fxx (0, 0) = −2 D √ , 0 > 0 and fxx √ , 0 < 0
2 2
and so there is a local maximum at (0, 0).
1
so the point √ , 0 is a local maximum.
8. fx = sin y 2
fy = x cos y 1 1
fxx = 0 D − √ , 0 > 0 and fxx − √ , 0 > 0
2 2
fyy = −x sin y
fxy = cos y 1
so the point − √ , 0 is a local minimum.
D = − cos2 y 2
Solving f = 0, 0 gives equations 2 2
12. fx = 2 x − x3 e−x −y
sin y = 0 and x cos y = 0. 2 2
This gives critical points at fy = −2x2 ye−x −y
2 2
(0, nπ) D(0, nπ) = −1 < 0 fxx = e−x −y (4x4 − 10x2 + 2)
2 2
and so there are saddles at (0, nπ). fyy = e−x −y (4x2 y 2 − 2x2 )
2 2
fxy = e−x −y (4x3 y − 4xy)
9. Domain = {(x, y) | x = 0, y = 0 } 2 2
D = e−2x −2y (−8x6 − 8x4 y 2 − 8x2 y 2
1 1 + 20x4 − 4x2 )
fx = y − , fy = x − 2
x2 y
2 2
fxx = 3 , fyy = 3 Solving f = 0, 0 gives equation gives x = 0
x y
fxy = 1 and x = ±1. So the critical points are (±1, 0)
4 and (0,y) for any y.
D = 3 3 −1
x y D (0, y) = 0, giving us no information.
But f (0, y) = 0 ≤ f (x, y) for all (x, y)
Solving f = 0, 0 gives critical point (1, 1)
D (1, 1) = 3,fxx = 2 Therefore f has local minima at (0,y)

and so there is local minimum at (1, 1). D (±1, 0) > 0 and fxx (±1, 0) < 0 so the points
(±1, 0) are both local maxima.
10. fx = 2xey , fy = ey x2 − y 2 − 2y 4y
fxx = 2ey , fyy = ey x2 − y 2 − 4y − 2 13. fx = 2x −
y2 +
1
fxy = 2xey 2
y −1
2 fy = −4x 2
D = 2e2y x2 − y 2 − 4y − 2 − (2xey ) (y + 1)2
Solving f = 0, 0 gives equation Solving fy = 0 gives x = 0 or y = ±1. Now


fx = 0 gives critical points (0, 0), (1, 1), and D(0, 0) < 0, so (0, 0) is a saddle point.
(−1, −1). 1 1
D(± √ , ± √ ) > 0, so all four of these
Using a CAS, we ﬁnd that: 2 2
D(0, 0) < 0 so (0, 0) is a saddle point. points are extrema. Using fxx we see
D(1, 1) > 0 and fxx (1, 1) > 0, so (1, 1) is a lo- 1 1
that ±( √ , √ ) are both local maxima, and
cal minimum. 2 2
1 1
D(−1, −1) > 0 and fxx (−1, −1) > 0, so ±( √ , − √ ) are both local minima.
(−1, −1) is a local minimum. 2 2
2 4
(x2 + y 2 + 1) − (x + y)(2x) 16. fx = y(1 − 2x2 )e−x −y
14. fx = 2 4
x2 + y 2 + 1 fy = x(1 − 4y 4 )e−x −y
(x2 + y 2 + 1) − (x + y)(2y) Solving f = 0, 0 gives equations
fy =
x2 + y 2 + 1 y(1 − 2x2 ) = 0
Solving f = 0, 0 gives equations x(1 − 4y 4 ) = 0
x2 + y 2 + 1 − 2x2 − 2xy = 0 Solving gives critical points
x2 + y 2 + 1 − 2y 2 − 2xy = 0 1 1 1 1
(0, 0), √ , √ , − √ , √ ,
Subtracting these equations gives 2 2 2 2
−2x2 + 2y 2 = 0 1 1 1 1
√ , − √ , and − √ , − √ .
x2 = y 2 2 2 2 2
x = ±y Using a CAS,
Substituting this result into our ﬁrst equation D(0, 0) = −1 and so there is a saddle at (0, 0)
gives 1 1
D √ ,√ = 8e−3/2 > 0
(±y)2 + y 2 + 1 − 2(±y)2 − 2(±x)y = 0 2 2
± 2y 2 = 1 1 1
fxx √ , √ = −2e−3/4 < 0
Notice that the equation −2y 2 = 1 does not 2 2
give any solutions, which means x = y and so there is a local maximum at
(x = −y can not hold). 1 1 1 1
√ ,√ D −√ , √ = 8e−3/2 > 0
1 2 2 2 2
Therefore, y = ± √
2 1 1
fxx − √ , √ = 2e−3/4 > 0
This gives critical points at 2 2
1 1 1 1 and so there is a local minimum at
√ ,√ and − √ , − √ 1 1 1 1
2 2 2 2 −√ , √ D √ , −√ = 8e−3/2
Using a CAS, 2 2 2 2
1 1 1 1 1
D √ ,√ = >0 fxx √ , − √ = 2e−3/4 > 0
2 2 2 2 2
1 1 1 and so there is a local minimum at
fxx √ , √ = −√ < 0 1 1 1 1
2 2 2 √ , −√ D −√ , −√ = 8e−3/2
and so there is a local maximum at 2 2 2 2
1 1 1 1
√ ,√ . fxx − √ , − √ = −2e−3/4
2 2 2 2
1 1 1 and so there is a local maximum at
D −√ , −√ = >0 1 1
2 2 2 −√ , −√
1 1 1 2 2
fxx − √ , − √ = √ >0
2 2 2 1
and so there is a local minimum at 17. fx = y 2 − 2x + x3
4
1 1 fy = 2xy − 1
−√ , −√ .
2 2 1
Equation fy = 0 gives y = .
2 2
2x
15. fx = y(1 − 2x2 )e−x −y Substituting this into fx = 0 and clearing de-
2
fy = x(1 − 2y 2 )e−x −y
2
nominators yields x5 − 8x3 + 1 = 0.
1 Numerically, this gives solutions at approxi-
Solving fx = 0 gives y = 0 or x = ± √ . mately (0.5054, 0.9892), (2.8205, 0.1773), and
2
Then fy = 0 gives critical points (0, 0) and (−2.8362, −0.1763).
1 1 D(0.5054, 0.9892) ≈ −5.7420, so this is a sad-
(± √ , ± √ ). dle point.
2 2


D(2.8205, 0.1773) ≈ 22.2488 and fxx (0.6102, 0) ≈ 4.885 > 0
fxx (2.8205, 0.1773) ≈ 3.966 so this is a relative and so there is a local minimum at
minimum. (0.6102, 0)
D(−2.8362, −0.1763) ≈ −23.0012, so this is a D(3.1662, 0) ≈ 4.16 × 10−8 > 0
saddle point. fxx (3.1662, 0) ≈ −9 × 10−4 < 0
and so there is a local maximum at
18. fx = 2y − 2x (3.1662, 0)
fy = 2x + 4 + 4y 3 − 18y
Solving f = 0, 0 gives equations 21. (a) The sum of the squares of the residuals is
2y − 2x = 0 given by
n
2x + 4 + 4y 3 − 18y = 0
f (a, b) = (axk + b − yk )2
The first equation gives y = x, substituting k=1
gives the equation The partial with respect to a is
n
2x + 4 + 4x3 − 18x = 0
− 16x + 4 + 4x3 = 0 fb = 2(axk + b − yk )
k=1
We can solve this last equation using a CAS and the partial with respect to b is
n
or graphing calculator to find that x ≈
−2.1149, 0.2541, 1.8608 fa = 2xk (axk + b − yk )
k=1
This gives critical points Setting these equal to 0, dividing by 2,
(−2.1149, −2.1149), (0.2541, 0.2541), and distributing sums, we get equations
n n n
(1.8608, 1.8608)
D(−2.1149, −2.1149) ≈ −75.348 < 0 a xk +b 1 − yk =0
and so there is a saddle at (−2.1149, −2.1149) k=1 k=1 k=1
and
D(0.2541, 0.2541) ≈ 30.4504 > 0 n n
fxx (0.2541, 0.2541) = −2 < 0 a x2
k +b xk
k=1 k=1
and so there is a local maximum at n
(0.2541, 0.2541) − xk yk =0
D(1.8608, 1.8608) ≈ −51.1024 < 0 k=1
and so there is a saddle at (1.8608, 1.8608) as desired.
n
2 3 −x2 −y 2 (b) Note that 1 = n. Multiply the second
19. fx = 2x(1 − x + y )e k=1
2 2
fy = y(2y 3 − 2x2 − 3y)e−x −y equation by n, multiply the first equation
n
Using a CAS to find and analyze the critical by xk , and subtract to get
points, we get: k=1
2
(0, 0) is a saddle point. n n
(±1, 0) are local maxima. a n x2 −
k xk
k=1 k=1
3 n n n
(0, ) is a local minimum. =n xk yk − xk yk
2
k=1 k=1 k=1
3 n n n
(0, − ) is a local maximum. n xk yk − xk yk
√ 2 k=1 k=1 k=1
57 −2 a= 2
(± , ) are both saddle points. n n
9 3 n x2 −
k xk
k=1 k=1
3 2 −x2 −y 2
20. fx = (−2x + 3 + 2x − 6x )e Using this to solve for b, we find that
2 2 n n
fy = −2xy(x − 3)e−x −y 1
Solving f = 0, 0 gives critical points b= yk − a xk
n
k=1 k=1
(−0.7764, 0), (0.6102, 0), = y − ax
(3.1662, 0) where x and y are the means of the x and
D(−0.7764, 0) ≈ 19.20 > 0 y values of the data.
fxx (−0.7764, 0) ≈ −5.984 < 0
and so there is a local maximum at 22. Let x be the number of years since 1970. Then
(−0.7764, 0) the sum of the squares of the residuals is given
D(0.6102, 0) ≈ 9.817 > 0 by


f (a, b) = (b − 0.34)2 + (5a + b − 0.59)2 f (a, b) = (15a + b − 4.57)2 + (35a + b − 3.17)2
+ (10a + b − 1.23)2 + (55a + b − 1.54)2
+ (15a + b − 1.11)2 + (75a + b − 0.24)2
fa = 700a + 60b − 63.8 + (95a + b + 1.25)2
fb = 60a + 8b − 6.54 Solving fa = fb = 0 gives us
Solving fa = fb = 0 gives us a ≈ −0.07285, b ≈ 5.66,
a ≈ 0.059, b ≈ 0.375, so the linear model is y = −0.07285x + 5.66
so the linear model is y = 0.059x + 0.375 This model predicts the average number of
This model predicts the gas prices in 1990 and points scored starting from the 60 yard line
1995 as will be y(60) = 1.29, and the average number
y(20) = $1.56 of points scored starting from the 40 yard line
y(25) = $1.85 will be y(40) = 2.75.
The forecasts are not accurate because they as- 25. (x0 , y0 ) = (0, −1)
sume that the prices will increase indefinitely. f = 2y − 4x, 2x + 3y 2
23. (a) The sum of the squares of the residuals is f (0, −1) = −2, 3
given by g(h) = f (0 − 2h, −1 + 3h)
f (a, b) = (68a+b−160)2 +(70a+b−172)2 g (h) = −2fx (−2h, −1 + 3h)
+ (70a + b − 184)2 + 3fy (−2h, −1 + 3h)
+ (71a + b − 180)2 = −2 [2(−1 + 3h)) − 4(−2h)]
fa = 38930a + 558b − 97160 + 3[2(−2h) + 3(−1 + 3h)2 ]
fb = 558a + 8b − 1392 = 13 − 94h + 81h2
The smallest positive solution to g (h) = 0 is
Solving fa = fb = 0 gives us
h ≈ 0.16049. This leads us to
a ≈ 7.16, b ≈ −325.26,
(x1 , y1 ) = (0, −1) + 0.16049(−2, 3)
so the linear model is y = 7.16x − 325.26
= (−0.32098, −0.51853)
This model predicts the weight of a 6 8” f (−0.32098, −0.51853) = 0.24686, 0.16466
person will be y(80) = 248 pounds, g(h) = f (−0.32098 + 0.24686h,
and the weight of a 5 0” person will be − 0.51853 + 0.16466h)
y(60) = 104 pounds. g (h) = 0.24686 · fx (−0.32098 + 0.24686h,
There are many other factors besides − 0.51853 + 0.16466h)
height that influence a persons weight. + 0.16466 · fy (−0.32098 + 0.24686h,
(b) The sum of the squares of the residuals is − 0.51853 + 0.16466h)
given by = 0.08805 − 0.16552h + 0.01339h2
2
f (a, b) = (68a + b − 160) Solving g (h) = 0 gives
2
+ (70a + b − 172) h ≈ 11.80144, 0.55709.
2 Using the first positive value means we arrive
+ (70a + b − 184)
+ (70a + b − 221)
2 at
+ (71a + b − 180)
2 (x2 , y2 ) = (−0.32098, −0.51853)
fa = 48730a + 698b − 128100 + 0.55709 0.24686, 0.16466
fb = 698a + 10b − 1834 ≈ (−0.18346, −0.42680).

26. f = 3y − 3x2 , 3x − 2y
Solving fa = fb = 0 gives us f (1, 1) = 0, 1
a ≈ 9.0417, b ≈ −447.71. g(h) = f (1 + 0 · h, 1 + 1 · h) = f (1, 1 + h)
So linear model is y = 9.0417x − 447.71. g (h) = 0 · fx (1, 1 + h) + 1 · fy (1, 1 + h)
This model predict the weight of 6’8” per- = 1 − 2h
son will be y (80) = 275.626 pounds, and 1
Solving g (h) = 0 gives h = and we arrive at
the weight of 5’0” person will be 2
1 3
y (60) = 94.792. (x1 , y1 ) = (1, 1) + 0, 1 = 1,
The additional point dramatically 2 2
3 3
changed the linear model! f 1, = ,0
2 2
24. The sum of the squares of the residuals is given 3 3
by g(h) = f 1 + · h, + 0 · h
2 2


3 3 29. Refer to exercise 25:
=f 1 + h,
2 2 f (0, 0) = (0, 0) g(h) = f (0 + 0h, 0 + 0h) = 0,
3 3 3 3 3 g (h) = 0, and there is no smallest positive so-
g (h) = ·fx 1 + h, +0·fy 1 + h, lution to g (h) = 0. Graphically, we started at
2 2 2 2 2
a point where the tangent plane was horizon-
81 2 27 9 tal (a saddle point in this case), so the gradient
=− h − h+
8 2 4 didn’t tell us which direction to move!
Solving g (h) = 0 gives h ≈ −1.4832, 0.1498.
Using the ﬁrst positive value means we arrive 30. Instead of using f , use − f which gives the
at direction of steepest descent. Then, apply the
3 3
(x2 , y2 ) = 1, + (0.1498) ,0 steepest ascent algorithm for ﬁnding local max-
2 2
ima.
≈ (1.2247, 1.5)
31. In the interior:
27. f = 1 − 2xy 4 , −4x2 y 3 + 2y fx = 2x − 3y fy = 3 − 3x
f (1, 1) = −1, −2 Solving fy = 0 gives x = 1. Then fx = 0 gives
g(h) = f (1 − h, 1 − 2h) 2 2
g (h) = −1 · fx (1 − h, 1 − 2h) y = , and we have one critical point 1, .
3 3
− 2 · fy (1 − h, 1 − 2h) Along y = x:
= 5 − 74h + 264h2 − 416h3 + 320h4 − 96h5 f (x, x) = g(x) = x2 + 3x − 3x2 = 3x − 2x2
The smallest positive solution to g (h) = 0 is 4
approximately h = 0.09563, and we arrive at g (x) = 3 − 4x = 0 at x = .
3
(x1 , y1 ) = (1, 1) + 0.09563 −1, −2 4 4
= (0.90437, 0.80874) This gives a local maximum at ,
3 3
f (0.90437, 0.80874) Along y = 0:
= 0.22623, −0.11305 f (x, 0) = x2 , which has a minimum at (0, 0).
g(h) = f (0.90437 + 0.22623h, Along x = 2:
0.80874 − 0.11305h) f (2, y) = h(y) = 4 + 3y − 6y = 4 − 3y.
g (h) = 0.22623 · fx (0.90437 + 0.22623h, This will give a minimum and maximum value
0.80874 − 0.11305h) at the intersection points along the boundary.
− 0.11305 · fy (0.90437 + 0.22623h, The intersection points of the boundaries are
0.80874 − 0.11305h) (0, 0), (2, 2), and (2, 0). The function values at
= 0.06396 + 0.09549h − 0.01337h the points of interest are:
− 0.00315h3 + 0.00086h4 f (0, 0) = 0
− 0.00005h5 f (2, 0) = 4
The smallest positive solution to g (h) = 0 is f (2, 2) = −2
approximately h = 10.56164, and we arrive at 2
(x2 , y2 ) = (0.90437, 0.80874) f 1, =1
3
+ 10.56164 0.22623, −0.11305 4 4 4
≈ (3.29373, −0.38525) f , =
3 3 9
The absolute maximum is 4 and the absolute
28. f = y 2 − 2x, 2xy − 1 minimum is −2.
f (1, 0) = −2, −1
g(h) = f (1 − 2 · h, 0 − 1 · h) = f (1 − 2h, −h) 32. In the interior:
g (h) = −2 · fx (1 − 2h, −h) − 1 · fy (1 − 2h, −h) fx = 2x − 4y fy = 2y − 4x
= −6h2 − 6h + 5 Solving fx = 0, fy = 0 gives equations
Solving g (h) = 0 gives h ≈ 0.5408 and we ar- 2x = 4y and 2y = 4x
rive at Solving these equations gives critical point
(x1 , y1 ) = (1, 0) + (0.5408) −2, −1 (0, 0).
= (−0.0817, −0.5408) Along y = x:
f (−0.0817, −0.5408) = 0.4559, −0.9117 f (x, x) = g(x) = 2x2 − 4x2 = −2x2
g(h) = f (−0.0817 − 0.4559h, which has a maximum at (0, 0).
− 0.5408 − 0.9117h) Along x = 3:
g (h) = 0.03889 + 0.3476h + 1.1366h2 f (3, y) = h(y) = 9 + y 2 − 12y
Solving g (h) = 0 has no real solutions. h (y) = 2y − 12


Solving h (y) = 0 gives (3, 6), which is not in Finding functional values at all these points:
the region. f (0, 0) = 0
Along y = −3: f (0, 3) = −3
f (x, −3) = k(x) = x2 + 9 + 12x f (3, 3) = 0
k (x) = 2x + 12. f (1, 2) = −5
Solving k (x) = 0 gives (−6, −3), which is not 3 3 9
f , =−
in the region. The intersection points of the 2 2 2
boundaries are f (1, 3) = −4
(−3, −3), (3, −3), (3, 3). f (0, 2) = −4
Finding functional values at all these points: Therefore the absolute maximum is 0 and the
f (0, 0) = 0 absolute minimum is −5.
f (−3, −3) = −18
f (3, −3) = 54 35. In the interior :
f (3, 3) = −18 x2 y2
− −
Therefore the absolute maximum is 54 and the fx = ye 2 2 1 − x2
absolute minimum is −18. x 2
y2
− −
fy = xe 2 2 1 − y2
fx = 2x fy = 2y Solving fx = 0 and fy = 0 gives critical points
Solving fx = 0, fy = 0 gives critical point (0,0), (1,-1),(-1,1),(-1,-1)and (1,1).
(0, 0). Along y = 0: f (x, 0) = 0
On the circle (x − 1)2 + y 2 = 4, we substitute This has no critical point.
y 2 = 4 − (x − 1)2 to get
f (x, y) = g(x) = x2 + 4 − (x − 1)2 = 3 + 2x. Along y = 2 :
This has no critical points, but is maximized x2
− −2
for the largest value of x and minimized for the f (x, 2) = h (x) = 2xe 2
smallest value of x. The point with the largest x2
value of x on the circle is (3, 0). The point with − −2
h (x) = 2e 2 1 − x2
the smallest value of x on the circle is (−1, 0).
Finding functional values at all these points:
f (0, 0) = 0 Solving for h (x) =0 give points (1,2) and
f (3, 0) = 9 (-1,2) but (-1,2) is not in the region.
f (−1, 0) = 1
Along x = 0 : f (0, y) = 0 This has no critical
Therefore the absolute maximum is 9 and the
point.
absolute minimum is 0.
Along x = 2 :
y2
fx = 2x − 2 fy = 2y − 4 −2−
Solving fx = 0, fy = 0 gives gives critical point f (2, y) = k (y) = 2ye 2
(1, 2). y2
−2−
Along y = x: k (y) = 2e 2 1 − y2
f (x, x) = g(x) = 2x2 − 6x
g (x) = 4x − 6 Solving for k (y)=0 give points (2,1) and (2,-1)
3 3 but (2,-1) is not in the region.
which gives us critical point , .
2 2
Along y = 3: The intersection points at the boundaries are
f (x, 3) = h(x) = x2 + 9 − 2x − 12 (0,0), (0,2), (2,2) and (2,0). Finding functional
h (x) = 2x − 2 values at all these points:
Solving k (x) = 0 gives the point (1, 3). f (0, 0) = 0
Along x = 0: f (1, −1) = −0.3679
f (0, y) = k(y) = y 2 − 4y f (−1, 1) = −0.3679
k (y) = 2y − 4 f (−1, −1) = 0.3679
Solving k (y) = 0 gives the point (0, 2). f (1, 1) = 0.3679
The intersection points of the boundaries are f (1, 2) = 0.1642
(0, 0), (0, 3), (3, 3) f (2, 1) = 0.1642


f (2, 2) = 0.0733 B = s(s − a)(s − b)(a + b − s)
f (0, 2) = 0 Treating s as a constant, we see that
f (2, 0) = 0 Ba = −s(s − b)(a + b − s) + s(s − a)(s − b)
Bb = −s(s − a)(a + b − s) + s(s − a)(s − b)
Therefore the absolute maximum is 0.3679 and Subtracting Bb = 0 from Ba = 0, we get
the absolute minimum is -0.3679. s(s − a)(a + b − s) − s(s − b)(a + b − s) = 0.
Note that the semi-perimeter s = 0, and that
36. In the interior: if a + b − s = 0 then c = 0, giving the triangle
2x 0 area. Therefore s − a = s − b, so a = b.
fx = 2 Substituting b = a into Ba yields
(x + y 2 + 1)
2y −s(s − a)(2a − s) + s(s − a)2 and this is 0 when
fy = 2 −y 2
(x + y 2 + 1) a = s or a = s.
3
Solving fx = 0 and fy = 0 gives critical points If a = b = s, then c = 0 and the area is 0.
2 2
(0,0), (0,1) and (0,-1). If a = b = s, then c = s, as well, and we
3 3
Along x2 + y 2 = 4: see that an equilateral triangle gives the max-
imum area for a fixed perimeter. (That this is
We substitute y 2 = 4 − x2 gives
indeed a maximum can be verified using The-
4 − x2 orem 7.2.)
h (x) = ln (5) −
2
h (x) = x. 38. The maximum of x2 + y 2 occurs at the points
of maximum distance from the origin. For the
square, this occurs at (1, 1), (−1, 1), (−1, −1),
Solving for h (x) =0 give points (0,2) and
and (1, −1). A computer graph over the square
(0,-2) shows peaks at these corner points instead of
We substitute x2 = 4 − y 2 gives a circular level curve.
y2 39. For the function f (x, y) = x2 y 2
k (y) = ln (5) −
2 fx = 2xy 2 fy = 2x2 y
k (y) = −y. fx = fy = 0 whenever x = 0 or y = 0, so we
have critical points at (x, 0) and (0, y) for all x
Solving for k (y)=0 give points (2,0) and (-2,0) and y.
So the critical points are (0,2) (0,-2), (2,0) and fxx = 2y 2
(-2,0). fxy = 4xy
fyy = 2x2
Finding functional values at all these points: D(x, y) = 4x2 y 2 − 16x2 y 2
f (0, 0) = 0 D(0, y) = D(x, 0) = 0 for all critical points, so
f (0, 1) = 0.1931 Theorem 7.2 fails to identify them.
f (0, −1) = 0.193 f (x, 0) = f (0, y) = 0 for all x and y.
f (0, 2) = −0.3906 If x and y are both not zero, then f (x, y) > 0.
f (0, −2) = −0.3906 This means that all the critical points are min-
f (2, 0) = 1.6094 ima.
f (−2, 0) = 1.6094 For the function
f (x, y) = x2/3 y 2
Therefore the absolute maximum is 1.6094 and 2
fx = x−1/3 y 2 fy = 2x2/3 y
the absolute minimum is −0.3906 3
fx = fy = 0 whenever y = 0,
37. We first simplify the calculations by noting so we have critical points at (x, 0) for all x.
that we may maximize B = A2 instead of A When x = 0, fx is undefined and fy = 0, so we
(since x2 is an increasing function for positive have critical points at (0, y) for all y.
x). −2 −4/3 2
1 fxx = x y
We solve s = (a + b + c) for c, and substitute 9
2 4
into B to find: fxy = x−1/3 y
3
c = 2s − a − b, and fyy = 2x2/3


−4 −2/3 2 16 −2/3 2 (b) Substituting y = kx gives
D(x, y) = x y − x y
9 9 g(x) = f (x, kx) = x2 − 3k 2 x3 + 4kx4
D(x, 0) = 0 for all critical points (x, 0) with
g (x) = 2x − 9k 2 x2 + 16kx3
x = 0, so Theorem 7.2 fails to identify them.
g (x) = 2 − 18k 2 x + 48kx2
Theorem 7.2 also fails to identify critical points
Since g (0) = 2 > 0, all traces have a
(0, y), since the second partial derivatives are
local minimum at the origin.
not continuous there.
f (x, 0) = f (0, y) = 0 for all x and y. 44. (a) fx = z − 1, fy = 3y 2 − 3, fz = x
If x and y are both not zero, then f (x, y) > 0. These all equal zero at (0, 1, 1), so this is
a critical point.
This means that all the critical points are min- f (∆x, 1 + ∆y, 1 + ∆z)
ima. = ∆x(1 + ∆z) − ∆x
+ (1 + ∆y)3 − 3(1 + ∆y)
40. (a) f (x, y) = (x + 1)2 + (y − 2)2 − 4 = ∆x∆z + 3∆y 2 + ∆y 3 − 2
f has a global (and local) minimum of −4 = ∆x∆z + 3∆y 2 + ∆y 3 + f (0, 1, 1)
at (−1, 2). So with ∆y = 0, as we move away from
(b) f (x, y) = (x2 − 3)2 + (y 2 + 1)2 − 11 the critical point with ∆x∆z > 0, f in-
f has global (and √
√ local) minimum of −10 creases, while with ∆x∆z < 0, f de-
at ( 3, 0) and (− 3, 0). creases. The critical point is therefore nei-
ther a local maximum or local minimum.
41. We substitute y = kx into z = x3 − 3xy + y 3 (b) fx = z − 1, fy = 3y 2 − 3, fz = x
and get z = x3 − 3x(kx) + (kx)3 These all equal zero at (0, −1, 1), so this
= (1 + k 3 )x3 − 3kx2 . is a critical point.
If we set f (x) = (1 + k 3 )x3 − 3kx2 , we find f (∆x, −1 + ∆y, 1 + ∆z)
f (x) = 3(1 + k 3 )x2 − 6kx = 0 when x = 0, so = ∆x∆z − 3∆y 2 + ∆y 3 + 2
this is a critical point. Then = ∆x∆z − 3∆y 2 + ∆y 3 + f (0, −1, 1).
f (x) = 6(1 + k 3 )x − 6k, so f (0) = −6k. So with ∆y = 0, as we move away from
The Second Derivative Test then shows f (x) the critical point with ∆x∆z > 0, f in-
has a local maximum if k > 0 and a local min- creases, while with ∆x∆z < 0, f de-
imum if k < 0. creases. The critical point is therefore nei-
ther a local maximum or local minimum.
42. We substitute y = kx into
z = 4xy − x4 − y 4 + 4 and get 45. False. The partial derivatives could be 0 or
f (x) = 4x(kx) − x4 − (kx)4 + 4 undefined.
= 4kx2 − (1 + k 4 )x4 + 4.
Then, f (x) = 8kx − 4(1 + k 4 )x3 , 46. False, we don’t know the type of extremum or
so f (x) has a critical point at x = 0. if the point is a saddle point.
f (x) = 8k − 12(1 + k 4 )x2 , 47. False. There do not have to be any local min-
so f (0) = 8k. ima.
The Second Derivative Test tells us that f (x)
has a minimum when k > 0 and a maximum 48. False, the number of extrema does not affect
when k < 0. the possible types of extrema.

43. (a) We substitute y = kx into 49. The extrema occur in the centers of the four
z = x3 − 2y 2 − 2y 4 + 3x2 y to get circles. The saddle points occur in the nine
f (x) = x3 − 2(kx)2 − 2(kx)4 + 3x2 (kx) crosses between the circles.
= (1 + 3k)x3 − 2k 2 x2 − 2k 4 x4 50. Extrema at approximately (-1,-1) and (1, 1).
f (x) = 3(1 + 3k)x2 − 4k 2 x − 8k 4 x3 , so Saddle at approximately (0,0).
f (x) has a critical point at x = 0.
f (x) = 6(1 + 3k)x − 4k 2 − 24k 4 x2 , so 51. fx = 5ey − 5x4 , fy = 5xey − 5e5y
f (0) = −4k 2 . fxx = −20x , fyy = 5xey − 25e5y
3

The Second Derivative Test shows that fxy = 0
f (x) has a local maximum for all k = 0. Solving fx = 0 gives ey = x4 . Substitute this
When k = 0 the graph looks like x3 , which into fy = 0 to see that 5x5 − 5x20 = 0. The
has an inflection point at x = 0. solution x = 0 is extraneous, leaving us with


solution x = 1 and y = 0. 54. (a) The closest point on the sphere will be
D(1, 0) = 400 > 0 below the xy plane. The distance from a
fxx (1, 0) = −20 < 0 point (x, y, − 9 − x2 − y 2 ) to the point
Therefore f (1, 0) = 3 is a local maximum. (2, 1, −3) squared is
Since f (−5, 0) = 3099, (for example) it is clear g(x, y) = (x − 2)2 + (y − 1)2
that (1, 0) is not a global maximum. + (− 9 − x2 − y 2 + 3)2
= −4x − 2y − 6 9 − x2 − y 2 + 23
52. fx = 8x3 − 8xey , fy = 4e4y − 4x2 ey 6x
fxx = 24x − 8e , fyy = 16e4y − 4x2 ey
2 y
gx = −4 +
fxy = −8xey 9 − x2 − y 2
6y
Solving fx = fy = 0 gives the points (1, 0) and gy = −2 +
(−1, 0). 9 − x2 − y 2
D(1, 0) = 128 > 0 54 − 6y 2
gxx =
fxx (1, 0) = 16 > 0 (9 − x2 − y 2 )3/2
D(−1, 0) = 128 > 0 54 − 6x2
gyy =
fxx (−1, 0) = 16 > 0 (9 − x2 − y 2 )3/2
Therefore, both these points are local minima. 6xy
gxy =
(9 − x2 − y 2 )3/2
53. (a) The distance from a point Solving gx = gy = 0 numerically yields
(x, y, 4 − x2 − y 2 ) to the point (3, −2, 1) (1.6, 0.8).
is D(1.6, 0.8) ≈ 9.4 and
d(x, y)
gxx (1.6, 0.8) ≈ 3.6, therefore this point
= (x − 3)2 + (y + 2)2 + (3 − x2 − y 2 )2 is a minimum. The closest point on the
To minimize this it is useful to note that sphere to the point (2, 1, −3) is approxi-
we can minimize g(x, y) = d(x, y)2 in- mately (1.6, 0.8, −2.4).
stead.
gx = 2(x − 3) − 4x(3 − x2 − y 2 ) 1
(b) d(x, y) = x2 + y 2 + (12 − 3x + 4y)2
gy = 2(y + 2) − 4y(3 − x2 − y 2 ) 9
gxx = −10 + 12x2 + 4y 2 1
g(x, y) = x + y + (12 − 3x + 4y)2
2 2

gyy = −10 + 12y 2 + 4x2 9
2
gxy = 8xy gx = 2x + (−3)(12 − 3x + 4y)
9
Solving gx = gy = 0 numerically yields 2
(1.55, −1.03). gy = 2y + (4)(12 − 3x + 4y)
9
D(1.55, −1.03) ≈ 297.5 and 50 8
gxx = 4, gyy = , gxy = −
gxx (1.55, −1.03) ≈ 23.1, therefore this 9 3
point is a minimum. The closest point on Solving gx = gy = 0 gives the point
the paraboloid to the point (3, −2, 1) is 18 24
,−
approximately (1.55, −1.03, 0.54). 17 17
18 24 136
(b) d(x, y) = (x − 2)2 + (y + 3)2 + x2 + y 2 D ,− = >0
17 17 9
Minimize: 18 24
g(x, y) = (x − 2)2 + (y + 3)2 + x2 + y 2 fxx ,− =4>0
17 17
gx = 2(x − 2) + 2x Therefore, the closest point is
gy = 2(y + 3) + 2y 18 24 72
gxx = 4, gyy = 4, gxy = 0 ,− , .
17 17 17
Solving gx = gy = 0 gives the point
3 55. Let x, y, z be the dimensions. In this case, the
1, − . amount of material used would be
2
3 2(xy + xz + yz) = 96
D 1, − = 16 > 0 Solving for z we get
2
48 − xy
3 z=
fxx 1, − =4>0 x+y
2 This gives volume
Therefore, the closest point is
√ V (x, y) = xyz
3 13 (48 − xy)xy 48xy − x2 y 2
1, − , . = =
2 2 x+y x+y


(48y − 2xy 2 )(x + y) − (48xy − x2 y 2 )] 32 − x2
Vx = Solving 32 − x2 − 2xy for y gives y = .
(x + y)2 2x
2
y (48 − 2xy − x2 )
2 Substituting this into 32 − y − 2xy = 0 gives
= 32 − x2
2
32 − x2
(x + y)2 0 = 32 − − 2x
(48x − 2x2 y)(x + y) − (48xy − x2 y 2 ) 2x 2x
Vy = 3x4 + 64x2 − 1024
(x + y)2 =
x2 (48 − 2xy − y 2 ) 4x2
= The only positive solution to this equation is
(x + y)2 4√
Solving Vx = 0 and Vy = 0 gives equations x= 6
3
y 2 (48 − 2xy − x2 ) = 0 Thus, our critical point is
x2 (48 − 2xy − y 2 ) = 0 4√ 4√ √
Since maximum volume can not occur when 6, 6, 2 6
3 3
x or y is 0, we assume both x and y are A quick check of the discriminant assures that
nonzero. Solving 48 − 2xy − x2 = 0 for y gives this gives the maximum volume of
48 − x2 4√ 4√ √ 64 √
y= . Substituting this into V 6, 6, 2 6 = 6
2x 3 3 3
2
48 − y − 2xy = 0 gives
2
48 − x2 48 − x2 57. If P is the total population we must have
0 = 48 − − 2x pP + qP + rP = P or
2x 2x
3x4 + 96x2 − 2304 p+q+r =1
= g(p, q) = f (p, q, 1 − p − q)
4x2
The only positive solution to this equation is = 2p − 2p2 + 2q − 2q 2 − 2pq
x=4 gp = 2 − 4p − 2q, gq = 2 − 4q − 2p
1 1
Thus, our critical point is (4, 4, 4) Solving gp = gq = 0 gives p = and q =
3 3
A quick check of the discriminant assures that 1
and therefore r = . Note that g(p, q) deﬁnes
this gives the maximum volume of V (4, 4, 4) = 3
a downward opening elliptic paraboloid, so this
64
is an absolute maximum.
1 1 1 2
56. Let x, y, z be the dimensions. In this case, the f , , = .
amount of material used would be 3 3 3 3
2(xy + xz + yz) + xy = 96 58. As in example 7.4, we take the equation to be
Solving for z we get y = ax2 + bx + c ,with constants a, b, c to be
3(32 − xy)
z= determined. For the given data residuals are
2(x + y) shown in the following table.
This gives volume as
3(32 − xy)xy x ax2 + bx + c y Residual
V (x, y) = xyz =
2(x + y) 0 c 179 c-179
3(32xy − x2 y 2 ) 1 a+b+c 203 a + b + c-203
= 2 4a + 2b + c 227 4a + 2b + c − 227
2(x + y)
3[(32y − 2xy 2 )(x + y) − (32xy − x2 y 2 )] 3 9a + 3b + c 249 9a + 3b + c − 249
Vx =
2(x + y)2 The sum of the squares of the residuals is then
3y 2 (32 − x2 − 2xy) given by
=
2(x + y)2
f (a, b, c) = (c − 179)2 + (a + b + c − 203)2
3[(32x − 2yx2 )(x + y) − (32xy − x2 y 2 )]
Vy = + (4a + 2b + c − 227)2
2(x + y)2 + (9a + 3b + c − 249)2
2 2
3x (32 − y − 2xy)
= df
2(x + y)2 = 196a + 72b + 28c − 6704
da
Solving Vx = 0 and Vy = 0 gives equations df
= 72a + 28b + 12c − 2808
3y 2 (32 − x2 − 2xy) = 0 db
3x2 (32 − y 2 − 2xy) = 0 df
= 28a + 12b + 8c − 1716
Since maximum volume can not occur when dc
x = y = 0, we assume both x and y are df df df
nonzero. Solving for = = =0
da db dc


We get a = −0.5, b = 24.9, c − 178.9 Substituting this into the constraint
The curve is given by y = 2x + 1
− x = 2x + 1
2
y = −0.5x2 + 24.9x + 178.9 2
x = −5, y=1 5
For the above values of a, b and c we get the
minimum function value. f (a, b, c)=0.2 and 3. f (x, y) = (x − 4)2 + y 2
g(x, y) = 2x + y − 3
From 7.4, f (a, b)=1.2 . Therefore, the sum of
f = 2(x − 4), 2y
squares of the residuals for quadratic model is
g = 2, 1 f =λ g
less than that for the linear model.
2(x − 4) = 2λ, 2y = λ
500 Eliminating λ we get y = x − 2
2
450 Substituting this into the constraint
400 y = 3 − 2x
x
350 2 − 2 = 3 − 2x
x = 2, y = −1
300

4. f (x, y) = x2 + (y − 2)2
250
g(x, y) = x − 2 − y
200
f = 2x, 2(y − 2)
g = 1, −1 f =λ g
0 5 10 15 20 25 30 35
x
2x = λ, 2(y − 2) = −λ
Eliminating λ we get y = 2 − x
From the graph, it can be seen that the curve
deviates faster for larger values of x and hence, Substituting this into the constraint
the linear model is more preferable than the y =x−2
quadratic model. This can also be observed 2−x=x−2
from the following table. For 100 years, x=10 x = 2, y=0

Let us ﬁnd Population in millions 5. f (x, y) = (x − 3)2 + y 2
Future Past g(x, y) = x2 − y
Using Quadratic model 377.4 58.9 f = 2(x − 3), 2y
Using Linear model 413.4 124.6 g = 2x, −1 f =λ g
2(x − 3) = 2xλ, 2y = −λ
Eliminating λ gives x − 3 = −2xy
12.8 Constrained Optimiza- Applying the constraint
tion and Lagrange Mul- y = x2
2x3 + x − 3 = 0
tipliers x = 1, y=1
1. f (x, y) = x2 + y 2 6. f (x, y) = x2 + (y − 2)2
g(x, y) = 3x − 4 − y g(x, y) = x2 − y
f = 2x, 2y f = 2x, 2(y − 2)
g = 3, −1 f =λ g g = 2x, −1 f =λ g
2x = 3λ, 2y = −λ 2x = 2xλ, 2(y − 2) = −λ
Eliminating λ we get y = − x
3
3
Therefore λ = 1 and y = 2 .
Substituting this into the constraint Substituting this into the constraint
y = 3x − 4 y = x2
− x = 3x − 4 3
3
6
x = 5, y = −2 x2 =
5 2
3 3
2. f (x, y) = x2 + y 2 x=± , y=
2 2
g(x, y) = 2x + 1 − y
f = 2x, 2y 7. f (x, y) = (x − 2)2 + (y − 1 )2
2
g = 2, −1 f =λ g g(x, y) = x2 − y
1
2x = 2λ, 2y = −λ f = 2(x − 2), 2(y − 2 )
Eliminating λ we get y = − x
2 g = 2x, −1 f =λ g

12.8. CONSTRAINED OPTIMIZATION AND LAGRANGE MULTIPLIERS 717

2(x − 2) = 2xλ, 2(y − 1 ) = −λ
2 4x2 + 4x2 − 8 = 0
1
Eliminating λ we get y = x . x2 = 1
Substituting this into the constraint x = ±1
y = x2 This gives the points
1 (1, 2), (1, −2), (−1, 2), (−1, −2).
= x2
x f (1, 2) = 8, maximum
x3 = 1
f (−1, −2) = 8, maximum
x = 1, y=1 f (1, −2) = −8, minimum
f (−1, 2) = −8, minimum
8. f (x, y) = (x − 1)2 + (y − 2)2
g(x, y) = x2 − 1 − y 11. Vertices of the triangle are
f = 2(x − 1), 2(y − 2) (0, 0, ) , (2, 0) and (0, 4).
g = 2x, −1 f =λ g
Equation of the sides are given by
2(x − 1) = 2xλ, 2(y − 2) = −λ
Eliminating λ we get y = 3x+1 . x = 0, y = 0 and2x + y = 4
2x
Substituting this into the constraint Consider the side 2x + y = 4, then
y = x2 − 1 g (x, y) = 2x + y − 4 = 0
3x+1 2
2x = x − 1
f = 8xy, 4x2
g = 2, 1
2x3 − 5x − 1 = 0
f =λ g
Solving this numerically we get 8xy = 2λ, 4x2 = λ
x ≈ −1.469, −0.203, 1.67. 4
Eliminating λ we have x = 0 or x =
Since the point is to be closest to (1, 2), 3
x ≈ −1.67, y ≈ 1.80 Substituting this value of x in the constraint
4
9. g(x, y) = x2 + y 2 − 8 = 0 we have y = 4 or y =
3
f = 4y, 4x 4 4
g = 2x, 2y This gives points (0, 4) , ,
3 3
f =λ g Consider the side x = 0, then g (x, y) = x = 0
4y = 2xλ f = 8xy, 4x2
4x = 2yλ g = 1, 0
Eliminating λ we get y = ±x. f =λ g
Substituting this into the constraint, 8xy = λ, 4x2 = 0
x2 + y 2 − 8 = 0 The above equations have no solution.
x2 + x2 − 8 = 0 Consider the side y = 0, then
x2 = 4
g (x, y) = y = 0
x = ±2
f = 8xy, 4x2
This gives the points g = 0, 1
(2, 2), (2, −2), (−2, 2), (−2, −2). f =λ g
f (2, 2) = 16, maximum 8xy = 0, 4x2 = λ
f (−2, −2) = 16, maximum This gives points (x, 0) for 0 ≤ x ≤ 2
f (2, −2) = −16, minimum 4 4 256
f , = ,maxima
f (−2, 2) = −16, minimum 3 3 27
10. g(x, y) = 4x2 + y 2 − 8 = 0 f (x, 0) = 0, minima
f = 4y, 4x f (0, 4) = 0 , minima
g = 8x, 2y 12. Vertices of the Rectangle are
f =λ g (−2, 1) , (1, 1) , (1, 3) and (−2, 3).
4y = 8xλ
Equation of the sides are given by
4x = 2yλ
Eliminating λ we get y = ±2x. x = −2 , y = 1 , x = 1 and y = 3
Substituting this into the constraint, Equations of the sides are of the form:
4x2 + y 2 − 8 = 0 y − k = 0, x − p = 0


where k = −2, 1 and p = 1, 3 e2x+y = 2yλ
1
First consider g(x, y) = y − k Eliminating λ we get y = x.
2
∆f = (6x2 y, 2x3 ) Substituting this into the constraint,
∆g = (0, 1) x2 + y 2 − 5 = 0
Now ∆f = λ∆g gives x2
x2 + −5=0
4
6x2 y, 2x3 = λ (0, 1) x = ±2
⇒x=0 This gives the points (2, 1), (−2, −1)
⇒ λ = 0, y = k
f (2, 1) = e5 , maximum
The points are (0, 1), (0, 3).
f (−2, −1) = e−5 , minimum
Consider g(x, y) = x − p
∆g = (1, 0) 15. g(x, y) = x2 + y 2 − 3 = 0
f = 2xey , x2 ey
Now ∆f = λ∆g gives g = 2x, 2y
(6x2 y, 2x3 ) = λ(1, 0) ⇒ x = 0
f =λ g
But x = 0 is not in g(x, y) = x − p = 0. 2xey = 2xλ
Therefore there is no maximum or minimum at x2 ey = 2yλ
the sides (open line segment) x = −2, x = 1. Eliminating λ we get either
Hence function can be maxima or minima at 1
x = 0 or y = x2 .
(0, 1) or (0, 3), or corner points. Since , 2
√
f (0, 1) = f (0, 3) = 0 If x = 0 the constraint gives y = ± 3.
1
f (−2, 1) = −16 If y = x2 , the constraint gives:
2
f (−2, 3) = −48 x2 + y 2 − 3 = 0
4
f (1, 1) = 2 x2 + x − 3 = 0
4
f (1, 3) = 6 (x2 + 6)(x2 − 2) = 0
√
x=± 2
Therefore Maximum value is 6 at (1, 3) and
minimum value is −48 at (−2, 3). This gives the points
√ √ √
(± 2, 1), (± 2, −1), and (0, ± 3).
13. g (x, y) = 4x2 + y 2 − 4 = 0 √
f (± 2, 1) = 2e, maximum
f = ey , xey √ 2
g = 8x, 2y f (± 2, −1) = , neither
√ e
f =λ g f (0, ± 3) = 0, minimum.
ey = λ8x; xey = λ2y
Eliminating λ we have, 16. g(x, y) = x2 + 4y 2 − 24 = 0
√
y f = 2xy 2 , 2x2 y
y = 4x2 or x = ± g = 2x, 8y
2
Substituting this value of y in the constraint f =λ g
we have, y 2 + y − 4 = 0 gives 2xy 2 = 2xλ
y = 1.5615 , x = ±0.6248 2x2 y = 8yλ
1
This gives points Eliminating λ we get y = ± x.
2
(0.6248, 1.5615),(−0.6248, 1.5615) Substituting this into the constraint,
f (0.3124, 1.5615) = 2.97798 , maxima x2 + 4y 2 − 24 = 0
x2 + x2√ 24 = 0
−
f (−0.3124, 1.5615) = −2.97798 , minima
x = ±2 3
14. g(x, y) = x2 + y 2 − 5 = 0 This gives the points
√ √ √ √ √ √
f = 2e2x+y , e2x+y (2 √ 3), (−2 3, 3), (2 3, − 3),
3, √
g = 2x, 2y (−2 3, − 3)
f =λ g Another set of solutions is when λ = 0.
2e2x+y = 2xλ This gives the points

√ √
(0, 6), (0, − √6)
√ f (0, 0, ±1) = 1, minima
(2 6, 0), (−2 6, 0) f (0, ±1, 0) = 1, minima
√ √ 1 1 √
f (±2 √ ± 3) = 36, maxima
3, f 0, ± 1 , ± 1 = 2
f (0, ± 6) = 0, minima
√ 24 24
f (±2 6, 0) = 0, minima f (±1, 0, 0) = 1, minima
2 1 √
17. g (x, y, z) = x4 + y 4 + z 4 − 1 = 0 f ± 1 , 0, ± 1 = 17
(17) 4 (17) 4
f = 8x, 2y, 2z √
2 1
g = 4x3 , 4y 3 , 4z 3 f ± 1 ,± 1 ,0 = 17
f =λ g (17) 4 (17) 4
8x = λ4x3 ; 2y = λ4y 3 ; 2z = λ4z 3 .....(eq.1)
For x, y, z = 0 18. g (x, y, z) = x2 − y 2 + z 2 − 1 = 0
Eliminating λ we get f = 1, −1, −1
g = 2x, −2y, 2z
x2 = 4z 2 , y 2 = z 2 gives
f =λ g
x= ± 2z , y = ±z
1 = λ2x; −1 = λ (−2y) ; −1 = λ (2z)
Substituting the values of x and y in the con-
1 Eliminating λ gives
1 4 x = y = −z
straint, we have z = ±
18 Substituting the values of x and y in the con-
This gives points straint, we have z 2 = 1 and y = −z, x = y
2 1 1 This gives points (1,1,-1), (-1,-1,1)
± ,± ,±
(18)
1
4
(18)
1
4
(18)
1
4 f (1, 1, −1) = 1 , maximum
For x = 0, we have from (1) y = 0 or y = ±z f (−1, −1, 1) = −1 , minimum

Substituting this value of y in the constraint 19. g (x, y, z) = x + 2y + z − 1 = 0
we have, f = 3z, 2y + 2z, 3x + 2y
1 g = 1, 2, 1
z = ±1 or z= ± 1 f =λ g
24
3z = λ; 2y + 2z = 2λ; 3x + 2y = λ
This gives points z
Eliminating λ gives y = 2z; x = −
1 1 3
(0, 0, ±1), 0, ± 1 , ± 1 Substituting these values in the constraint,
24 24
For x = 0, we have from (1) z = 0 or z = ±y we have
3 1 3
Substituting this value of y in the constraint z= ,x = − ,y =
14 14 7
we have, 1 3 3 9
1 f − , , = , minima
y = ±1 or y = ± 1 14 7 14 28
24
This gives points x2 y2
20. g (x, y, z) = + + z2 = 1
4 9
1 1 f = 1, 1, 1
(0, ±1, 0), 0, ± 1 , ± 1
24 24 x 2y
g= , , 2z
Similar case for y = 0 gives points 2 9
f =λ g
2 1 x 2y
(±1, 0, 0) , ± 1 , 0, ± 1 1 = λ, 1 = λ , 1 = 2zλ
(17) (17) 4 4 2 9
Eliminating λ we get x = 4z; y = 9z
And for z = 0 gives points
Substituting the values of x and y in the con-
2 1 straint we have,
(0, ±1, 0) , ± 1 ,± 1 ,0
(17) 4 (17) 4 1 4 9
z = ±√ , x = ±√ , y = ±√
√ 14 14 14
2 1 1
f ± 1 ,± 1 ,± 1 = 18, This gives points
(18) 4
(18) 4
(18) 4
4 9 1 4 9 1
maxima −√ , −√ , −√ , √ ,√ ,√
14 14 14 14 14 14


4 9 1 √ ab2 1
f √ ,√ ,√ = 14, maximum −√ , −√
14 14 14 1 + a2 b2 1 + a2 b2
4 9 1 √ ab2 1
f −√ , −√ , −√ = − 14, minimum f √ ,√
14 14 14 1+a 2 b2 1 + a2 b2
x2 = sin 1 + a2 b2 , maximum
21. g (x, y) = + y 2 − 1 = 0.......(a > 0)
a2 ab2 1
f = 1, 1 f −√ , −√
2x 1+a 2 b2 1 + a2 b2
g= , 2y
a2 = − sin 1 + a2 b2 , minimum
f =λ g
2x 23. g (x, y) = x2 + y 2 − 1 = 0
1, 1 = λ , 2y
a2 f = axa−1 y b , bxa y b−1
2λ g = 2x, 2y
1 = 2 x ; 1 = 2λy
a f =λ g
x = a2 y
axa−1 y b , bxa y b−1 = λ 2x, 2y
Substituting the value of x in the constraint
axa−1 y b = 2λx; bxa y b−1 = 2λy
we have b
1 Eliminating λ we get y = ± x
y = ±√ a
1 + a2
Substituting this value of y in the constraint
This gives points a
a2 1 we have, x = ±
√ ,√ , a+b
1+a 2 1 + a2 This gives points
a2 1
−√ , −√ a b a b
1+a 2 1 + a2 , , − ,− ,
a+b a+b a+b a+b
a2 1
f √ ,√ = a2 + 1 ,
1 + a2 1 + a2 a b a b
,− , − ,
a+b a+b a+b a+b
maximum
a2 1 Another solutions is when x = 0 or y = 0, this
f −√ , −√ =− a2 + 1 , gives points (0, ±1) or (±1, 0)
1+a 2 1 + a2
If a and b both are even then for all points
minimum
a b
x2 a b
22. g (x, y) = 2 + y 2 − 1 = 0...... (a, b > 0) f= , maxima
b a+b a+b
f = a cos (ax + y) , cos (ax + y)
2x f (0, ±1) = f (±1, 0) = 0 ,minima
g= , 2y
b2 If a and b both are odd then
f =λ g
a b
2x f ± ,±
a cos (ax + y) , cos (ax + y) = λ , 2y a+b a+b
b2
b
2λ a
a
b
a cos (ax + y) = 2 x; cos (ax + y) = 2λy = ,
b a+b a+b
Eliminating λ we get x = ab2 y
maxima
Substituting the value of x in the constraint, a b
we have f ± ,
a+b a+b
1 b
y=± a
(1 + a2 b2 ) a b
=− ,
a+b a+b
This gives points
minima
ab2 1
√ ,√ , If a is odd and b is even then
1 + a2 b2 1 + a 2 b2


a b 25. On the boundary, x2 + y 2 = 3
f ,− g(x, y) = x2 + y 2 − 3 = 0
a+b a+b
a b f = 8xy, 4x2
a b g = 2x, 2y
= ,
a+b a+b
f =λ g
maxima 8xy = 2xλ
4x2 = 2yλ
a b 1
f − , Eliminating λ we get y = ± √ x.
a+b a+b 2
a b
a a b Substituting this into the constraint,
= (−1) , x2 + y 2 − 3 = 0
a+b a+b
1
x2 + x2 − 3 = 0
minima 2√
x=± 2
If a is even and b is odd then √ √
This gives the points (± 2, 1), (± 2, −1)
a b
f − , In the interior, solving f = 0, 0
a+b a+b
gives the critical points (0, y).
a b √
a b f (±√2, 1) = 8, maxima
=
a+b a+b f (± 2, −1) = −8, minima
f (0, y) = 0
maxima
a b 26. On the boundary, x2 + y 2 = 4
f ,− g(x, y) = x2 + y 2 − 4 = 0
a+b a+b
a b f = 6x2 y, 2x3
b a b g = 2x, 2y
= (−1) ,
a+b a+b
f =λ g
minima 6x2 y = 2xλ
2x3 = 2yλ
24. g (x, y) = xn + y n − 1 = 0 (n ≥ 2) 1
Eliminating λ we get y = ± √ x.
f = 1, 1 3
g = nxn−1 , ny n−1 Substituting this into the constraint,
f =λ g x2 + y 2 − 4 = 0
1, 1 = λ nxn−1 , ny n−1 1
x2 + x2 = 4
1 = nλxn−1 ; 1 = nλy n−1 3√
Eliminating λ we get x=± 3
xn−1 = y n−1 ....(1) This gives √ points
√ the √ √
( 3, 1), (− 3, 1), ( 3, −1), (− 3, −1)
If n is odd then (1) gives x = −y or x = y
In the interior, solving f = 0, 0
But x = −y does not satisfy the constraint
gives the critical points (0, y).
Hence x = y √ √ √
f ( √ 1) = f (−√3, −1) = 6 √ maxima
3, 3,
Also, if n is even then (1) gives x = y, which
1 f (− 3, 1) = f ( 3, −1) = −6 3, minima
1 n f (0, y) = 0
gives y =
2
1 1 27. g (x, y) = x4 + y 4 − 1 = 0
1 n
1 n
f = 3x2 , 3y 2
This gives points ,
2 2 g = 4x3 , 4y 3
1 1 1 1 f =λ g
1 n
1 n
1 n
1 n
3x2 , 3y 2 = λ 4x3 , 4y 3
f , = +
2 2 2 2 3x2 = 4λx3 ; 3y 2 = 4λy 3
1
For x, y = 0
= 21− n , maxima
Eliminating λ we get x = y


Substituting this value of x in the constraint, 29. The triangle is bounded by the lines
we get x = 1, y = 0, x + y = 5
1
1 4 Let R: region inside the triangle
y=±
2
To ﬁnd the critical points:
This gives points
1 1 f (x, y) = 3 − x + xy − 2y
1 4
1 4
f (x, y) = (0, 0)
, ,
2 2 ⇒ (y − 1, x − 2) = (0, 0)
1 1 ⇒ (x, y) = (2, 1)
1 1
− 2
4
,− 2
4
Solving, we get
When y = 0, we get x = ±1 this gives points (2, 1) is a critical point.
(1, 0), (-1, 0)
On the boundary x = 1, The function is
When x = 0, we get y = ±1 this gives points
g (y) = f (1, y) = 2 − y, 0 ≤ y ≤ 4
(0,1), (0, -1)
In the interior, solving f = 0, 0 gives the Hence the function has no critical points on
critical point (0, 0) x = 1.
f (1, 0) = 1 On the boundary y = 0, The function is
f (−1, 0) = −1 g (x) = f (x, 0) = 3 − x, 1 ≤ x ≤ 5
f (0, 0) = 0 Hence the function has no critical points on
1 1 3 3 y=0
1 4
1 4
1 4
1 4
f , = + On the boundary x + y = 5, The function is
2 2 2 2
h (x) = f (x, 5 − x) = 6x − x2 − 7, 1 ≤ x ≤ 5
1
= 2 ≈ 1.1892,
4 h (x) = 6 − 2x
⇒ h (x) = 0, when x = 3
maximum Hence, we get (3, 2) is a critical point.
1 1 3 3
1 4
1 4
1 4
1 4
Now evaluate f at all critical point and the
f − ,− =− −
2 2 2 2 boundary points.
1 f (1, 0) = 2
= −2 4 ≈ −1.189, f (5, 0) = −2
minimum f (1, 4) = −2
f (3, 2) = 2
28. On the boundary, 4x2 + y 2 = 8 f (2, 1) = 1
g(x, y) = 4x2 + y 2 − 8 = 0 The points of maxima are (1, 0) and (3, 2)
f = 4y, 4x
The points of minima are (5, 0) and (1, 4)
g = 8x, 2y
f =λ g 30. g (x, y) = x2 + y 2 − 3 = 0
4y = 8xλ f = y 2 , 2xy
4x = 2yλ g = 2x, 2y
Eliminating λ we get y = ±2x. f =λ g
Substituting this into the constraint, y 2 , 2xy = λ 2x, 2y
4x2 + y 2 − 8 = 0 y 2 = λ2x; 2xy = λ2y
4x2 + 4x2 − 8 = 0 If y = √ substituting in the constraints we get
0,
x = ±1 x=± 3
√
This gives the points Since x ≥ 0 this gives point 3, 0
(1, 2), (1, −2), (−1, 2), (−1, −2) Eliminating λ we get
In the interior, solving f = 0, 0 2x2 = y√
2
gives the critical point (0, 0). y = ±x 2
f (1, 2) = f (−1, −2) = 8, maxima Substituting the values of y in the constraint,
f (−1, 2) = f (1, −2) = −8, minima have x = ±1 but x, y ≥ 0 gives the point
we √
f (0, 0) = 0 1, 2


In the interior, solving f = 0, 0 gives the And for z = 0 gives points
critical point (x, 0) 1 1
√ (0, ±1, 0) , ± 1 , ± 1 , 0
f 1, 2 = 2 , maximum 24 24
√ In the interior, solving f = 0, 0, 0 gives the
f 3, 0 = 0 , minimum critical point (0, 0, 0)
f (x, 0) = 0 , minimum 1 1 1
1 4 1 4 1 4 √
f ± ,± ,± = 3
3 3 3
31. g (x, y, z) = x4 + y 4 + z 4 − 1 = 0 , maxima
f = 2x, 2y, 2z f (0, 0, ±1) = 1
g = 4x3 , 4y 3 , 4z 3 f (0, ±1, 0) = 1
f =λ g 1 1 √
2x, 2y, 2z = λ 4x3 , 4y 3 , 4z 3 f 0, ± 1 , ± 1 = 2
24 24
2x = 4λx3 ; 2y = 4λy 3 ; 2z = 4λz 3 f (±1, 0, 0) = 1,
For x, y, z = 0 1 1 √
f ± 1 , 0, ± 1 = 2
Eliminating λ we get z 2 = x2 , z 2 = y 2 24 24
1 1 √
Or z = ±x , z = ±y f ± 1 ,± 1 ,0 = 2
24 24
Substituting value of z in the constraint, f (0, 0, 0) = 0, minima
we get
1
1 4 32. g (x, y, z) = x2 + y 2 + z 2 − 1 = 0
z=± , which gives f = 2xy 2 , 2x2 y, 2z
3
1 1 g = 2x, 2y, 2z
1 4
1 4
x=± ; y=± f =λ g
3 3 2xy 2 = λ2x , 2x2 y = λ2y , 2z = λ2z
⇒ z = 0 or λ = 1
This gives points
1 1 1
For z = 0:
1 4
1 4
1 4
If x = 0, y = 0
± ,± ,±
3 3 3
Eliminating λ we get y 2 = x2 or y = ±x
For x = 0, we have from (1) y = 0 or y = ±z Substituting value of y in the constraint,
Substituting this value of y in the constraint we get
we have, 1

1 1 2

z = ±1 or z = ± 1 x=±
2
24
This gives points 1 1
This gives points ±√ , ±√ , 0
1 1 2 2
(0, 0, ±1), 0, ± 1 ,± 1 If x = 0 or y = 0
2 24 4

For x = 0, we have from (1) We get points (0, 0, 0) , (0 ± 1, 0) and (±1, 0, 0)
z = 0 or z = ±y The point (0, 0, 0), does not satisfy the con-
straint.
Substituting this value of y in the constraint
we have, For λ = 1, We get points ((0, 0, ±1)
1 In the interior, solving f = 0, 0, 0 gives the
y = ±1 or y = ± 1 critical points (0, y, 0) and (x, 0, 0)
24
This gives points 1 1 1
f ±√ , ±√ , 0 =
1 1 2 2 4
(0, ±1, 0), 0, ± 1 ,± 1 f (0, 0, 0) = 0, minima
2 24 4
f (0, 0, ±1) = 1 , maxima
Similar case for y = 0 gives points f (0, ±1, 0) = 0, minima
1 1 f (±1, 0, 0) = 0, minima
(±1, 0, 0) , ± 1 , 0, ± 1 f (0, y, 0) = 0, minima
2 4 24


f (x, 0, 0) = 0, minima On the boundary, 2x2 + y 2 + 4z 2 = 8800
g(x, y) = 2x2 + y 2 + 4z 2 − 8800 = 0
g = 4x, 2y, 8z
33. We use constraint
P =λ g
g(t, u) = u2 t − 11, 000 = 0, so
3 = 4xλ
g = u2 , 2ut remains the same as in Exam-
6 = 2yλ
ple 8.2.
6 = 8zλ
128 Solving these three equations gives
As in the example, u = .
3 3 6 6
Applying the constraint gives x= ,y= ,z=
4λ 2λ 8λ
11, 000 Substituting into the constraint gives,
t= = 6.04.
(128/3)2 2x2 + y 2 + 4z 2 − 8800 = 0
2 2 2
3 6 6
34. Using the equation from exercise 21, 2 + +4 = 8800
4λ 2λ 8λ
(u − 32)t 9
λ= λ2 =
u2 6400
(42.67 − 32)(6.04) 3
= ≈ 0.035 Using λ = , we get (20, 80, 20).
(42.67)2 80
3
Using λ = − , we get (−20, −80, −20).
80
∆z 195 − 161 Of course, the production levels cannot be
= = 0.034
∆g 1000 negative (this would give a minimum of the
10, 000 profit function), so the maximum profit is
35. Substituting t = into f (t, u) gives P (20, 80, 20) = 660
u2
2
1 10, 000 38. P = 3z, 6, 3x , so there are no critical points
f (u) = (u − 32)
2 u2 in the interior.
1 32
= 50, 000, 000 − 4 On the boundary, x2 + 2y 2 + z 2 = 6
u3 u g(x, y) = x2 + 2y 2 + z 2 − 6 = 0
−3 128 g = 2x, 4y, 2z
f (u) = 50, 000, 000 + 5
u4 u P =λ g
12 640 3z = 2xλ
f (u) = 50, 000, 000 − 6 6 = 4yλ
u5 u
3x = 2zλ
f (u) is undefined when u = 0, but this clearly So from the first and third equations we get
does not lead to a maximum. 3z 2zλ
128 x= =
f (u) = 0 when u = . 2λ 3
3 9
Solving for λ gives us λ2 = .
This is a maximum because 4
128 3
f ( ) = −1.06 < 0 For λ = , z = x and y = 1.
3 2
Substituting into the constraint gives,
36. Using the constraint u2 t = k we can write
k x2 + 2y 2 + z 2 − 6 = 0
t = 2 . The results of Example 8.2 show x2 + 2 + x2 = 6
u √
(u − 32)t (u − 32) k x2 = ± 2
λ= 2
= and
u u2 u2 This gives us points √
2 √ √ √
1 1 k ( 2, 1, 2), (− 2, 1, − 2)
h(k) = (u − 32)t2 = (u − 32)
2 2 u2 3
Differentiating with respect to k yields For λ = − , z = −x and y = −1.
2
(u − 32)k Substituting into the constraint gives,
h (k) = =λ
u4
x2 + 2y 2 + z 2 − 6 = 0
37. P = 3, 6, 6 , so there are no critical points x2 + 2 + x2 = 6
√
in the interior. x2 = ± 2


This gives √ points
√ us √ √ g = yz, xz, xy
( 2, −1, − 2), (− 2, −1, 2)
f =λ g
We are only interested √
√ when x, y, z ≥ 0, which 2y + 2z =yzλ
leaves us with f ( 2, 1, 2) = 12
2y + 2z
λ=
yz
3
39. In exercise 37, the value of λ is λ = . 2x + 2z =xzλ
80
Following the work in exercise 37, we see that 2x + 2z
the constraint equation gives us λ=
xz
792 792 2x + 2y =xyλ
λ2 = , and λ = .
64k 64k 2x + 2y
We use the positive square root so that x, y, λ=
and z are all positive, and write the profit func- xy
tion as a function of k. Equating the first two expressions for λ we get
√ 2 2 2 2
3 6 6 198 k + = +
P (x, y, z) = P , , = √ z y z x
4λ 2λ 8λ 792 and therefore x = y. Similarly z = x and there-
Differentiating this function of k yields fore x = y = z and the minimum surface area
99 is a cube.
P (k) = √ ,
792k 43. Minimize the function f (x, y) = y − x
3
and P (8800) = = λ. subject to the constraint
80
g(x, y) = x2 + y 2 − 1 = 0.
3 f = −1, 1
40. From exercise 38, λ = , so a change of 1 in the g = 2x, 2y
2
3 f = λ g gives equations
production constraint results in a change of
2 −1 = 2xλ
3
in the profit. The new profit is 12 + = 13.5. 1 = 2yλ
2
Eliminating λ yields y = −x.
41. A rectangle with sides x and y has perimeter Substituting this into the constraint gives
√
P (x, y) = 2x + 2y and area xy. If we are given 2
x2 + (−x)2 = 1, so that x = ± .
area c, we get constraint g(x, y) = xy − c = 0 2
√ √
2 2 √
P = 2, 2 f( ,− ) = − 2 is a minimum.
g = y, x 2 2
√ √
2 2 √
P = λ g gives equations f (− , ) = 2 is a maximum.
2 2
2 = yλ
2 = xλ 44. g(x, y) = x2 + y 2 − 2 = 0
Eliminating λ gives y = x. f = ex+y , ex+y
g = 2x, 2y
This gives the minimum perimeter. f =λ g
For a given area, the rectangle with the small- ex+y = 2xλ
est perimeter is a square. ex+y = 2yλ
Eliminating λ we get y = x.
42. Place the box with one face in the xy-plane Substituting this into the constraint,
and opposite vertices of x2 + y 2 − 2 = 0
(0, 0, 0) and (x, y, z) with x, y, z > 0. x2 = 1
x = ±1
Minimize f (x, y, z) = 2xy + 2xz + 2yz subject This gives the points
to xyz = c. (1, 1), (−1, −1)
g(x, y, z) = xyz − c = 0 f (1, 1) = e2 , maximum
f = 2y + 2z, 2x + 2z, 2x + 2y f (−1, −1) = e−2 , minimum


45. Find the extreme values of f (x, y) = xy 2 sub- 2x = λ + µ
ject to the constraint g(x, y) = x + y = 0. 2x − x2 = λ − µ
f = y 2 , 2xy Equating the two expressions for λ + µ gives
g = 1, 1 critical point x = 1, y = 1 and z = 2.
f = λ g gives equations The maximum value of f (x, y, z) = 2.
y2 = λ
2xy = λ 50. f (x, y, z) = 3x + y + 2z
g(x, y, z) = y 2 + z 2 − 1 = 0
Eliminating λ yields y(y − 2x) = 0. Substitut- h(x, y, z) = x + y − z − 1 = 0
ing either y = 0 or y = 2x into the constraint
Setting f = λ g + µ h gives the equations:
yields x = y = 0, so (0, 0) is identified as a
3=µ
critical point.
1 = 2λy + µ
Graphically, this is seen to be a saddle point. 2 = 2λz − µ
y2 + z2 − 1 = 0
46. When y = −x, f (x, −x) = g(x) = x3 .
x+y−z−1=0
g (x) = 3x2
1
g (x) = 0 at x = 0, so x = 0 is a critical point. The second equation gives us λ = − y .
5
Since g (x) ≥ 0 for all x and g (0) = 0, x = 0 The third equation then gives us z = − 2 y
is an inflection point. The Lagrange multiplier Substituting these into the fourth equation
4
method fails because the constraint curve goes gives us y 2 = 29
through the saddle point of the function. Finally, using the last equation gives the points
7 2 5
√ + 1, − √ , √
47. f (x, y, z) = x2 + y 2 + z 2 29 29 29
g(x, y, z) = x + 2y + 3z − 6 = 0 7 2 5
− √ + 1, √ , − √
h(x, y, z) = y + z = 0 29 29 29
Setting f = λ g + µ h gives the equations: Maximum:
7 2 5 √
2x = λ f √ + 1, − √ , √ = 29 + 3
2y = 2λ + µ 29 29 29
2z = 3λ + µ Minimum:
7 2 5 √
The first and second equations give λ = 2x and f − √ + 1, √ , − √ = − 29 + 3
µ = 2y − 4x. Then the third equation yields 29 29 29
z = x + y. 51. We find the extreme values of
Substituting this into h(x, y, z) gives x = −2y, f (x, y, z) = x2 + y 2 + z 2
and using these relations in g(x, y, z) then subject to the constraints
shows y = −2, z = 2, and x = 4. g(x, y, z) = x2 + y 2 − 1 = 0 and
The minimum value of f (x, y, z) = 24. h(x, y, z) = x2 + z 2 − 1 = 0.
f = λ g + µ h gives the equations:
48. Our two planes are not parallel, so they inter-
2x = 2xλ + 2xµ
sect in a line, and minimizing the distance from
2y = 2yλ
the origin to the two planes simultaneously is
2z = 2zµ
equivalent to minimizing the distance to the
line which is their intersection. If y and z are not equal to zero we are led to
solution λ = µ = 1 and x = 0. The constraints
49. f (x, y, z) = xyz then show that y = ±1 and z = ±1.
g(x, y, z) = x + y + z − 4 = 0 We also have a solution when y = 0. The con-
h(x, y, z) = x + y − z = 0 straints then give x = ±1 and z = 0. (We get
Setting f = λ g + µ h gives the equations: the same solutions if we start with z = 0.)
yz = λ + µ f (±1, 0, 0) = 1 are minima.
xz = λ + µ
xy = λ − µ f (0, ±1, ±1) = 2 are maxima.

Subtracting h from g shows that z = 2 and 52. f (x, y, z) = x2 + y 2 + z 2
y = 2 − x. The above equations become g(x, y, z) = x + 2y + z − 2 = 0
4 − 2x = λ + µ h(x, y, z) = x − y = 0


Setting f = λ g + µ h gives the equations: f = λ g gives equations:
2x = λ + µ 8 = 8yλ
2y = 2λ − µ 6 = 4zλ
2z = λ 2
Eliminating λ yields y = z, and the con-
x + 2y + z − 2 = 0 3
straint becomes
x−y =0 2
2
4 z + 2z 2 = 800
Solving this system of equations gives the point 3
closest to the origin: 60 40
Therefore z = √ , y = √ , x = 0.
6 6 4 17 17
, ,
11 11 11 This gives production P ≈ 164.9.
On the boundary y = 0, we maximize
53. We minimize the square of the distance from
f (x, z) = 4x + 6z subject to
(x, y) to (0, 1), f (x, y) = x2 + (y − 1)2 , subject
x2 + 2z 2 ≤ 800.
to the constraint g(x, y) = xn − y = 0.
Again there are no critical points, so the max-
f = λ g leads to equations: imum must occur on the boundary
2x = λnxn−1 x2 + 2z 2 = 800.
2y − 2 = −λ
f = λ g gives equations:
Eliminating λ yields:
4 = 2xλ
2x + 2nx2n−1 − 2nxn−1 = 0
6 = 4zλ
We always have solution (0, 0). 4
Eliminating λ yields x = z, and the con-
In order to tell whether this is a minimum or 3
straint becomes
a maximum, we notice that by substituting 4
2
y = xn into f (x, y), we get z + 2z 2 = 800
3
f (x, xn ) = x2 + (xn − 1)2 , and 60 80
f (x) = 2x + 2nx2n−1 − 2nxn−1 Therefore z = √ , x = √ , y = 0.
17 17
(Not surprisingly, f (x) = 0 is the relation we This also gives production P ≈ 164.9.
were led to by the method of Lagrange multi-
pliers.) On the boundary z = 0, we maximize
f (x) = 2 + 2n(2n − 1)x2n−2 − 2n(n − 1)xn−2 f (x, y) = 4x + 8y subject to
f (0) = 2 if n > 2 and this is a local minimum. x2 + 4y 2 ≤ 800.
f (0) = −2 if n = 2 and this is a local maxi- Again there are no critical points, so the max-
mum. imum must occur on the boundary
x2 + 4y 2 = 800.
The last part of this question is best explored
f = λ g gives equations:
with a CAS. The function f (x) has absolute
4 = 2xλ
minimum at its largest critical value. As n in-
8 = 8yλ
creases, this critical value approaches x = 1. 1
At x = 1 the distance to the point (0, 1) is one, Eliminating λ yields y = x, and the con-
2
the same as the distance at x = 0. Therefore straint becomes
2
the diﬀerence between the absolute minimum 1
x2 + 4 x = 800
value and the local minimum at x = 0 goes to 2
0. Therefore x = 20, y = 10, z = 0.
This also gives production P = 160.
54. As in Example 8.4, there are no critical points
As expected, the global maximum occurs at
in the interior, so the maximum must occur on
P (16, 8, 12) = 200.
the boundary, and on the boundary
x2 + 4y 2 + 2z 2 = 800, we get maximum 55. We minimize the square of the distance,
P (16, 8, 12) = 200. f (x, y) = (x − 1)2 + y 2 + z 2
On the boundary x = 0, we maximize subject to the constraint
f (y, z) = 8y + 6z subject to g(x, y, z) = x2 + y 2 − z = 0
4y 2 + 2z 2 ≤ 800. f = λ g leads to equations:
Again there are no critical points, so the max- 2x − 2 = 2xλ
imum must occur on the boundary 2y = 2yλ
4y 2 + 2z 2 = 800. 2z = −λ


If y = 0 then λ = 1 and the first equation is g = 1, 1
inconsistent.
f =λ g
If y = 0, the constraint gives z = x2 , and com- 3x2 − 5y = λ
bining the first and third equation above yields 3y 2 − 5x = λ
4x3 + 2x − 2 = 0 This gives us 3y 2 + 5y − 3x2 − 5x = 0
which we numerically solve to find (y − x)(3y + 3x + 5) = 0
x = 0.5898, y = 0, z = 0.3478. 5
Therefore y = x or y = −x − .
3
56. We minimize the square of the distance,
Substituting y = x into the constraint,
f (x, y) = x2 + (y − 2)2 + z 2
x+y−k =0
subject to the constraint
x+x−k =0
g(x, y, z) = x2 + y 2 − z 2 − 1 = 0 k
x= .
f = λ g leads to equations: 2
2x = 2xλ k k
This gives the point ,
2y − 4 = 2yλ 2 2
2z = −2zλ 5
Substituting y = −x − into the constraint,
If x = 0 then λ = 1 and the second equation is 3
inconsistent. Therefore x = 0. x+y−k =0
5
If z = 0, then λ = −1 and the second equation x−x− −k =0
3
gives y = 1. However, if x = 0 and y = 1, 5
k=−
then the constraint gives z = 0, a contradic- 3
tion. Therefore z = 0. This gives us no additional points.
3 3
Thus x = z = 0 and the constraint forces k k k k k k
f , = + −5
y = ±1. The closest point is therefore (0, 1, 0). 2 2 2 2 2 2
2k 3 5k 2
57. The angles α, β, and θ sum to the angle be- = −
8 4
tween due east and due north, so k 2 (k − 5)
π =
α+β+θ = . 4
2
df 1
We maximize f (α, β, θ) = sin α sin β sin θ = (3k 2 − 10k)
subject to the constraint dk 4
2
π k k
g(α, β, θ) = α + β + θ − . λ = 3x2 − 5y = 3 −5
2 2 2
f = cos α sin β sin θ, sin α cos β sin θ, 3k 2 − 10k df
= =
sin α sin β cos θ 4 dk
g = 1, 1, 1
f = λ g gives equations 59. C = 25, 100
cos α sin β sin θ = λ On the interior, there are no critical points.
sin α cos β sin θ = λ
On the boundary,
sin α sin β cos θ = λ
g(L, K) = 60L2/3 K 1/3 − 1920 = 0
Using these equations in pairs, we get 40K 1/3 20L1/3
tan α = tan β = tan θ g= ,
π L1/3 K 2/3
Since these are angles between 0 and , they
2 C = λ g gives equations
must all be equal. 40K 1/3
π 25 = λ
α=β=θ= L1/3
6
and the maximum northward component of 20L2/3
100 = λ
force is K 2/3
π π π 1
f( , , ) = . Eliminating λ yields L = 8K.
6 6 6 8 Substituting this into the constraint gives
58. f = 3x2 − 5y, 3y 2 − 5x 60(8K)2/3 K 1/3 = 1920, so K = 8 and L = 64
gives minimum cost.
Maximize f subject to the constraint
g(x, y) = x + y − k = 0 where k > 5. The minimum cost is C(64, 8) = 2400.

12. REVIEW EXERCISES 729

∂C ∂P ∂C ∂P K or L is zero, but this gives production zero
60. =λ , =λ
∂L ∂L ∂K ∂K (not a maximum).
Dividing these equations gives
∂C/∂L ∂P/∂L On the boundary,
= g(L, K) = 2L + 5K − 150 = 0
∂C/∂K ∂P/∂K
which is what we were to show. g = 2, 5
P = λ g gives equations
61. f (c, d) = 10c0.4 d0.6
400K 1/3
g(c, d) = 10c + 15d − 300 = 0 = 2λ
3L1/3
4d0.6 6c0.4 200L 2/3
f= , = 5λ
c0.6 d0.4 3K 2/3
g = 10, 15 Eliminating λ yields L = 5K.
f = λ g gives the equations: Substituting this into the constraint gives
4d0.6 2(5K) + 5K = 150, so K = 10 and L = 50.
= 10λ
c0.6
6c 0.4 Production is maximized when K = 10 and
= 15λ L = 50.
d0.4
Eliminating λ gives c = d.
Using the constraint, we find that 64. Since the profit function P is linear and the
10c + 15c = 300, so that c = d = 12 maximizes constraint region is convex, the maximum of P
the utility function. is on the boundary of the constraint region.
g(x, y) = 2x2 + 5y 2 − 32500 = 0
62. f (x, y) = xp y 1−p f = 4, 5
g(x, y) = ax + by − k = 0 g = 4x, 10y
f = pxp−1 y 1−p , (1 − p)xp y −p
f =λ g
g = a, b
4 = 4xλ
Setting f = λ g gives the equations: 5 = 10yλ
pxp−1 y 1−p = aλ 1
Eliminating λ gives us y = x.
(1 − p)xp y −p = bλ 2
ax + by − k = 0 Substituting this into the constraint,
Solving for λ in the first and second equations 2x2 + 5y 2 − 32500 = 0
gives 5
2x2 + x2 − 32500 = 0
bpxp−1 y 1−p = a(1 − p)xp y −p 4
x = ±100
Dividing through by y −p xp−1 and collecting This gives us the points (100, 50), (−100, −50)
gives P (100, 50) = 650, maximum
a(1 − p) aq P (−100, 50) = −650, minimum
y= x= x
bp bp
Substituting this into the last equation gives
us 12. Review Exercises
aq
ax + b x=k
bp 1.
q
ax 1 + =k
p
1−p
ax 1 + =k 8
p
p 4

x=k -3
a -2
-1 0 -1
-2
-3

q 1
00
1
y=k 3
2
x -4 y
2
3
b
-8

400K 1/3 200L2/3
63. P = ,
3L1/3 3K 2/3
On the interior, the critical points occur where


2. −5.0

5.0
y
−2.5
2.5
00.0 x
4 z 1 0.0
2 −2.5
3
4 −5.0
3
2.5

2
5.0

1

-2 -3
00-1
-3 x 1
-2 2 -1
3 0 1 2 3
y

7.
3.

x y 1
-3 -3
-2 -1
1 0 -1 -2
1
3 2 2 3
0
-2 0.5
-4 -1
00 2
0 1
-8 -2 -1
x 1 y
-0.5 2
-12

-1
-16

4. 8.

2
1

1
-3 0.5
-2
-1
-2 0 -3
-1 -2 00 -2
0 -1
2 1 0 1
1 0 1-1 y
2 2
3 2
-1 x -0.5 3

-2 -1

5. 9.

x 5 y
-1 -0.5 -2
0 -1
10 2 1 0.5 1
0

-5
10.0 5
7.5 -10
5.0
5.0
2.5 2.5 -15
z 0
0.00.0
-20
−2.5
−5.0 −5 -25
x y

−10

6. 10.


17. Along the line x = 0, we have
y2
-1
x
-0.5
y
-1 -2 lim =1
2 1 00
0 0.5 (0,y)→(0,0) y 2
1

-10 Along the curve y = x, we have
-20 x2 + x2 2
lim 2 + x2 + x2
=
-30 (x,x)→(0,0) x 3
-40 Since these limits are different,the limit does
-50 not exist.
0
lim =0
(0,y)→(0,0) y 2

11. a. Surface D Along the curve y = x, we have
x2 1
b. Surface B lim =
(x,x)→(0,0) x 2 + x2 + x2 3
c. Surface C
Since these limits are different,the limit does
d. Surface A not exist.
e. Surface F
19. We use Theorem 2.1.
f. Surface E x3 + xy 2 x3 xy 2
≤ 2 + 2
x2 + y 2 x + y2 x + y2
12. a. Contour D by the triangle inequality. We make the de-
b. Contour A nominators smaller in both terms to get,
c. Contour B x3 xy 2
≤ 2 + 2
x y
d. Contour C = |x| + |x| = 2|x|
13. a. Contour C lim 2|x| = 0, therefore
(x,y)→(0,0)
b. Contour A x3 + xy 2
lim = 0.
c. Contour D (x,y)→(0,0) x2 + y 2

d. Contour B 20. We use Theorem 2.1.
3x 0
14. a. lim = =0 3y 2 ≤x2 + 3y 2
y2 + 1
(x,y)→(0,2) 5
xy − 1 π−1 3y 2 | ln(x + 1)| ≤(x2 + 3y 2 )| ln(x + 1)|
b. lim = = −π + 1 3y 2 | ln(x + 1)|
(x,y)→(1,π) cos xy −1 ≤| ln(x + 1)|
x2 + 3y 2
0
lim =0
(0,y)→(0,0) 0 + y 2
Since lim | ln(x + 1)| = 0,
(x,y)→(0,0)
Along the curve y = x2 , we have 2
3x4 3 3y | ln(x + 1)|
lim = lim =0
(x,x2 )→(0,0) x4 + x4 2 (x,y)→(0,0) x2 + 3y 2
Since these limits are different,the limit does 21. f (x, y) is continuous unless x = 0.
not exist.
22. 4−4x2 −y 2 is continuous for all x, y. Therefore
16. Along the line x = 0, we have we only need the radicand to be positive, so f
0 is continuous inside the ellipse:
lim =0
(0,y)→(0,0) 0 + y 3 4 − 4x2 − y 2 ≥ 0
y2
Along the curve y = x2/3 , we have x2 + ≤1
3/2 4
2x x2/3
lim =1 4
(x,x2/3 )→(0,0) x2 + x2 23. fx = + xyexy + exy
y
Since these limits are different,the limit does −4x
not exist. fy = 2 + x2 exy
y


24. fx = exy + xyexy 4 3
L(x, y) = ln 22 + (x − 4) + (x − 4)
fy = x2 exy + 6y 11 22
1 35. fx = 8x3 y + 6xy 2
25. fx = 6xy cos y − √
2 x fy = 2x4 + 6x2 y
fy = 3x2 cos y − 3x2 y sin y fxx = 24x2 y + 6y 2
3√ fyy = 6x2
26. fx = xy + 3 fxy = 8x3 + 12xy
2
√
x3
fy = √ 36. fx = 2xe3y
2 y
fy = 3x2 e3y − cos y
27. fx = fxx = ex sin y fxx = 2e3y
fy = ex cos y fyy = −ex sin y fyy = 9x2 e3y + sin y
Therefore, fyyx = 18xe3y
fxx + fyy = ex sin y − ex sin y = 0
37. fx = 2xy + 2, fy = x2 − 2y
28. fxx = ex cos y
fyy = −ex cos y f (1, −1) = 0
Therefore fxx + fyy = 0. fx (1, −1) = 0, fy (1, −1) = 3

∂f 1.6 − 2.4 3(y + 1) − z = 0
29. (0, 0) ≈ = −0.04
∂x 10 − (−10) x y
∂f 2.6 − 1.4 38. fx = , fy =
(0, 0) ≈ = 0.06 x2 + y 2 x2 + y 2
∂y 10 − (−10)
∂f 1.2 − 2.0 f (3, −4) = 5
30. (10, 0) ≈ = −0.04 3 4
∂x 20 − 0 fx (3, −4) = , fy (3, −4) = −
∂f 2.2 − 1.0 5 5
(10, 0) ≈ = 0.06 3 4
∂y 10 − (−10) (x − 3) − (y + 4) − (z − 5) = 0
5 5
3xy
31. fx = √ , fy = 3 x2 + 5
x2 + 5 39. f (x, y, z) = x2 + 2xy + y 2 + z 2 = 5
f (−2, 5) = 45
f = 2x + 2y, 2x + 2y, 2z
fx (−2, 5) = −10, fy (−2, 5) = 9
f (0, 2, 1) = 4, 4, 2
L(x, y) = 45 − 10(x + 2) + 9(y − 5)
4(x − 0) + 4(y − 2) + 2(z − 1) = 0
1 4(x + 2)
32. fx = , fy = −
4y − 2 (4y − 2)2 40. f (x, y, z) = x2 z − y 2 x + 3y − z = −4
2 f = 2xz − y 2 , −2xy + 3, x2 − 1
f (2, 3) =
5 f (1, −1, 2) = 3, 5, 0
1 4
fx (2, 3) = , fy (2, 3) = −
10 25 3(x − 1) + 5(y + 1) + 0(z − 2) = 0
2 1 4
L(x, y) = + (x − 2) − (y − 3)
5 10 25 41. g (t) = fx (x(t), y(t))x (t)
2
33. fx = sec (x + 2y), 2
fy = 2 sec (x + 2y) + fy (x(t), y(t))y (t)
π
f π, =0 fx = 2xy fy = x2 + 2y
2
π π
fx π, = 1, fy π, =2 x (t) = 4e4t y (t) = cos t
2 2
π g (t) = 2e4t sin t(4e4t ) + (e8t + 2 sin t) cos t
L(x, y) = (x − π) + 2 y −
2 = 8e8t sin t + (e8t + 2 sin t) cos t
2x 3
34. fx = fy = 2 ∂f ∂f
x2 + 3y x + 3y 42. = 8x, = −1
f (4, 2) = ln 22 ∂x ∂y
4 3 ∂x ∂x
fx (4, 2) = fy (4, 2) = = 3u2 v + cos u, = u3
11 22 ∂u ∂v


∂y ∂y 50. f = 2x + y 2 , 2xy
= 0, = 8v
∂u ∂v f (2, 1) = 5, 4
3 2
∂g ∂f ∂x ∂f ∂y u = √ , −√
= + 13 13
∂u ∂x ∂u ∂y ∂u
3 2
=8x(3u2 v + cos u) + (−1)0 Du f (2, 1) = 5, 4 · √ , −√
13 13
=8(u3 v + sin u)(3u2 v + cos u) =√
7
∂g ∂f ∂x ∂f ∂y 13
= +
∂v ∂x ∂v ∂y ∂v 51. f = 3ye3xy , 3xe3xy − 2y
=8xu3 − 8v f (0, −1) = −3, 2
=8(u3 v + sin u)u3 − 8v 1 −2
u= √ ,√
5 5
43. g (t) = fx (x(t), y(t), z(t), w(t))x (t) 1 −2
+ fy (x(t), y(t), z(t), w(t))y (t) Du f (0, −1) = −3, 2 · √ , √
5 5
+ fz (x(t), y(t), z(t), w(t))z (t) −7
+ fw (x(t), y(t), z(t), w(t))w (t) Du f (0, −1) = √
5
∂g ∂f ∂x ∂f ∂y 2x + y 2 xy
44. = + 52. f= ,
∂u ∂x ∂u ∂y ∂u
∂g ∂f ∂x ∂f ∂y 2 x2
+ xy 2 x2 + xy 2
= + 5 2
∂v ∂x ∂v ∂y ∂v f (2, 1) = √ ,√
2 6 6
45. F (x, y, z) = x2 + 2xy + y 2 + z 2 1 2
u = √ , −√
Fx = 2x + 2y Fy = 2x + 2y 5 5
Fz = 2z 5 2 1 2
Du f (2, 1) = √ ,√ · √ , −√
∂z Fx x+y 2 6 6 5 5
=− =− 3
∂x Fz z =− √
∂z Fy x+y 2 30
=− =−
∂y Fz z 53. f = 3x2 y, x3 − 8y
−32
f (−2, 3) = 36,√
46. F (x, y, z) = x2 z − y 2 x + 3y − z
f (−2, 3) = 4 145
Fx = 2xz − y 2 , Fy = −2xy + 3,
Fz = x2 − 1 The direction of maximum change is
∂z Fx −2xy + y 2 36, −32 . √
=− = The maximum rate of change is 4 145.
∂x Fz x2 + 1
∂z Fx 2xy − 3 The direction of minimum change is
=− = 2 −36, 32 . √
∂y Fz x +1
The minimum rate of change is −4 145.
√ √
y x
47. f = 3 sin 4y − √ , 12x cos 4y − √ 54. f = 2x + y 2 , 2xy
2 x 2 y f (2, 1) = 5,√
4
1 1 f (2, 1) = 41
f (π, π) = − , 12π −
2 2 The direction of maximum change is 5, 4 .
√
48. 2
f = 4z + 3 sin x, 8y, 8xz The maximum rate of change is 41.

f (0, 1, −1) = 4, 8, 0 The direction of minimum change is
−5, −4 . √
49. f = 3x2 y, x3 − 8y The minimum rate of change is − 41.
f (−2, 3) = 36, −32
2x3 2y 3
3 4 55. f= ,
Du f (−2, 3) = 36, −32 · , x4 + y 4 x4 + y 4
5 5
−20 f (2, 0) = 4, 0
= = −4 f (2, 0) = 4
5


The direction of maximum change is 4, 0 . The first equation gives us x = 0 or 4x2 = y.
The maximum rate of change is 4. If x = 0 then the second equation gives us
The direction of minimum change is −4, 0 . y = 0 and we have the critical point (0, 0).
The minimum rate of change is −4.
If 4x2 = y and x2 = 3y 2 , substitution gives us
1
56. f = 2x + y 2 , 2xy y = 0 or y = and we get the critical points
f (1, 2) = 6, 4 12
√ 1 1 1 1
f (1, 2) = 2 13 √ , , − √ , .
4 3 12 4 3 12
The direction of maximum change is 6, 4 .
√ D(0, 0) = 0, no information. But, along the
The maximum rate of change is 2 13. trace x = 0, f (0, y) = y 3 , which shows that
The direction of minimum change is this point must be a saddle point.
−6, −4 . √ 1 1 1
The minimum rate of change is −2 13. D √ , = >0
4 3 12 12
57. f = −8x, −2 1 1 1
fxx √ , = >0
f (2, 1) = −16, −2 4 3 12 3
1 1 1
The rain will run in the direction 16, 2 . f √ , =− , local minimum
4 3 12 3456
2 2 1 1 1
58. T = 20e−z , −15y −2 e−z , D − √ , = >0
2
4 3 12 12
−10z(4x + 3y −1 )e−z 1 1 1
fxx − √ , = >0
15 −1 4 3 12 3
T (1, 2, 1) = e , −55e−1
20e−1 , − 1 1 1
4 f − √ , =− , local minimum
≈ 7.358, −1.380, 20.233 4 3 12 3456

The direction of most rapid temperature de- 61. f = 4y − 3x2 , 4x − 4y
crease will be fxx = −6x, fyy = −4, fxy = 4
15
− T (1, 2, 1) = −20e−1 , e−1 , 55e−1 Solving f = 0, 0 gives equations
4
4y − 3x2 = 0
59. f = 8x3 − y 2 , −2xy + 4y 4x = 4y
fxx = 24x2 , fyy = −2x + 4, fxy = −2y The second equation gives us x = y.
Solving f = 0, 0 gives equations The first equation then becomes
8x3 = y 2 4
x(4 − 3x) = 0, so that x = 0 or x = .
2y(2 − x) = 0 3
The second equation gives us y = 0 or x = 2. 4 4
We get the critical points , , (0, 0).
If y = 0 then the first equation gives us x = 0 3 3
and we have the critical point (0, 0). D(0, 0) = −16 < 0, so this is a saddle point.
3 2 4 4
If x = 2 and 8x = y , then y = ±8 and we
D , = 16 > 0
get the critical points (2, ±8) 3 3
D(0, 0) = 0, so Theorem 7.2 provides no infor- 4 4
fxx , = −8 < 0,
mation. But, along every trace y = cx, 3 3
f (x, cx) = 2x4 − c2 x3 + 2c2 x2 , and the second 4 4 32
so f , = is a local maximum.
derivative test shows this to be a minimum. 3 3 27
D(2 ± 8) = −256 < 0 so these are both saddle 62. f = 3y − 3x2 y, 3x − x3 + 2y − 1
points. fxx = −6xy, fyy = 2, fxy = 3 − 3x2
60. f = 8x3 − 2xy, 3y 2 − x2 Solving f = 0, 0 gives equations
fxx = 24x2 − 2y, fyy = 6y, fxy = −2x 3y − 3x2 y = 0
Solving f = 0, 0 gives equations 3x − x3 + 2y − 1 = 0
2x(4x2 − y) = 0 The first equation gives us 3y(1 − x2 ) = 0 and
x2 = 3y 2 so y = 0 or x = ±1.


If y = 0 then the second equation gives us ∂g
= 13, 088a + 312b − 22, 880
3x − x3 − 1 = 0 which we solve using a CAS to ∂a
∂g
get the critical points = 312a + 8b − 524
(1.532, 0), (−1.879, 0), (0.347, 0) ∂b
∂g ∂g
If x = 1 then the second equation gives us Solving = = 0 we get
∂a ∂b
1 1 611
y = − and we get the critical point 1, − a= ≈ 2.657
2 2 230
4382
If x = −1 then the second equation gives us b=− ≈ −38.104
2 2 115
y = and we get the critical point −1, y = 2.657x − 38.104
3 3
y(20) ≈ $15, 026
D (1.532, 0) ≈ −16 < 0, saddle. y(60) ≈ $121, 287
D (−1.879, 0) ≈ −56 < 0, saddle. 65. f = 8x3 − y 2 , −2xy + 4y
D (0.347, 0) ≈ −7 < 0, saddle. fxx = 24x2 , fyy = −2x + 4, fxy = −2y
1 Solving f = 0, 0 gives equations
D 1, − =6>0 y 2 = 8x3
2
1 2y(2 − x) = 0
fxx 1, − =3>0
2 The second equation gives us x = 2 or y = 0.
1 1 If y = 0 then the ﬁrst equation gives us x = 0
f 1, − = − , local minimum
2 4 and we have the critical point (0, 0).
2 If x = 2 and y 2 = 8x2 , then y = ±8 and we
D −1, = 18 > 0
3 get the critical points
2 (2, ±8), neither of which are in the region.
fxx −1, =9>0
3 Along y = 0, f (x, 0) = 2x4 which has a critical
2 9 point at x = 0 which gives us the critical point
f −1, = − , local minimum
3 4 (0, 0) (we already had this point).
Along y = 2, f (x, 2) = 2x4 − 4x + 8, which has
63. The residuals are, 1
64a + b − 140, 66a + b − 156 a critical point at x = √ and the only critical
3
70a + b − 184, 71a + b − 190 2
1
g(a, b) = (64a + b − 140)2 + (66a + b − 156)2 point in the region is √ , 2 .
3
2
+ (70a + b − 184)2 + (71a + b − 190)2 2
Along x = 0, f (0, y) = 2y which has a critical
∂g point at y = 0 and we get the same critical
= 36786a + 542b − 91252
∂a point of (0, 0).
∂g
= 542a + 8b − 1340 Along x = 4, f (4, y) = 512 − 2y 2 , which has
∂b
a critical point at y = 0 and we get the same
∂g ∂g
Solving = = 0 we get critical point of (0, 0).
∂a ∂b
934 In addition, the intersection points of our
a= ≈ 7.130
131 boundaries are (0, 0), (4, 0), (0, 2), (4, 2).
41336
b=− ≈ −315.542 f (0, 0) = 0, minimum
131 1
y = 7.130x − 315.542 f √ , 2 ≈ 5.619,
3
2
y(74) ≈ 212 f (0, 2) = 8
y(60) ≈ 112 f (4, 0) = 512, maximum
f (4, 2) = 504.
64. The residuals are, (income in thousands of dol-
lars) 66. f = 8x3 − 2xy, 3y 2 − x2
28a + b − 36, 32a + b − 34 fxx = 24x2 − 2y, fyy = 6y, fxy = −2x
40a + b − 88, 56a + b − 104 Solving f = 0, 0 gives equations
g(a, b) = (28a + b − 36)2 + (32a + b − 34)2 2x(4x2 − y) = 0
+ (40a + b − 88)2 + (56a + b − 104)2 x2 = 3y 2


The ﬁrst equation gives us x = 0 or 4x2 = y. f =λ g
If x = 0 then the second equation gives us 4xy = 2xλ
y = 0 and we have the critical point (0, 0). 2x2 = 2yλ
x
The second equation gives y = ± √ .
If 4x2 = y and x2 = 3y 2 , substitution gives us 2
1
y = 0 or y = and we get the critical points Substituting this into the constraint,
12 x2 + y 2 − 4 = 0
1 1
√ , , in our region. x2
4 3 12 x2 + −4=0
2√
1 1 2 2
− √ , , not in the region. x=± √
4 3 12 3
Along y = 0, f (x, 0) = 2x4 which has a critical This gives the points
point at x = 0 which gives us the critical point √ √
2 2 2 2 2 2
(0, 0) (we already had this point). √ ,√ , √ , −√
3 3 3 3
Along x = 2, f (2, y) = 32 + y 3 − 4y, which √ √
2 2 2 2 2 2 2
has a critical point at y = ± √ and the only − √ , −√ , −√ ,√
3 3 3 3 3
2 Maxima:
critical point in the region is 2, √ . √ √
3 2 2 2 2 2 2 32
4
Along y = x, f (x, x) = 2x which has a criti- f √ ,√ =f −√ ,√ = √
3 3 3 3 3 3
cal point at x = 0 and we get the same critical
point of (0, 0). Minima:
√ √
2 2 2 2 2 2
In addition, the intersection points of our f √ , −√ =f − √ , −√
boundaries are (0, 0), (2, 0), (2, 2). 3 3 3 3
f (0, 0) = 0 32
=− √
1 1 3 3
f √ , ≈ −0.00029, minimum
4 3 12
2 69. g(x, y) = x2 + y 2 − 1 = 0
f 2, √ ≈ 28.9 f = y, x
3
f (2, 0) = 32, maximum g = 2x, 2y
f (2, 2) = 32, maximum f =λ g
y = 2xλ
67. g(x, y) = x2 + y 2 − 5 = 0 x = 2yλ
f = 1, 2 Eliminating λ we see that y = ±x.
g = 2x, 2y Substituting this into the constraint yields
f =λ g 1
x2 + x2 = 1, so that x = ± √
1 = 2xλ 2
2 = 2yλ Therefore our critical points are
Eliminating λ gives y = 2x. 1 1 1 1
( √ , √ ), (− √ , √ ),
Substituting this into the constraint, 2 2 2 2
1 1 1 1
x2 + y 2 − 5 = 0 ( √ , − √ ), (− √ , − √ ).
x2 + 4x2 − 5 = 0 2 2 2 2
x = ±1 1 1 1 1 1
f ( √ , √ ) = f (− √ , − √ ) = , maximum.
2 2 2 2 2
This gives the points 1 1 1 1 −1
(1, 2) , (1, −2) f ( √ , − √ ) = f (− √ , √ ) = ,
2 2 2 2 2
(−1, −2) , (−1, 2)
minimum.
Maximum: f (1, 2) = 5
70. g(x, y) = x2 + y 2 − 1 = 0
Minimum: f (−1, −2) = −5 f = 2x − 2, 4y
68. g(x, y) = x2 + y 2 − 4 = 0 g = 2x, 2y
f = 4xy, 2x2 f =λ g
g = 2x, 2y 2x − 2 = 2xλ


4y = 2yλ Substituting this into our constraint gives
x−1 4−x
The ﬁrst equation gives us λ = . = x3 or
x 3x2
The second equation is y(2−λ) = 0 which gives 3x5 + x − 4 = 0
us
x−1 With the aid of a CAS, we ﬁnd x = 1 is the only
y 2− =0 real solution. This gives closest point (1, 1).
x
x+1
y =0 72. We want to minimize
x
so y = 0 or x = −1. f (x, y) = (x − 2)2 + (y − 1)2
subject to the constraint y = x3 .
If y = 0, our constraint gives us x = ±1.
If x = −1, our constraint gives us y = 0. g(x, y) = x3 − y = 0
Therefore our critical points are f = 2(x − 2), 2(y − 1)
(1, 0), (−1, 0). g = 3x2 , −1
f (1, 0) = −1, minimum f =λ g
f (−1, 0) = 3, maximum 2(x − 2) = 3x2 λ
2(y − 1) = −λ
71. We want to minimize
Eliminating λ gives
f (x, y) = (x − 4)2 + y 2 −x + 2 + 3x2
subject to the constraint y = x3 . y=
3x2
g(x, y) = x3 − y = 0
Substituting this into our constraint gives
f = 2(x − 4), 2y −x + 2 + 3x2
g = 3x2 , −1 = x3 or
3x2
f =λ g 3x5 − 3x2 + x − 2 = 0
2(x − 4) = 3x2 λ
We solve this using a CAS and get the closest
2y = −λ
point (there is only one real solution to this
Eliminating λ gives
4−x equation):
y= (1.081, 1.262)
3x2

Ism et chapter_12

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