Sol Purcell Ingles

Derivatives for Functions of
CHAPTER 12 Two or More Variables
12.1 Concepts Review 4. a. 6

1. real-valued function of two real variables b. 12

2. level curve; contour map c. 2

3. concentric circles d. (3cos 6)1/ 2 + 1.44 ≈ 3.1372
4. parallel lines
5. F (t cos t , sec2 t ) = t 2 cos 2 t sec2 t = t 2 , cos t ≠ 0

Problem Set 12.1 6. F ( f (t ), g (t )) = F (ln t 2 , et / 2 )
1. a. 5 = exp(ln t 2 ) + (et / 2 ) 2 = t 2 + et , t ≠ 0
b. 0
7. z = 6 is a plane.
c. 6

d. a6 + a 2

e. 2 x2 , x ≠ 0
f. Undefined
The natural domain is the set of all (x, y) such
that y is nonnegative.

2. a. 4 8. x + z = 6 is a plane.
b. 17
17
c.
16

d. 1 + a2 , a ≠ 0

e. x 3 + x, x ≠ 0
f. Undefined
The natural domain is the set of all (x, y) such 9. x + 2y + z = 6 is a plane.
that x is nonzero.

3. a. sin(2π ) = 0

⎛π⎞
b. 4sin ⎜ ⎟ = 2
⎝6⎠

⎛π⎞
c. 16sin ⎜ ⎟ = 16
⎝2⎠

d. π2 sin(π2 ) ≈ –4.2469

744 Section 12.1 Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be
reproduced, in any form or by any means, without permission in writing from the publisher.

10. z = 6 – x 2 is a parabolic cylinder. 15. z = exp[–( x 2 + y 2 )]

11. x 2 + y 2 + z 2 = 16 , z ≥ 0 is a hemisphere. x2
16. z = ,y>0
y

x2 y 2 z 2
12. + + = 1 , z ≥ 0 is a hemi-ellipsoid.
4 16 16 17. x 2 + y 2 = 2 z; x 2 + y 2 = 2k

13. z = 3 – x 2 – y 2 is a paraboloid.
18. x = zy, y ≠ 0 ; x = ky, y ≠ 0

14. z = 2 – x – y 2
19. x 2 = zy, y ≠ 0; x 2 = ky, y ≠ 0

Instructor’s Resource Manual Section 12.1 745

20. x 2 = –( y – z ); x 2 = –( y – k ) 24. ( x – 2)2 + ( y + 3)2 =
16
V2

25. a. San Francisco and St. Louis had a
x2 + 1 temperature between 70 and 80 degrees
21. z = , k = 1, 2, 4
x2 + y 2 Fahrenheit.
k = 1: y 2 = 1 or y = ±1 ; b. Drive northwest to get to cooler
two parallel lines temperatures, and drive southeast to get
k = 2: 2 x 2 + 2 y 2 = x 2 + 1 warmer temperatures.
x2 y 2 c. Since the level curve for 70 runs southwest
+ = 1 ; ellipse
1 1 to northeast, you could drive southwest or
2
northeast and stay at about the same
k = 4: 4 x 2 + 4 y 2 = x 2 + 1 temperature.
x2 y2 The lowest barometric pressure, 1000
+ = 1; ellipse 26. a.
1 1
3 4 millibars and under, occurred in the region
of the Great Lakes, specifically near
Wisconsin. The highest barometric
22. y = sin x + z; y = sin x + k pressure, 1025 millibars and over, occurred
on the east coast, from Massachusetts to
South Carolina.

b. Driving northwest would take you to lower
barometric pressure, and driving southeast
would take you to higher barometric
pressure.

c. Since near St. Louis the level curves run
southwest to northeast, you could drive
23. x = 0, if T = 0:
southwest or northeast and stay at about the
⎛1 ⎞
y 2 = ⎜ – 1⎟ x 2 , if y ≠ 0 . same barometric pressure.
⎝T ⎠
27. x 2 + y 2 + z 2 ≥ 16 ; the set of all points on and
outside the sphere of radius 4 that is centered at
the origin

28. The set of all points inside (the part containing
the z-axis) and on the hyperboloid of one sheet;
x2 y 2 z 2
+ – = 1.
9 9 9

x2 y 2 z 2
29. + + ≤ 1 ; points inside and on the
9 16 1
ellipsoid


30. Points inside (the part containing the z-axis) or on x2 y2
37. 4 x 2 – 9 y 2 = k , k in R; – = 1, if k ≠ 0;
x2 y 2 z 2 k k
the hyperboloid of one sheet, + – = 1, 4 9
9 9 16
2x
excluding points on the coordinate planes planes y = ± (for k = 0) and all hyperbolic
3
31. Since the argument to the natural logarithm cylinders parallel to the z-axis such that the ratio
function must be positive, we must have ⎛1⎞ ⎛1⎞
a:b is ⎜ ⎟ : ⎜ ⎟ or 3:2 (where a is associated
x 2 + y 2 + z 2 > 0 . This is true for all ( x, y , z ) ⎝ 2⎠ ⎝3⎠
except ( x, y, z ) = ( 0, 0, 0 ) . The domain consists with the x-term)
3
all points in except the origin. 2 + y2 + z2
38. e x = k, k > 0
32. Since the argument to the natural logarithm x 2 + y 2 + z 2 = ln k
function must be positive, we must have xy > 0 .
concentric circles centered at the origin.
This occurs when the ordered pair ( x, y ) is in the
first quadrant or the third quadrant of the 39. a. All ( w, x, y, z ) except ( 0, 0, 0, 0 ) , which
xy-plane. There is no restriction on z. Thus, the would cause division by 0.
domain consists of all points ( x, y, z ) such that x
and y are both positive or both negative. b. All ( x1 , x2 ,… , xn ) in n-space.

33. x 2 + y 2 + z 2 = k , k > 0; set of all spheres c. All ( x1 , x2 ,… , xn ) that satisfy
centered at the origin 2 2
x1 + x2 + 2
+ xn ≤ 1 ; other values of

34. 100 x 2 + 16 y 2 + 25 z 2 = k , k > 0;
( x1 , x2 ,… , xn ) would lead to the square
root of a negative number.
x2 y2 z2
+ + = 1; set of all ellipsoids centered
k
100
k
16
k
25 40. If z = 0, then x = 0 or x = ± 3 y .
at origin such that their axes have ratio
⎛ 1 ⎞ ⎛1⎞ ⎛1⎞
⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 2:5:4.
⎝ 10 ⎠ ⎝ 4 ⎠ ⎝ 5 ⎠

x2 y2 z2
35. + – = k ; the elliptic cone
1 1 1
16 4
2 2
x y z2
+ = and all hyperboloids (one and two
9 9 16
sheets) with z-axis for axis such that a:b:c is 41. a. AC is the least steep path and BC is the most
steep path between A and C since the level
⎛1⎞ ⎛1⎞ ⎛1⎞ curves are farthest apart along AC and
⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ or 3:3:4.
⎝ 4⎠ ⎝ 4⎠ ⎝ 3⎠ closest together along BC.

x2 y2 z2 AC ≈ (5750) 2 + (3000)2 ≈ 6490 ft
36. – – = k ; the elliptical cone b.
1 1 1
9 4
BC ≈ (580) 2 + (3000) 2 ≈ 3060 ft
2 2
y z x2
+ = and all hyperboloids (one and two
9 36 4
sheets) with x-axis for axis such that a:b:c is
⎛1⎞ ⎛1⎞
⎜ ⎟ : ⎜ ⎟ :1 or 2:3:6
⎝3⎠ ⎝ 2⎠

42. Completing the squares on x and y yields the
equivalent equation
f ( x, y ) + 25.25 = ( x – 0.5) 2 + 3( y + 2)2 , an
elliptic paraboloid.


43. (2 x – y 2 ) exp(– x 2 – y 2 )

sin 2x 2 + y 2

46.

44.
sin x sin y
(1 + x 2 + y 2 )

sin( x 2 + y 2 )
x2 + y2

12.2 Concepts Review

[( f ( x0 + h, y0 ) – f ( x0 , y0 )]
1. lim ; partial
h →0 h
derivative of f with respect to x

2. 5; 1

∂2f
3.
45. ∂ y∂ x

4. 0

Problem Set 12.2

1. f x ( x, y ) = 8(2 x – y )3 ; f y ( x, y ) = –4(2 x – y )3

2. f x ( x, y ) = 6(4 x – y 2 )1/ 2 ;
f y ( x, y ) = –3 y (4 x – y 2 )1/ 2


( xy )(2 x) – ( x 2 – y 2 )( y ) x2 + y2 f s ( s, t ) = –2set
2 – s2
; f s ( s, t ) = 2tet
2 – s2
3. f x ( x, y ) = = 14.
( xy )2 x2 y
( xy )(−2 y ) − ( x 2 − y 2 )( x) 15. Fx ( x, y ) = 2 cos x cos y; Fy ( x, y ) = –2sin x sin y
f y ( x, y ) =
( xy )2
16. f r (r , θ ) = 9r 2 cos 2θ ; fθ (r , θ ) = –6r 3 sin 2θ
( x2 + y2 )
=−
xy 2 17. f x ( x, y ) = 4 xy 3 – 3x 2 y5 ;
f xy ( x, y ) = 12 xy 2 – 15 x 2 y 4
4. f x ( x, y ) = e x cos y; f y ( x, y ) = – e x sin y
f y ( x, y ) = 6 x 2 y 2 – 5 x 3 y 4 ;
5. f x ( x, y ) = e y cos x; f y ( x, y ) = e y sin x
f yx ( x, y ) = 12 xy 2 – 15 x 2 y 4

⎛ 1⎞
6. f x ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (6 x) 18. f x ( x, y ) = 5( x3 + y 2 ) 4 (3 x 2 );
⎝ 3⎠
f xy ( x, y ) = 60 x 2 ( x3 + y 2 )3 (2 y )
= –2 x(3 x 2 + y 2 ) –4 / 3 ;
⎛ 1⎞ = 120 x 2 y ( x3 + y 2 )3
f y ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (2 y )
⎝ 3⎠ f y ( x, y ) = 5( x3 + y 2 ) 4 (2 y );
⎛ 2y ⎞ 2 2 –4 / 3
= ⎜– ⎟ (3 x + y ) f yx ( x, y ) = 40 y ( x3 + y 2 )3 (3x 2 )
⎝ 3 ⎠
= 120 x 2 y ( x3 + y 2 )3
2 2 −1/ 2
7. f x ( x, y ) = x ( x − y ) ;
f y ( x, y ) = – y ( x 2 – y 2 ) –1/ 2 19. f x ( x, y ) = 6e2 x cos y; f xy ( x, y ) = –6e2 x sin y
f y ( x, y ) = –3e2 x sin y; f yx ( x, y ) = –6e2 x sin y
8. fu (u , v) = ve ; f v (u , v) = ue
uv uv

20. f x ( x, y ) = y (1 + x 2 y 2 ) –1;
– xy – xy
9. g x ( x, y ) = – ye ; g y ( x, y ) = – xe
f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2

10. f s ( s, t ) = 2 s ( s 2 – t 2 ) –1 ; f x ( x, y ) = x(1 + x 2 y 2 ) –1;

ft ( s, t ) = −2t ( s 2 − t 2 )−1 f xy ( x, y ) = (1 − x 2 y 2 )(1 + x 2 y 2 ) −2

11. f x ( x, y ) = 4[1 + (4 x – 7 y ) 2 ]–1; ( xy )(2) – (2 x – y )( y ) y2 1
21. Fx ( x, y ) = = = ;
2 2 2
f y ( x, y ) = –7[1 + (4 x – 7 y ) ] 2 –1 ( xy ) x y x2
1
Fx (3, − 2) =
1 ⎛1⎞ –1 ⎛ w ⎞
9
12. Fw ( w, z ) = w ⎜ ⎟ + sin ⎜ ⎟ ( xy )(–1) – (2 x – y )( x) –2 x 2 2
1– ( )
w 2
z
⎝ z⎠ ⎝z⎠ Fy ( x, y ) =
( xy ) 2
=
2 2
x y
=–
x2
;
w
⎛ w⎞ 1
= z
+ sin –1 ⎜ ⎟ ; Fy (3, − 2) = −
2
( w) ⎝z⎠
2
1– z
22. Fx ( x, y ) = (2 x + y )( x 2 + xy + y 2 ) –1;
( w)
2
1 ⎛ w⎞ –
Fz = ( w, z ) = w z 2
⎜– 2 ⎟ = Fx (–1, 4) = ≈ 0.1538
( ) w 2 ⎝ z ⎠
1– ( w)
2
1– 13
z z
Fy ( x, y ) = ( x + 2 y )( x 2 + xy + y 2 ) –1 ;
13. f x ( x, y ) = –2 xy sin( x 2 + y 2 ); Fy (–1, 4) =
7
≈ 0.5385
2 2 2 2 2 13
f y ( x, y ) = –2 y sin( x + y ) + cos( x + y )


23. f x ( x, y ) = – y 2 ( x 2 + y 4 ) –1; 31. P (V , T ) =
kT
V
fx ( )
5, – 2 = –
4
21
≈ –0.1905 k
P (V , T ) = ;
T
V
f y ( x, y ) = 2 xy ( x 2 + y 4 ) –1;
k
P (100, 300) = lb/in.2 per degree
( ) 4 5 T
fy 5, – 2 = – ≈ –0.4259 100
21
32. V [ P (V , T )] + T [ P (V , T )]
V T
24. f x ( x, y ) = e y sinh x;
= V (– kTV –2 ) + T (kV –1 ) = 0
f x (–1, 1) = e sinh(–1) ≈ –3.1945
⎛ kT ⎞ ⎛ k ⎞ ⎛ V ⎞ kT PV
f y ( x, y ) = e y cosh x; P VT TP = ⎜ –
V ⎟⎜ ⎟⎜ ⎟=– =– = –1
⎝ V 2 ⎠⎝ P ⎠⎝ k ⎠ PV PV
f y (–1, 1) = e cosh(–1) ≈ 4.1945

x2 y 2 33. f x ( x, y ) = 3 x 2 y – y 3 ; f xx ( x, y ) = 6 xy;
25. Let z = f ( x, y ) = + .
9 4 f y ( x, y ) = x3 – 3xy 2 ; f yy ( x, y ) = –6 xy

f y ( x, y ) =
y Therefore, f xx ( x, y ) + f yy ( x, y ) = 0.
2
The slope is f y (3, 2) = 1.
34. f x ( x, y ) = 2 x( x 2 + y 2 ) –1 ;
f xx ( x, y ) = –2( x 2 – y 2 )( x 2 + y 2 ) –1
26. Let z = f ( x, y ) = (1/ 3)(36 – 9 x 2 – 4 y 2 )1/ 2 .
⎛ 4⎞ f y ( x, y ) = 2 y ( x 2 + y 2 ) –1 ;
f y ( x, y ) = ⎜ − ⎟ y (36 − 9 x 2 − y 2 ) −1/ 2
⎝ 3⎠ f yy ( x, y ) = 2( x 2 − y 2 )( x 2 + y 2 ) −1
8
The slope is f y (1, – 2) = ≈ 0.8040.
3 11 35. Fy ( x, y ) = 15 x 4 y 4 – 6 x 2 y 2 ;

⎛1⎞ Fyy ( x, y ) = 60 x 4 y 3 − 12 x 2 y;
27. z = f ( x, y ) = ⎜ ⎟ (9 x 2 + 9 y 2 − 36)1/ 2
⎝2⎠ Fyyy ( x, y ) = 180 x 4 y 2 – 12 x 2
9x
f x ( x, y ) =
2(9 x + 9 y 2 – 36)1/ 2
2
36. f x ( x, y ) = [– sin(2 x 2 – y 2 )](4 x)
f x (2, 1) = 3
= –4 x sin(2 x 2 – y 2 )
⎛5⎞
28. z = f ( x, y ) = ⎜ ⎟ (16 – x 2 )1/ 2 . f xx ( x, y ) = (–4 x)[cos(2 x 2 – y 2 )](4 x)
⎝4⎠
⎛ 5⎞ + [sin(2 x 2 – y 2 )](–4)
f x ( x, y ) = ⎜ – ⎟ x(16 – x 2 ) –1/ 2
⎝ 4⎠ f xxy ( x, y ) = –16 x 2 [– sin(2 x 2 – y 2 )](–2 y )
5
f x (2, 3) = – ≈ –0.7217 – 4[cos(2 x 2 – y 2 )](–2 y )
4 3
= –32 x 2 y sin(2 x 2 – y 2 ) + 8 y cos(2 x 2 – y 2 )
29. Vr (r , h) = 2πrh;
Vr (6, 10) = 120π ≈ 376.99 in.2 ∂3f
37. a.
∂ y3
2
30. Ty ( x, y ) = 3 y ; Ty (3, 2) = 12 degrees per ft
∂ 3y
b.
∂ y∂ x 2

∂ 4y
c.
∂ y 3∂ x


38. a. f yxx 45. Domain: (Case x < y)
The lengths of the sides are then x, y – x, and
b. f yyxx 1 – y. The sum of the lengths of any two sides
must be greater than the length of the remaining
side, leading to three inequalities:
c. f yyxxx 1
x + (y – x) > 1 – y ⇒ y >
2
39. a. f x ( x, y, z ) = 6 xy – yz 1
(y – x) + (1 – y) > x ⇒ x <
2
b. f y ( x, y, z ) = 3 x 2 – xz + 2 yz 2 ; 1
x + (1 – y) > y – x ⇒ y < x +
f y (0, 1, 2) = 8 2

c. Using the result in a, f xy ( x, y, z ) = 6 x – z.

40. a. 12 x 2 ( x3 + y 2 + z )3

b. f y ( x , y , z ) = 8 y ( x 3 + y 2 + z )3 ;
f y (0, 1, 1) = 64

c. f z ( x, y, z ) = 4( x3 + y 2 + z )3 ; The case for y < x yields similar inequalities
2
f zz ( x, y, z ) = 12( x + y + z )2 2 (x and y interchanged). The graph of DA , the
domain of A is given above. In set notation it is
f x ( x, y, x) = – yze – xyz – y ( xy – z 2 ) –1 ⎧ 1 1 1⎫
41. D A = ⎨ ( x, y ) : x < , y > , y < x + ⎬
⎩ 2 2 2⎭

⎛ 1 ⎞⎛ xy ⎞
–1/ 2
⎛ y⎞ ⎧ 1 1 1⎫
f x ( x, y, z ) = ⎜ ⎟⎜ ⎟ ∪ ⎨ ( x, y ) : x > , y < , x < y + ⎬ .
42. ⎜ ⎟; ⎩ 2 2 2⎭
⎝ 2 ⎠⎝ z ⎠ ⎝z⎠
–1/ 2 Range: The area is greater than zero but can be
⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1⎞ 1 arbitrarily close to zero since one side can be
f x (–2, – 1, 8) = ⎜ ⎟ ⎜ ⎟ ⎜– ⎟ = –
⎝ 2 ⎠⎝ 4 ⎠ ⎝ 8⎠ 8 arbitrarily small and the other two sides are
bounded above. It seems that the area would be
43. If f ( x, y ) = x 4 + xy3 + 12, f y ( x, y ) = 3 xy 2 ; largest when the triangle is equilateral. An
1
f y (1, – 2) = 12. Therefore, along the tangent line equilateral triangle with sides equal to has
3
Δy = 1 ⇒ Δz = 12, so 0, 1, 12 is a tangent 3 ⎛ 3⎤
area . Hence the range of A is ⎜ 0,
⎜ ⎥ . (In
vector (since Δx = 0). Then parametric equations 36 ⎝ 36 ⎦
⎧ x =1 ⎫ Sections 8 and 9 of this chapter methods will be
⎪ ⎪
of the tangent line are ⎨ y = –2 + t ⎬ . Then the presented which will make it easy to prove that
⎪ z = 5 + 12t ⎪ the largest value of A will occur when the triangle
⎩ ⎭
is equilateral.)
point of xy-plane at which the bee hits is
(1, 0, 29) [since y = 0 ⇒ t = 2 ⇒ x = 1, z = 29].
46. a. u = cos (x) cos (ct):
u x = – sin( x ) cos(ct ) ; ut = – c cos( x) sin(ct )
44. The largest rectangle that can be contained in the
circle is a square of diameter length 20. The edge u xx = – cos( x) cos(ct )
of such a square has length 10 2, so its area is utt = – c 2 cos( x) cos(ct )
200. Therefore, the domain of A is
Therefore, c 2 u xx = utt .
2 2
{( x, y ) : 0 ≤ x + y < 400}, and the range is
u = e x cosh(ct ) :
(0, 200].
u x = e x cosh(ct ), ut = ce x sinh(ct )
u xx = e x cosh(ct ), utt = c 2 e x cosh(ct )
Therefore, c 2u xx = utt .


b. u = e – ct sin( x) : 48. a. sin( x + y 2 )
u x = e – ct cos x
u xx = – e – ct sin x
ut = – cect sin x
Therefore, cu xx = ut .
2
u = t –1/ 2 e – x / 4ct :
2 / 4ct ⎛ x ⎞
u x = t –1/ 2 e – x ⎜– ⎟ b. Dx (sin( x + y 2 ))
⎝ 2ct ⎠
( x 2 – 2ct )
u xx =
2 / 4ct
(4c 2 t 5 / 2 e x )
2
( x – 2ct )
ut =
2
(4ct 5 / 2 e x / 4ct )
Therefore, cu xx = ut .

47. a. Moving parallel to the y-axis from the point c. D y (sin( x + y 2 ))
(1, 1) to the nearest level curve and
Δz
approximating , we obtain
Δy
4–5
f y (1, 1) = = –4.
1.25 – 1

b. Moving parallel to the x-axis from the point
(–4, 2) to the nearest level curve and
Δz d. Dx ( D y (sin( x + y ) 2 ))
approximating , we obtain
Δx
1– 0 2
f x (–4, 2) ≈ = .
–2.5 – (–4) 3

c. Moving parallel to the x-axis from the point
(–5, –2) to the nearest level curve and
Δz
approximately , we obtain
Δx
1– 0 2 49. a. f y ( x, y , z )
f x (–4, – 5) ≈ = .
–2.5 – (–5) 5 f ( x , y + Δy , z ) − f ( x , y , z )
= lim
Δy →0 Δy
d. Moving parallel to the y-axis from the point
(0, –2) to the nearest level curve and
b. f z ( x, y , z )
Δz
approximating , we obtain f ( x , y , z + Δz ) − f ( x , y , z )
Δy = lim
Δz →0 Δz
0 –1 8
f y (0, 2) ≈ = .
–19 – (–2) 3
8 c. Gx ( w, x, y, z )
G ( w, x + Δx, y , z ) − G ( w, x, y, z )
= lim
Δx →0 Δx


∂ 4. The limit does not exist because of Theorem A.
d. λ ( x, y , z , t ) The function is a rational function, but the limit
∂z
of the denominator is 0, while the limit of the
λ ( x , y , z + Δz , t ) − λ ( x , y , z , t )
= lim numerator is -1.
Δz →0 Δz
5
∂ 5. –
e. S (b0 , b1 , b2 ,… , bn ) = 2
∂b2
⎛ S (b0 , b1 , b2 + Δb2 ,… , bn ) ⎞ 6. −1
⎜ ⎟
− S (b0 , b1 , b2 ,… , bn ) ⎟ 7. 1
= lim ⎜
Δb2 →0 ⎜ Δb2 ⎟
⎜ ⎟ tan( x 2 + y 2 )
⎝ ⎠ 8. lim
( x, y )→(0, 0) ( x2 + y2 )
∂ sin( x 2 + y 2 )
50. a. ( sin w sin x cos y cos z ) = lim
1
∂w
( x, y )→(0, 0) ( x + y ) cos( x + y 2 )
2 2 2
= cos w sin x cos y cos z = (1)(1) = 1

∂ wyz 9. The limit does not exist since the function is not
b. ⎡ x ln ( wxyz ) ⎤ = x ⋅
⎣ ⎦ + 1 ⋅ ln ( wxyz )
∂x wxyz defined anywhere along the line y = x. That is,
there is no neighborhood of the origin in which
= 1 + ln ( wxyz ) the function is defined everywhere except
possibly at the origin.
c. λt ( x, y, z , t )
(1 + xyzt ) cos x − t ( cos x ) xyz ( x 2 + y 2 )( x 2 – y 2 )
= 10. lim
( x, y )→(0, 0) x2 + y 2
(1 + xyzt )2
cos x = lim ( x2 – y 2 ) = 0
= ( x, y )→(0, 0)
(1 + xyzt )2
11. Changing to polar coordinates,
xy r cos θ ⋅ r sin θ
lim = lim
12.3 Concepts Review ( x, y )→(0,0) x 2 + y 2 r →0 r

1. 3; (x, y) approaches (1, 2). = lim r cos θ ⋅ sin θ = 0
r →0
2. lim f ( x, y ) = f (1, 2)
( x, y )→(1, 2) 12. If ( x, y ) approaches (0, 0) along the line y = x ,

3. contained in S x2 1
lim = lim = +∞
2 2 2
( x, x )→(0,0) 4 x 2
+x )
( x, x )→(0,0) ( x
4. an interior point of S; boundary points Thus, the limit does not exist.

13. Use polar coordinates.
Problem Set 12.3
r 7 / 3 ( cos θ )
7/3
x7 / 3
= r1/ 3 ( cos θ )
7/3
=
1. –18 x +y2 2
r 2

2. 3 r 1/ 3
( cos θ ) 7/3
→ 0 as r → 0 , so the limit is 0.

⎡ 2 ⎛ xy ⎞ ⎤ 14. Changing to polar coordinates,
3. lim ⎢ x cos xy – sin ⎜ 3 ⎟ ⎥
( x, y )→(2, π) ⎣ ⎝ ⎠⎦ r 2 cos 2 θ − r 2 sin 2 θ
lim r 2 cos θ sin θ ⋅
⎛ 2π ⎞ 3 r →0 r2
= 2 cos 2 2π – sin ⎜ ⎟ = 2 – ≈ 1.1340
⎝ 3 ⎠ 2 = lim r 2 cos θ sin θ cos 2θ = 0
r →0


r 4 cos 2 θ sin 2 θ 26. Require 4 − x 2 − y 2 − z 2 > 0;
15. f ( x, y ) =
r 2 cos 2 θ + r 4 sin 4 θ x 2 + y 2 + z 2 < 4. S is the space in the interior
⎛ cos 2 θ sin 2 θ ⎞ of the sphere centered at the origin with radius 2.
= r2 ⎜ ⎟
⎜ cos 2 θ + r 2 sin 4 θ ⎟
⎝ ⎠
27. The boundary consists of the points that form the
If cos θ = 0 , then f ( x, y ) = 0 . If cos θ ≠ 0 , outer edge of the rectangle. The set is closed.
hen this converges to 0 as r → 0 . Thus the
limit is 0.

16. As ( x, y ) approaches (0,0) along x = y2,
y4 1
lim = . Along the x-axis,
4 24
y +y
( x, x )→(0,0)

0
however, lim = 0. Thus, the limit does
( x,0)→(0,0) x 2
not exist.

17. f ( x, y ) is continuous for all ( x, y ) since
28. The boundary consists of the points of the circle
for all ( x, y ), x 2 + y 2 + 1 ≠ 0. shown. The set is open.

18. f ( x, y ) is continuous for all ( x, y ) since
for all ( x, y ), x 2 + y 2 + 1 > 0.

19. Require 1 – x 2 – y 2 > 0; x 2 + y 2 < 1. S is the
interior of the unit circle centered at the origin.

20. Require 1 + x + y > 0; y > − x − 1. S is the set
of all ( x, y ) above the line y = − x − 1.
29. The boundary consists of the circle and the
2
21. Require y – x ≠ 0. S is the entire plane except origin. The set is neither open (since, for
example, (1, 0) is not an interior point), nor
the parabola y = x 2 . closed (since (0, 0) is not in the set).

22. The only points at which f might be
discontinuous occur when xy = 0.
sin( xy )
lim = 1 = f (a, 0) for all nonzero
( x, y )→( a , 0) xy
a in , and then
sin( xy )
lim = 1 = f (0, b) for all b in .
( x, y )→(0, b ) xy
Therefore, f is continuous on the entire plane.

23. Require x – y + 1 ≥ 0; y ≤ x + 1. S is the region
below and on the line y = x + 1.

24. Require 4 – x 2 – y 2 > 0; x 2 + y 2 < 4. S is the
interior of the circle of radius 2 centered at the
origin.

25. f ( x, y, z ) is continuous for all ( x, y, z ) ≠ ( 0, 0, 0 )
since for all ( x, y, z ) ≠ ( 0, 0, 0 ) , x 2 + y 2 + z 2 > 0.


30. The boundary consists of the points on the line 0
x = 1 along with the points on the line x = 4. The 35. Along the x-axis (y = 0): lim =0.
( x, y )→(0, 0) x 2 +0
set is neither closed nor open.
Along y = x:
x2 1 1
lim = = .
lim
( x, y )→(0, 0) 2 x 2
( x, y )→(0, 0) 2 2
Hence, the limit does not exist because for some
points near the origin f(x, y) is getting closer to 0,
1
but for others it is getting closer to .
2

0
36. Along y = 0: lim = 0. Along y = x:
x→ 0 x 2 +0
31. The boundary consists of the graph of
⎛1⎞ x 2 + x3 1+ x 1
y = sin ⎜ ⎟ along with the part of the y-axis for lim = lim = .
x→ 0 x 2 +x 2 x→ 0 2 2
⎝ x⎠
which y ≤ 1. The set is open.
x 2 (mx) mx3
37. a. lim = lim
x→ 0 x 4 + (mx)2 x→ 0 x 4 + m2 x 2
mx
= lim =0
x→ 0 x 2 + m2

x2 ( x2 ) x4 1 1
b. lim = lim = lim =
x→ 0 x 4 + (x )2 2 x→ 0 2 x 4 x→ 0 2 2

32. The boundary is the set itself along with the x2 y
origin. The set is neither open (since none of its c. lim does not exist.
( x, y )→( 0, 0) x 4 +y
points are interior points) nor closed (since the
origin is not in the set).
38. f is discontinuous at each overhang. More
interesting, f is discontinuous along the
Continental Divide.

39. a. {( x, y, z ) : x 2 + y 2 = 1, z in [1, 2]} [For
x 2 + y 2 < 1, the particle hits the hemisphere
and then slides to the origin (or bounds
toward the origin); for x 2 + y 2 = 1, it

2 2
bounces up; for x 2 + y 2 > 1, it falls straight
x – 4y ( x + 2 y )( x – 2 y )
33. = = x + 2 y (if x ≠ 2y) down.]
x – 2y x – 2y
If x = 2 y , x + 2 y = 2 x . Take g ( x ) = 2 x . b. {( x, y, z ) : x 2 + y 2 = 1, z = 1} (As one moves
at a level of z = 1 from the rim of the bowl
34. Let L and M be the latter two limits. toward any position away from the bowl
[ f ( x, y ) + g ( x, y )] – [ L + M ] there is a change from seeing all of the
interior of the bowl to seeing none of it.)
ε ε
≤ f ( x, y ) – L + f ( x , y ) – M ≤ +
2 2 c. {(x, y, z): z = 1} [f(x, y, z) is undefined
for (x, y) in some δ-neighborhood of (a, b). (infinite) at (x, y, 1).]
Therefore,
lim [ f ( x, y ) + g ( x, y )] = L + M . d. φ (Small changes in points of the domain
( x , y ) →( a , b ) result in small changes in the shortest path
from the points to the origin.)


40. f is continuous on an open set D and P0 is in D implies that there is neighborhood of P0 with radius r on which f
is continuous. f is continuous at P0 ⇒ lim f ( P) = f ( P0 ). Now let ε = f ( P0 ) which is positive. Then there is a δ
P → P0
such that 0 < δ < r and f ( p ) – f ( P0 ) < f ( P0 ) if P is in the δ-neighborhood of P0 . Therefore,
– f ( P0 ) < f ( p ) – f ( P0 ) < f ( P0 ), so 0 < f(p) (using the left-hand inequality) in that δ-neighborhood of P0 .

⎧( x 2 + y 2 )1/ 2 + 1 if y ≠ 0 ⎫
⎪ ⎪
41. a. f ( x, y ) = ⎨ ⎬ . Check discontinuities where y = 0.
⎪
⎩ x –1 if y = 0 ⎪
⎭
As y = 0, ( x 2 + y 2 )1/ 2 + 1 = x + 1, so f is continuous if x + 1 = x – 1 . Squaring each side and simplifying
yields x = – x, so f is continuous for x ≤ 0. That is, f is discontinuous along the positive x-axis.
b. Let P = (u, v) and Q = (x, y).
⎧
⎪ OP + OQ if P and Q are not on same ray from the origin and neither is the origin ⎫
⎪
f (u, v, x, y ) = ⎨ ⎬.
⎪ PQ
⎩ otherwise ⎪
⎭
This means that in the first case one travels from P to the origin and then to Q; in the second case one travels
directly from P to Q without passing through the origin, so f is discontinuous on the set
{(u, v, x, y ) : u, v = k x, y for some k > 0, u , v ≠ 0, x, y ≠ 0}.

⎛ hy ( h2 – y 2 ) ⎞
⎜ 2 2 –0⎟ 2 2
42. a. f x (0, y ) = lim ⎜
h +y ⎟ = lim y (h − y ) = − y
h →0 ⎜ h ⎟ h →0 h 2 + y 2
⎜ ⎟
⎝ ⎠
⎛ xh ( x 2 – h 2 ) ⎞
⎜ –0⎟
x 2 + h2 y( x2 – h2 )
b. f y ( x, 0) = lim ⎜ ⎟ = lim =x
h →0 ⎜ h ⎟ h →0 x 2 + y 2
⎜ ⎟
⎝ ⎠
f y (0 + h, y ) – f y (0, y ) h–0
c. f yx (0, 0) = lim = lim =1
h →0 h h →0 h

f x ( x, 0 + h) – f x ( x, 0) –h – 0
d. f xy (0, 0) = lim = lim = –1
h →0 h h →0 h
Therefore, f xy (0, 0) ≠ f yx (0, 0).

43. b.

44. a. 45.


46. A function f of three variables is continuous at a
point (a, b, c ) if f (a, b, c) is defined and equal to
the limit of f ( x, y, z ) as ( x, y, z ) approaches
(a, b, c) . In other words,
lim f ( x, y, z ) = f (a, b, c) .
( x , y , z ) → ( a ,b , c )
A function of three variables is continuous on an
open set S if it is continuous at every point in the
interior of the set. The function is continuous at
a boundary point P of S if f (Q) approaches
f (P) as Q approaches P along any path through
points in S in the neighborhood of P.

47. If we approach the point ( 0, 0, 0 ) along a straight
path from the point ( x, x, x ) , we have
x ( x )( x) x3 1
lim = lim =
3 3 3 3 3
x +x +x
( x, x, x )→(0,0,0) ( x , x, x )→(0,0,0) 3 x
Since the limit does not equal to f (0, 0, 0) , the
function is not continuous at the point (0, 0, 0) .

48. If we approach the point (0, 0, 0) along the
x-axis, we get
( x 2 − 02 ) x2
lim (0 + 1) = lim =1
( x,0,0)→(0,0,0) ( x 2 + 02 ) ( x,0,0)→(0,0,0) x 2
Since the limit does not equal f (0, 0, 0) , the
function is not continuous at the point (0, 0, 0).

1. gradient
2. locally linear

∂f ∂f
3. (p)i + (p) j; y 2 i + 2 xyj
∂x ∂y

4. tangent plane

Problem Set 12.4

1. 2 xy + 3 y, x 2 + 3 x

2. 3 x 2 y , x3 – 3 y 2

3. ∇f ( x, y ) = ( x)(e xy y ) + (e xy )(1), xe xy x = e xy xy + 1, x 2

4. 2 xy cos y, x 2 (cos y – y sin y )

5. x( x + y ) –2 y ( x + 2), x 2

6. ∇f ( x, y ) = 3[sin 2 ( x 2 y )][cos( x 2 y )](2 xy ), 3[sin 2 ( x 2 y )][cos( x 2 y )]( x 2 ) = 3x sin 2 ( x 2 y ) cos( x 2 y ) 2 y, x


7. ( x 2 + y 2 + z 2 ) –1/ 2 x, y, z

8. 2 xy + z 2 , x 2 + 2 yz , y 2 + 2 xz

9. ∇f ( x, y ) = ( x 2 y )(e x – z ) + (e x – z )(2 xy ), x 2 e x – z , x 2 ye x – z (–1) = xe x – z y ( x + 2), x, – xy

10. xz ( x + y + z ) –1 + z ln( x + y + z ), xz ( x + y + z ) –1 , xz ( x + y + z ) –1 + x ln( x + y + z )

11. ∇f ( x, y ) = 2 xy – y 2 , x 2 – 2 xy ; ∇f (–2, 3) = –21, 16
z = f (–2, 3) + –21, 16 ⋅ x + 2, y – 3 = 30 + (–21x – 42 + 16 y – 48)
z = –21x + 16y – 60

12. ∇f ( x, y ) = 3x 2 y + 3 y 2 , x3 + 6 xy , so ∇f (2, – 2) = (–12, – 16).
Tangent plane:
z = f (2, – 2) + ∇ (2, – 2) ⋅ x – 2, y + 2 = 8 + –12, – 16 ⋅ x – 2, y + 2 = 8 + (–12x + 24 – 16y – 32)
z = –12x – 16y

13. ∇f ( x, y ) = – π sin(πx) sin(πy ), π cos(πx) cos(πy ) + 2π cos(2πy )
⎛ 1⎞
∇f ⎜ –1, ⎟ = 0, – 2π
⎝ 2⎠
⎛ 1⎞ 1
z = f ⎜ –1, ⎟ + 0, – 2π ⋅ x + 1, y – = –1 + (0 – 2πy + π);
⎝ 2⎠ 2
z = –2 π y + ( π – 1)

2x x2
14. ∇f ( x, y ) = ,− ; ∇f (2, − 1) = −4, − 4
y y2
z = f (2, – 1) + −4, – 4 ⋅ x – 2, y + 1
= –4 + (–4x + 8 –4y – 4)
z = –4x – 4y

15. ∇f ( x, y, z ) = 6 x + z 2 , – 4 y, 2 xz , so ∇f (1, 2, – 1) = 7, – 8, – 2
Tangent hyperplane:
w = f (1, 2, – 1) + ∇f (1, 2, – 1) ⋅ x – 1, y – 2, z + 1 = –4 + 7, – 8, – 2 ⋅ x – 1, y – 2, z + 1
= –4 + (7x – 7 – 8y + 16 – 2z – 2)
w = 7x – 8y – 2z + 3

16. ∇f ( x, y, z ) = yz + 2 x, xz , xy ; ∇f (2, 0, – 3) = 4, – 6, 0
w = f (2, 0, – 3) + 4, – 6, 0 ⋅ x – 2, y, z + 3 = 4 + (4x – 8 – 6y + 0)
w = 4x – 6y – 4

⎛ f ⎞ gf x − fg x , gf y − fg y , gf z − fg z g fx , f y , fz – f gx , g y , gz g ∇f – f ∇g
17. ∇ ⎜ ⎟ = = =
⎝g⎠ g2 g2 g2

18. ∇( f r ) = rf r –1 f x , rf r –1 f y , rf r –1 f z = rf r –1 f x , f y , f z = rf r –1∇f


19. Let F ( x, y, z ) = x 2 − 6 x + 2 y 2 − 10 y + 2 xy − z = 0 respectively. Again we find the cross product
of the normal vectors:
∇F ( x, y, z ) = 2 x − 6 + 2 y, 4 y − 10 + 2 x, −1
12,10, −1 × 1, 0, 0 = 0,10,10
The tangent plane will be horizontal if
Thus, parametric equations for the desired
∇F ( x, y, z ) = 0, 0, k , where k ≠ 0 . Therefore, tangent line are x = 2
we have the following system of equations: y = 1 + 10t
2x + 2 y − 6 = 0
z = 9 + 10t
2 x + 4 y − 10 = 0
c. Using the equation for the tangent plane from
Solving this system yields x = 1 and y = 2 .
the first part, we now want the vertical plane
Thus, there is a horizontal tangent plane at to be parallel to the line y = x , but still pass
( x, y ) = (1, 2 ) . through the projected point (2,1, 0) . The
vertical plane now has equation y − x + 1 = 0 .
20. Let F ( x, y, z ) = x3 − z = 0 The normal equations are given by
∇F ( x, y , z ) = 3 x 2 , 0, −1 12,10, −1 and 1, −1, 0 for the tangent and

The tangent plane will be horizontal if vertical planes respectively. Again we find
the cross product of the normal vectors:
∇F ( x, y, z ) = 0, 0, k , where k ≠ 0 . Therefore,
12,10, −1 × 1, −1, 0 = −1, −1, −22
we need only solve the equation 3x 2 = 0 . There Thus, parametric equations for the desired
is a horizontal tangent plane at ( x, y ) = (0, y ). tangent line are x = 2 − t
(Note: there are infinitely many points since y y = 1− t
can take on any value).
z = 9 − 22t
21. a. The point (2,1,9) projects to (2,1, 0) on the xy
plane. The equation of a plane containing this 22. a. The point (3, 2, 72) on the surface is the point
point and parallel to the x-axis is given by (3, 2, 0) when projected into the xy plane. The
y = 1 . The tangent plane to the surface at the equation of a plane containing this point and
point (2,1,9) is given by parallel to the x-axis is given by y = 2 . The
z = f (2,1) + ∇f (2,1) ⋅ x − 2, y − 1 tangent plane to the surface at the point
(3, 2, 72) is given by
= 9 + 12,10 x − 2, y − 1
z = f (3, 2) + ∇f (3, 2) ⋅ x − 3, y − 2
= 12 x + 10 y − 25
= 72 + 48,108 x − 3, y − 2
The line of intersection of the two planes is
the tangent line to the surface, passing = 48 x + 108 y − 288
through the point (2,1,9) , whose projection in The line of intersection of the two planes is
the xy plane is parallel to the x-axis. This line the tangent line to the surface, passing
of intersection is parallel to the cross product through the point (3, 2, 72) , whose projection
of the normal vectors for the planes. The in the xy plane is parallel to the x-axis. This
normal vectors are 12,10, −1 and 0,1, 0 for line of intersection is parallel to the cross
the tangent plane and vertical plane product of the normal vectors for the planes.
respectively. The cross product is given by The normal vectors are
12,10, −1 × 0,1, 0 = 1, 0,12 48,108, −1 and 0, 2, 0 for the tangent plane
Thus, parametric equations for the desired and vertical plane respectively. The cross
tangent line are x = 2 + t product is given by
48,108, −1 × 0, 2, 0 = 2, 0,96
y =1
Thus, parametric equations for the desired
z = 9 + 12t
tangent line are
b. Using the equation for the tangent plane from x = 3 + 2t
the previous part, we now want the vertical y=2
plane to be parallel to the y-axis, but still pass
z = 72 + 96t
through the projected point (2,1, 0) . The
vertical plane now has equation x = 2 . The
normal equations are given by 12,10, −1 and
1, 0, 0 for the tangent and vertical planes

Instructor's Resource Manual Section 12.4 759

b. Using the equation for the tangent plane from c. Using the equation for the tangent plane from
the previous part, we now want the vertical the first part, we now want the vertical plane
plane to be parallel to the y-axis, but still pass to be parallel to the line x = − y , but still pass
through the projected point (3, 2, 72) . The through the projected point (3, 2, 72) . The
vertical plane now has equation x = 3 . The vertical plane now has equation y + x − 5 = 0 .
normal equations are given by The normal equations are given by
48,108, −1 and 3, 0, 0 for the tangent and 48,108, −1 and 1,1, 0 for the tangent and
vertical planes respectively. Again we find vertical planes respectively. Again we find
the cross product of the normal vectors: the cross product of the normal vectors:
48,108, −1 × 3, 0, 0 = 0, −3, −324 48,108, −1 × 1,1, 0 = 1, −1, −60
Thus, parametric equations for the desired Thus, parametric equations for the desired
tangent line are tangent line are
x=3 x = 3+t
y = 2 − 3t y = 2−t
z = 72 − 324t z = 72 − 60t

⎛ 1 xy ⎞ ⎛ 1 xy ⎞ –5 xy ⎡ a a ⎤
23. ∇f ( x, y ) = –10 ⎜ y ⎟ , – 10 ⎜ x⎟ = y, x ⎢ Note that = .⎥
⎜ 2 xy xy ⎟ ⎜ 2 xy xy ⎟ xy
3/ 2
⎢
⎣ a a ⎥
⎦
⎝ ⎠ ⎝ ⎠
∇f (1, – 1) = –5, 5
Tangent plane:
z = f (1, – 1) + ∇f (1, – 1) ⋅ x – 1, y + 1 = –10 + –5, 5 ⋅ x – 1, y + 1 = –10 + (–5 x + 5 + 5 y + 5)
z = –5x + 5y

24. Let a be any point of S and let b be any other 26. f(b) – f(a) = f 2, 6 − f 0, 0 = 0 − 2 = −2
point of S. Then for some c on the line segment
between a and b: −x
∇f ( x , y ) = , 0 ; b – a = 2, 6
f (b) − f (a) = ∇f (c) ⋅ (b − a) = 0 ⋅ (b − a) = 0, so
4 − x2
f(b) = f(a) (for all b in S).
The value c = cx , c y will be the solution to
25. f(a) – f(b) = f 2,1 − f 0, 0 = 4 − 9 = −5
−c x
∇f ( x, y ) = −2 x, −2 y ; b – a = 2,1 −2 = ,0 2, 6
4 − cx 2
The value c = cx , c y will be a solution to
−2cx
−5 = −2cx , −2c y 2,1 −2 = ⇒ cx = 2
4 − cx 2
{
c ∈ cx , c y : 4cx + 2c y = 5 } Since c must be between a and b, c must lie on
In order for c to be between a and b, c must lie the line y = 3x. Since cx = 2, c y = 3 2.
on the line y = 1 x . Consequently, c will be the
2 Thus, c = 2,3 2 .
solution to the following system of equations:
4cx + 2c y = 5 and c y = 1 c x . The solution is
2
27. ∇f (p) = ∇g (p) ⇒ ∇[ f (p) – g (p)] = 0
⇒ f (p) – g (p) is a constant.
c = 1, 1 .
2


28. ∇f (p) = p ⇒ ∇f ( x, y ) = x, y 31. a. (i)
⇒ f x ( x , y ) = x, f y ( x, y ) = y ∂( f + g ) ∂( f + g ) ∂( f + g )
∇[ f + g ] = i+ j+ k
∂x ∂y ∂z
1 2
⇒ f ( x, y ) = x + α ( y ) for any function of y, ∂f ∂g ∂f ∂g ∂f ∂g
2 = i+ i+ j+ j+ k + k
∂x ∂x ∂y ∂y ∂z ∂z
1
and f ( x, y ) = y 2 + β ( x) for any function of x. ∂f ∂f ∂f ∂g ∂g ∂g
2 = i+ j+ k + i+ j+ k
1 ∂x ∂y ∂z ∂x ∂y ∂z
⇒ f ( x, y ) = ( x 2 + y 2 ) + C for any C in . = ∇f + ∇g
2

29. ∂[α f ] ∂[α f ] ∂[α f ]
(ii) ∇[α f ] = i+ j+ k
∂x ∂y ∂z
∂[ f ] ∂[ f ] ∂[ f ]
=α i +α j+α k
∂x ∂y ∂z
= α∇f

(iii)
− xy ∂ ( fg ) ∂ ( fg ) ∂ ( fg )
∇[ fg ] = i+ j+ k
∂x ∂y ∂z
⎛ ∂g ∂f ⎞ ⎛ ∂g ∂f ⎞
=⎜ f + g ⎟i + ⎜ f + g ⎟j
⎝ ∂x ∂x ⎠ ⎝ ∂y ∂y ⎠
⎛ ∂g ∂f ⎞
+⎜ f + g ⎟k
⎝ ∂z ∂z ⎠
⎛ ∂g ∂g ∂g ⎞
= f ⎜ i+ j+ k ⎟
⎝ ∂x ∂y ∂z ⎠
⎛ ∂f ∂f ∂f ⎞
+g ⎜ i + j+ k ⎟
a. The gradient points in the direction of ⎝ ∂x ∂y ∂z ⎠
greatest increase of the function. = f ∇g + g ∇f

b. No. If it were, 0 + h – 0 = 0 + h δ (h) where b. (i)
δ (h) → 0 as h → 0, which is possible. ∂( f + g ) ∂( f + g )
∇[ f + g ] = i1 + i2
∂x1 ∂x2
30. ∂( f + g )
+ + in
∂xn
∂f ∂g ∂f ∂g
= i1 + i1 + i2 + i2
∂x1 ∂x1 ∂x2 ∂x2
∂f ∂g
+ + in + in
∂xn ∂xn
sin(x) + sin(y) – sin(x + y) ∂f ∂f ∂f
= i1 + i2 + + in
∂x1 ∂x2 ∂xn
∂g ∂g ∂g
+ i1 + i2 + + in
∂x1 ∂x2 ∂xn
= ∇f + ∇ g


(ii) 4. Du f ( x, y )
∂[α f ] ∂[α f ] ∂[α f ] ⎡⎛ 1 ⎞ ⎤
∇[α f ] = i1 + i2 + + in
∂x1 ∂x2 ∂xn = 2 x – 3 y , – 3 x + 4 y ⋅ ⎢⎜ ⎟ 2, – 1 ⎥ ;
⎣⎝ 5 ⎠ ⎦
∂[ f ] ∂[ f ] ∂[ f ]
=α i1 + α i2 + +α in 27
∂x1 ∂x2 ∂xn Du f (–1, 2) = – ≈ –12.0748
5
= α∇f
⎡⎛ 1 ⎞ ⎤
(iii) 5. Du f ( x, y ) = e x sin y, cos y ⋅ ⎢⎜ ⎟ 1, 3 ⎥ ;
⎣ ⎝2⎠ ⎦
∂ ( fg ) ∂ ( fg ) ∂ ( fg )
∇[ fg ] =
∂x1
i1 +
∂x2
i2 + +
∂xn
in
⎛ π⎞
Du f ⎜ 0, ⎟ =
( 2+ 6 ) ≈ 0.9659
⎛ ∂g ∂f ⎞ ⎛ ∂g ∂f ⎞ ⎝ 4⎠ 4
=⎜ f +g ⎟ i1 + ⎜ f +g ⎟ i2
⎝ ∂x1 ∂x1 ⎠ ⎝ ∂x2 ∂x2 ⎠
–1, 3
⎛ ∂g ∂f ⎞ 6. Du f ( x, y ) = – ye – xy – xe – xy ⋅
+ +⎜ f +g ⎟ in 2
⎝ ∂xn ∂xn ⎠
–1, 3
⎛ ∂g ∂g ∂g ⎞ Du f (1, –1) = e, – e ⋅ =
–e – e 3
= f⎜ i1 + i2 + + in ⎟
⎝ ∂x1 ∂x2 ∂xn ⎠ 2 2
≈ –3.7132
⎛ ∂f ∂f ∂f ⎞
+g ⎜ i1 + i2 + + in ⎟
⎝ ∂x1 ∂x2 ∂xn ⎠ 7. Du f ( x, y, z ) =
= f ∇ g + g ∇f ⎡⎛ 1 ⎞ ⎤
= 3 x 2 y, x3 – 2 yz 2 , – 2 y 2 z ⋅ ⎢⎜ ⎟ 1, – 2, 2 ⎥ ;
⎣⎝ 3 ⎠ ⎦
52
Du f (–2, 1, 3) =
12.5 Concepts Review 3

[ f (p + hu) – f (p)] ⎡⎛ 1 ⎞ ⎤
1. 8. Du f ( x, y, z ) = 2 x, 2 y, 2 z ⋅ ⎢⎜ ⎟ 2, –1, –1 ⎥ ;
h ⎣⎝ 2 ⎠ ⎦
Du f (1, – 1, 2) = 2 – 1 ≈ 0.4142
2. u1 f x ( x, y ) + u2 f y ( x, y )
9. f increases most rapidly in the direction of the
3. greatest increase
gradient. ∇f ( x, y ) = 3 x 2 , – 5 y 4 ;
4. level curve ∇f (2, – 1) = 12, – 5
12, – 5
Problem Set 12.5 is the unit vector in that direction. The
13
rate of change of f(x, y) in that direction at that
3 4 8
1. Du f ( x, y ) = 2 xy, x 2 ⋅ , − ; Du f (1, 2) = point is the magnitude of the gradient.
5 5 5 12, – 5 = 13

⎡⎛ 1 ⎞ ⎤
2. Du f ( x, y ) = x –1 y 2 , 2 y ln x ⋅ ⎢⎜ ⎟ 1, – 1 ⎥ ; 10. ∇f ( x, y ) = e y cos x, e y sin x ;
⎣⎝ 2 ⎠ ⎦
Du f (1, 4) = 8 2 ≈ 11.3137 ⎛ 5π ⎞ 3 1
∇f ⎜ , 0 ⎟ = – , , which is a unit vector.
⎝ 6 ⎠ 2 2
⎛ a⎞ The rate of change in that direction is 1.
3. Du f ( x, y ) = f ( x, y ) ⋅ u ⎜ where u = ⎟
⎜ a⎟
⎝ ⎠
1, – 1
= 4 x + y, x – 2 y ⋅ ;
2
1, – 1 3
Du f (3, – 2) = 10, 7 ⋅ = ≈ 2.1213
2 2


11. ∇f ( x, y, z ) = 2 xyz , x 2 z , x 2 y ; 16. At (2, 1), x 2 + 4 y 2 = 8 is the level curve.
∇f ( x , y ) = 2 x , 8 y
f (1, – 1, 2) = –4, 2, – 1
A unit vector in that direction is ∇f (2, 1) = 4 1, 2 , which is perpendicular to the
⎛ 1 ⎞ level curve at (2, 1).
⎜ ⎟ −4, 2, − 1 . The rate of change in that
⎝ 21 ⎠
direction is 21 ≈ 4.5826.

12. f increases most rapidly in the direction of the
gradient. ∇f ( x, y, z ) = e yz , xze yz , xye yz ;
∇f (2, 0, – 4) = 1, – 8, 0
1, – 8, 0
is a unit vector in that direction.
65
2 2 1
1, – 8, 0 = 65 ≈ 8.0623 is the rate of change 17. u = ,– ,
3 3 3
of f(x, y, z) in that direction at that point. 2 2 1
Du f ( x, y, z ) = y, x, 2 z ⋅ ,– ,
13. –∇f ( x, y ) = 2 x, y ; –∇f (–1, 2) = 2 –1, 2 is 3 3 3
2
the direction of most rapid decrease. A unit Du f (1, 1, 1) =
⎛ 1 ⎞ 3
vector in that direction is u = ⎜ ⎟ –1, 2 .
⎝ 5⎠ ⎛ π⎞
18. ⎜ 0, ⎟ is on the y-axis, so the unit vector toward
⎝ 3⎠
14. – ∇f ( x, y ) = –3cos(3 x – y ), cos(3 x – y ) ;
the origin is –j.
⎛π π⎞ ⎛ 1 ⎞ Du ( x, y ) = – e – x cos y , – e – x sin y ⋅ 0, – 1
– ∇f ⎜ , ⎟ = ⎜ ⎟ –3, 1 is the direction of
⎝6 4⎠ ⎝ 2⎠
most rapid decrease. A unit vector in that = e – x sin y;
⎛ 1 ⎞ ⎛ π⎞ 3
direction is ⎜ ⎟ –3, 1 . Du ⎜ 0, ⎟ =
⎝ 10 ⎠ ⎝ 3⎠ 2

y 19. a. Hottest if denominator is smallest; i.e., at the
15. The level curves are = k . For p = (1, 2), origin.
2
x
y
k = 2, so the level curve through (1, 2) is =2 –200 2 x, 2 y, 2 z
x2 b. ∇T ( x , y , z ) = ;
(5 + x 2 + y 2 + z 2 )2
or y = 2 x (x ≠ 0).
2
⎛ 25 ⎞
∇T (1, – 1, 1) = ⎜ – ⎟ 1, – 1, 1
∇f ( x, y ) = –2 yx –3 , x –2 ⎝ 4 ⎠
∇f (1, 2) = –4, 1 , which is perpendicular to the −1,1, −1 is one vector in the direction of
parabola at (1, 2). greatest increase.

c. Yes

20. – ∇V ( x, y, z )
2 + y2 + z 2 )
= –100e –( x –2 x, – 2 y, – 2 z
–( x 2 + y 2 + z 2 )
= 200e x, y, z is the direction of
greatest decrease at (x, y, z), and it points away
from the origin.


21. ∇ f ( x, y , z ) 24. The unit vector from (2, 4) toward (5, 0) is
1 3 4
( )
− , – . Then
= x x2 + y 2 + z 2 2 cos x 2 + y 2 + z 2 , 5 5
3 4
Du f (2, 4) = –3, 8 ⋅ , – = –8.2.
1 5 5
( )
−
y x2 + y 2 + z 2 2 cos x 2 + y 2 + z 2 ,
25. The climber is moving in the direction of
1
⎛ 1 ⎞
( )
−
z x2 + y2 + z2 2 cos x 2 + y 2 + z 2 u=⎜ ⎟ −1, 1 . Let
⎝ 2⎠
2 + 2 y 2 ) /100
⎛ 1 ⎞ f ( x, y ) = 3000e−( x .
( )
−
= ⎜ x 2 + y 2 + z 2 2 cos x 2 + y 2 + z 2 ⎟ x, y, z 2 + 2 y 2 ) /100 x y
⎜ ⎟ ∇f ( x, y ) = 3000e –( x – ,– ;
⎝ ⎠ 50 25
which either points towards or away from the
origin. f (10, 10) = –600e –3 1, 2
She will move at a slope of
22. Let D = x + y + z be the distance. Then we
2 2 2 ⎛ 1 ⎞
Du (10, 10) = –600e –3 1, 2 ⋅ ⎜ ⎟ –1, 1
have ⎝ 2⎠

∇T =
∂T ∂T ∂T
, , =
dT ∂D dT ∂D dT ∂D
, , ( )
= –300 2 e –3 ≈ –21.1229 .
∂x ∂y ∂z dD ∂x dD ∂y dD ∂z She will descend. Slope is about –21.
1 1
=
dT
dD
x x2 + y 2 + ( z 2 −2
) ,
dT
dD
(
y x2 + y2 + z )
2 −2
,
26.
dx
dt
=
dy
dt
;
dx dy
= ; ln x = – ln y + C
2x –2 y x –y
1
dT
dD
z x +y +z
2 2
(
2 −2
) At t = 0: ln –2 = – ln 1 + C ⇒ C = ln 2.
2 2
ln x = – ln y + ln 2 = ln ; x = ; xy = 2
⎛ dT ⎞ 1 y y
=⎜
⎜ dD
x2 + y2 + ( ⎟ x, y , z
⎟
z 2 −2
) Since the particle starts at (–2, 1) and neither x
nor y can equal 0, the equation simplifies
⎝ ⎠
which either points towards or away from the to xy = −2. ∇T (–2, 1) = –4, – 2 , so the particle
origin. moves downward along the curve.
23. He should move in the direction of
1 1
–∇f (p) = – f x (p), f y (p) = – – , –
2 4
⎛1⎞
= ⎜ ⎟ 2, 1 . Or use 2, 1 . The angle α formed
⎝4⎠
⎛1⎞
with the East is tan –1 ⎜ ⎟ ≈ 26.57° (N63.43°E).
⎝2⎠

27. ∇T ( x, y ) = –4 x, – 2 y
dx dy
= –4 x, = –2 y
dt dt
dx dy
dt
= dt has solution x = 2 y 2 . Since the
–4 x –2 y
particle starts at (–2, 1), this simplifies to
x = –2 y 2 .


28. f(1, –1) = 5 -1, 1
D u , u f (1, –1) = u1 , u2 ⋅ –5, 5 = –5u1 + 5u2
1 2

⎛ 1 ⎞
a. -1, 1 (in the direction of the gradient); u = ⎜ ⎟ –1, 1 .
⎝ 2⎠

⎛ 1 ⎞
b. ± 1, 1 (direction perpendicular to gradient); u = ⎜ ± ⎟ 1, 1
⎝ 2⎠

3 4
c. Want Du f (1, – 1) = 1 where u = 1. That is, want –5u1 + 5u2 = 1 and u1 + u2 = 1. Solutions are u =
2 2
,
5 5
4 3
and − , − .
5 5

10(2 x) 10(2 y ) 10(2 z )
29. a. ∇T ( x , y , z ) = − ,− ,−
(x + y + z )
2 2 2 2
(x + y + z )
2 2 2 2
( x + y 2 + z 2 )2
2

20
=– x, y , z
( x + y 2 + z 2 )2
2

r (t ) = t cos π t , t sin π t, t , so r(1) = -1, 0, 1 . Therefore, when t = 1, the bee is at (–1, 0, 1), and
∇T (–1, 0, 1) = –5 –1, 0, 1 .
r ′(t ) = cos πt – πt sin πt , sin πt + πt cos πt , 1 , so r ′(1) = –1, – π, 1 .
r ′(1) –1, – π, 1
U= = is the unit tangent vector at (–1, 0, 1).
r ′(1) 2 + π2
DuT (–1, 0, 1) = u ⋅∇T (–1, 0, 1)
–1, – π, 1 ⋅ 5, 0, – 5 10
= =– ≈ –2.9026
2+π 2
2 + π2
Therefore, the temperature is decreasing at about 2.9°C per meter traveled when the bee is at (–1, 0, 1); i.e.,
when t = 1 s.

b. Method 1: (First express T in terms of t.)
10 10 10 5
T= = = =
x +y +z
2 2 2
(t cos πt ) + (t sin πt ) + (t )
2 2 2
2t 2
t2
T (t ) = 5t −2 ; T ′(t ) = −10t −3 ; t ′(1) = −10
Method 2: (Use Chain Rule.)
10 ⎛ 2⎞
= ( DuT ) ( r '(t ) ) , so Dt T (t ) = [ DuT (−1, 0, 1)] ( r '(1) ) = −
dT ds
Dt T (t ) = ⎜ 2 + π ⎟ = −10
ds dt 2 ⎝ ⎠
2+π
Therefore, the temperature is decreasing at about 10°C per second when the bee is at (–1, 0, 1); i.e., when
t = 1 s.

3 4
30. a. Du f = , – ⋅ f x ⋅ f y = –6, so
5 5
3 f x – 4 f y = –30.
4 3
Dv f = , ⋅ f x , f y = 17, so
5 5
4 f x + 3 f y = 85.
The simultaneous solution is
f x = 10, f y = 15, so ∇f = 10, 15 .


b. Without loss of generality, let Therefore, f is not differentiable at the origin. But
u = i and v = j . If θ and φ are the angles Du f (0, 0) exists for all u since
between u and ∇f , and between v and ∇f , f (0 + h, 0) – f (0, 0) 0–0
f x (0, 0) = lim = lim
then: h →0 h h →0 h
π = lim (0) = 0, and
1. θ + φ = (if ∇f is in the 1st quadrant). h →0
2
π f (0, 0 + h) – f (0, 0) 0–0
2. θ = + φ (if ∇f is in the 2nd quadrant). f y (0, 0) = lim = lim
h →0 h h →0 h
2
3π = lim (0) = 0, so ∇f (0, 0) = 0, 0 = 0. Then
3. φ + θ = (if ∇f is in the 3rd quadrant). h →0
2 Du f (0, 0) = ∇f (0, 0) ⋅ u = 0 ⋅ u = 0.
π
4. φ = + θ (if ∇f is in the 4th quadrant).
2 33. Leave: (–0.1, –5)
In each case cos φ = sin θ or cos φ = –sin θ,
so cos 2 φ = sin 2 θ . Thus,
( Du f ) 2 + ( Dv f )2 = (u ⋅∇f )2 + ( v ⋅∇f )2
2 2
= ∇f cos 2 θ + ∇f cos 2 φ
2
= ∇f (cos 2 θ + cos 2 φ )
2 2
= ∇f cos2 θ + sin 2 θ = ∇f .
x2 – y 2
31.

a. A′(100, 120)

b. B ′(190, 25) 34. Leave (–2, –5)

20 – 30 1
c. f x (C ) ≈ = – ; f y ( D) = 0;
230 – 200 3
40 – 30 2
Du f ( E ) ≈ =
25 5

32. Graph of domain of f
⎧0, in shaded region ⎫ x – x3
f ( x, y ) = ⎨ ⎬
⎩1, elsewhere ⎭ 9 – y2

lim f ( x, y ) does not exist since
( x, y )→(0, 0) 35. Leave: (3, 5)
( x, y ) → (0, 0) :
36. (4.2, 4.2)
along the y-axis, f(x, y) = 0, but along the y = x 4
curve, f(x, y) = 1.



∂ z dx ∂ z dy
1. +
∂ x dt ∂ y dt

2. y 2 cos t + 2 xy (– sin t )
= cos3 t – 2sin 2 t cos t

∂z∂x ∂z ∂y
3. +
∂ x ∂t ∂ y ∂t

4. 12

Problem Set 12.6

dw
1. = (2 xy 3 )(3t 2 ) + (3x 2 y 2 )(2t )
dt
= (2t 9 )(3t 2 ) + (3t10 )(2t ) = 12t11

dw
2. = (2 xy – y 2 )(– sin t ) + ( x 2 – 2 xy )(cos t )
dt
= (sin t + cos t )(1 − 3sin t cos t )

dw
3. = (e x sin y + e y cos x)(3) + (e x cos y + e y sin x)(2) = 3e3t sin 2t + 3e2t cos 3t + 2e3t cos 2t + 2e2t sin 3t
dt

dw ⎛ 1 ⎞ 2 ⎛ 1 ⎞ sec2 t sec2 t – 2 tan 2 t 1 – tan 2 t
4. = ⎜ ⎟ sec t + ⎜ – ⎟ (2sec2 t tan t ) = – 2 tan t = =
dt ⎝ x ⎠ ⎝ y⎠ tan t tan t tan t

dw
5. = [ yz 2 (cos( xyz 2 )](3t 2 ) + [ xz 2 cos( xyz 2 )](2t ) + [2 xyz cos( xyz 2 )](1)
dt
= (3 yz 2 t 2 + 2 xz 2 t + 2 xyz ) cos( xyz 2 ) = (3t 6 + 2t 6 + 2t 6 ) cos(t 7 ) = 7t 6 cos(t 7 )

dw
6. = ( y + z )(2t ) + ( x + z )(−2t ) + ( y + x)(−1) = 2t (2 – t – t 2 ) – 2t (1 – t + t 2 ) – (1) = –4t 3 + 2t –1
dt

∂w
7. = (2 xy )( s) + ( x 2 )(–1) = 2st ( s – t ) s – s 2t 2 = s 2 t (2s – 3t )
∂t

∂w ⎡ ⎛ s ⎞⎤
8. = (2 x – x –1 y )(– st –2 ) + (– ln x)( s 2 ) = s 2 ⎢1 – 2t –3 – ln ⎜ ⎟ ⎥
∂t ⎣ ⎝ t ⎠⎦

∂w 2 2 2 2 2 2
9. = e x + y (2 x)( s cos t ) + e x + y (2 y )(sin s) = 2e x + y ( xs cos t + y sin s )
∂t
= 2( s 2 sin t cos t + t sin 2 s) exp( s 2 sin 2 t + t 2 sin 2 s )

∂w 2e s (t +1) ( st –1)
10. = [( x + y ) −1 − ( x − y )−1 ](e s ) + [( x + y )−1 + ( x − y )−1 ]( se st ) =
∂t t 2 e2 s – e2 st


∂w x(– s sin st ) y ( s cos st ) z(s2 )
11. = + + = s 4 t (1 + s 4 t 2 ) –1/ 2
∂ t ( x 2 + y 2 + z 2 )1/ 2 ( x 2 + y 2 + z 2 )1/ 2 ( x 2 + y 2 + z 2 )1/ 2

∂w 2
12. = (e xy + z y )(1) + (e xy + z x)(−1) + (e xy + z )(2t ) = e xy + z ( y − x + 2t ) = e s (0) = 0
∂t

∂z ⎛∂ z⎞
13. = (2 xy )(2) + ( x 2 )(–2 st ) = 4(2t + s )(1 − st 2 ) − 2 st (2t + s )2 ; ⎜ ⎟ = 72
∂t ⎝ ∂ t ⎠ (1, − 2)

∂z ⎛∂ z ⎞
14. = ( y + 1)(1) + ( x + 1)(rt ) = 1 + rt (1 + 2s + r + t ); ⎜ ⎟ =5
∂s ⎝ ∂ s ⎠ (1, –1, 2)

dw
15. = (2u – tan v)(1) + (– u sec2 v)(π) = 2 x – tan πx – πx sec2 πx
dx
dw ⎛1⎞ ⎛π⎞ 1+ π
= ⎜ ⎟ –1 – ⎜ ⎟ = – ≈ –2.0708
dx x = 1 ⎝ 2 ⎠ ⎝ 2⎠ 2
4

∂w
16. = (2 xy )(– ρ sin θ sin φ ) + ( x 2 )( ρ cosθ sin φ ) + (2 z )(0) = ρ 3 cos θ sin 3 φ (–2sin 2 θ + cos 2 θ );
∂θ
⎛∂w⎞
⎜ ⎟ = –8
⎝ ∂θ ⎠ (2, π, π )
2

dr 19. The stream carries the boat along at 2 ft/s with
17. V (r , h) = πr 2 h, = 0.5 in./yr, respect to the boy.
dt
dh
= 8 in./yr
dt
dV ⎛ dr ⎞ ⎛ dh ⎞
= (2πrh) ⎜ ⎟ + (πr 2 ) ⎜ ⎟ ;
dt ⎝ dt ⎠ ⎝ dt ⎠
⎛ dV ⎞ dx dy
⎜ ⎟ = 11200π in.3/yr = 2, = 4, s 2 = x 2 + y 2
⎝ dt ⎠ (20, 400) dt dt
11200π in.3 1 board ft ⎛ ds ⎞ ⎛ dx ⎞ ⎛ dy ⎞
2s ⎜ ⎟ = 2 x ⎜ ⎟ + 2 y ⎜ ⎟
= × ≈ 244.35 board ft/yr ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠
1 yr 144 in.3
ds (2 x + 4 y )
=
18. Let T = e – x –3 y . dt s
dT dx dy When t = 3, x = 6, y = 12, s = 6 5. Thus,
= e – x –3 y (–1) + e – x –3 y (–3)
dt dt dt ⎛ ds ⎞
⎜ ⎟ = 20 ≈ 4.47 ft/s
= e – x –3 y (–1)(2) + e – x –3 y (–3)(2) = –8e – x –3 y ⎝ dt ⎠ t =3
dT
= –8, so the temperature is decreasing ⎛1⎞ dh
dt (0, 0) 20. V (r , h) = ⎜ ⎟ πr 2 h, = 3 in./min,
⎝3⎠ dt
at 8°/min.
dr
= 2 in./min
dt
dV ⎛ 2 ⎞ ⎛ dr ⎞ ⎛ 1 ⎞ ⎛ dh ⎞
= ⎜ ⎟ πrh ⎜ ⎟ + ⎜ ⎟ πr 2 ⎜ ⎟ ;
dt ⎝ 3 ⎠ ⎝ dt ⎠ ⎝ 3 ⎠ ⎝ dt ⎠
⎛ dV ⎞ 20,800π
⎜ ⎟ = ≈ 21, 782 in.3/min
⎝ dt ⎠ (40, 100) 3


21. Let F ( x, y ) = x3 + 2 x 2 y – y3 = 0 . ⎛1⎞
29. y = ⎜ ⎟ [ f (u ) + f (v)],
∂F ⎝2⎠
dy − ∂x (3 x 2 + 4 xy ) 3x 2 + 4 xy
Then = =− = . where u = x – ct, v = x + ct.
∂F
dx
∂y 2 x2 − 3 y 2 3 y2 − 2 x2 ⎛1⎞ ⎛1⎞
y x = ⎜ ⎟ [ f ′(u )(1) + f ′(v)(1)] = ⎜ ⎟ [ f ′(u ) + f ′(v)]
⎝ 2⎠ ⎝2⎠
22. Let F ( x, y ) = ye – x + 5 x – 17 = 0. ⎛1⎞
y xx = ⎜ ⎟ [ f ′′(u )(1) + f ′′(v)(1)]
dy (– ye – x + 5) ⎝2⎠
=– = y – 5e x
dx e– x ⎛ 1 ⎞ ′′
= ⎜ ⎟ [ f (u ) + f ′′(v)]
⎝2⎠
23. Let F(x, y) = x sin y + y cos x = 0. ⎛1⎞
dy (sin y – y sin x) y sin x – sin y yt = ⎜ ⎟ [ f ′(u )(– c) + f ′(v)(c)]
=– = ⎝2⎠
dx x cos y + cos x x cos y + cos x
⎛ c⎞
= ⎜ – ⎟ [ f ′(u ) – f ′(u )]
⎝ 2⎠
24. Let F ( x, y ) = x 2 cos y – y 2 sin x = 0.
⎛ c⎞
∂F ytt = ⎜ − ⎟ [ f ′′(u )(−c) − f ′′(v)(c)]
dy –(2 x cos y – y 2 cos x) ⎝ 2⎠
Then = – ∂x =
∂F
dx
∂y – x 2 sin y – 2 y sin x ⎛ c2 ⎞
= ⎜ ⎟ [ f ′′(u ) + f ′′(v )] = c 2 y xx
2 x cos y – y 2 cos x ⎜ 2 ⎟
= . ⎝ ⎠
x 2 sin y + 2 y sin x
30. Let w = f(x, y, z) where x = r – s, y = s – t,
z = t – r. Then
25. Let F ( x, y, z ) = 3 x 2 z + y 3 − xyz 3 = 0. wr + ws + wt = ( wx xr + wx xs ) + ( wy ys + wy yt )
∂z (6 xz – yz 3 ) yz 3 – 6 xz + ( wz zt + wz zr )
=– =
∂x 3x 2 – 3xyz 2 3 x 2 – 3 xyz 2 = [ wx (1) + wx (−1)] + [ wy (1) + wy (−1)]
+ [ wz (1) + wz (−1)]
26. Let f ( x, y, z ) = ye – x + z sin x = 0.
=0
∂x – sin x sin x
= =
∂ z – ye + z cos x ye – z cos x
–x –x y x
31. Let w = ∫ f (u )du = – ∫ f (u )du, where x = g(t),
x y
∂T ∂T ∂ x ∂T ∂ y ∂T ∂ z ∂T ∂ w y = h(t).
27. = + + +
∂s ∂x ∂s ∂ y ∂s ∂z ∂s ∂w ∂s Then
dw ∂w dx ∂w dy
= + = − f ( x) g ′(t ) + f ( y )h′(t )
∂z dt ∂x dt ∂y dt
28. We use the zr notation for .
∂r = f (h(t ))h′(t ) – f ( g (t )) g ′(t ) .
zr = z x xr + z y yr = z x cos θ + z y sin θ
Thus, for the particular function given
zθ = z x xθ + z y yθ = z x (– r sin θ ) + z y (r cos θ ), so
F ′(t ) = 9 + (t 2 )4 (2t )
r zθ = – z x sin θ + z y cos θ . Thus,
–1

( ) ( 2π cos )
4
( zr ) + ( r
2 –2
)( zθ ) = ( z x cos θ + z y sin θ )
2 2 – 9 + sin 2πt 2πt ;

+(– z x sin θ + z y cos θ )2 F ′ ( 2 ) = (5) ( 2 2 ) – (3) ( 2π )

= ( z x ) 2 + ( z y )2 (expanding and using = 10 2 – 3 2π ≈ 0.8135.

sin 2 θ + cos 2 θ = 1). 32. If f (tx, ty ) = tf ( x, y ), then
d d
[ f (tx, ty )] = [tf ( x, y )].
dt dt
That is,
[ ftx (tx, ty )][ x] + [ fty (tx, ty )][ y ] = f ( x, y ).
Letting t = 1 yields the desired result.


33. c 2 = a 2 + b 2 – 2ab cos 40° (Law of Cosines) where a, b, and c are functions of t.
aa ′ + bb′ – (a ′b + ab′) cos 40°
2cc′ = 2aa ′ + 2bb′ – 2(a ′b + ab′) cos 40° so c′ = .
c

When a = 200 and b = 150, c 2 = (200) 2 + (150)2 – 2(200)(150) cos 40° = 62,500 – 60,000 cos 40°.
It is given that a ′ = 450 and b′ = 400, so at that instant,
(200)(450) + (150)(400) − [(450)(150) + (200)(400)]cos 40°
c′ = ≈ 288 .
62,500 – 60, 000 cos 40°
Thus, the distance between the airplanes is increasing at about 288 mph.

2
34. r = x, y, z , so r 2 = r = x 2 + y 2 + z 2 .
GMm
F= , so
x + y2 + z2
2

F ′(t ) = Fm m′(t ) + Fx x ′(t ) + Fy y ′(t ) + Fz z ′(t )
GMm′(t ) 2GMmxx′(t )
= –
x +y +z
2 2
( x2 + y 2 + z 2 )2
2

2GMmyy ′(t ) 2GMmzz ′(t )
– +
(x + y + z )
2 2 2 2
( x 2 + y 2 + z 2 )2
GM [( x 2 + y 2 + z 2 )m′(t ) – 2m( xx′(t ) + yy ′(t ) + zz ′(t )]
= .
( x 2 + y 2 + z 2 )2

12.7 Concepts Review 2. ∇F ( x, y, z ) = 2 8 x, y, 8 z ;

1. perpendicular ⎛ 2⎞
∇F ⎜ 1, 2,
⎜ ⎟ = 4 4, 1, 2 2
⎝ 2 ⎟⎠
2. 3,1, −1 Tangent Plane:
⎛ 2⎞
3. x − +4( y − 1) + 6( z − 1) = 0 4( x – 1) + 1( y – 2) + 2 2 ⎜ z –
⎜ ⎟ , or
⎝ 2 ⎟
⎠
∂f ∂f 4 x + y + 2 2 z = 8.
4. dx + dy
∂x ∂y
3. Let F ( x, y , z ) = x 2 – y 2 + z 2 + 1 = 0.
Problem Set 12.7 ∇F ( x , y , z ) = 2 x , – 2 y , 2 z = 2 x , – y , z
1. ∇F ( x, y, z ) = 2 x, y, z ; ( )
∇F 1, 3, 7 = 2 1, – 3, 7 , so 1, – 3, 7 is

( )
∇F 2, 3, 3 = 2 2, 3, 3 normal to the surface at the point. Then the
tangent plane is
( )
Tangent Plane:
1( x – 1) – 3( y – 3) + 7 z – 7 = 0, or more
(
2( x – 2) + 3( y – 3) + 3 z – 3 = 0, or )
simply, x − 3 y + 7 z = –1.
2 x + 3 y + 3 z = 16


4. ∇f ( x, y, z ) = 2 x, y, – z ; ⎛1⎞
11. dz = 2 x −1dx + y −1dy = (−1)(0.02) + ⎜ ⎟ (−0.04)
∇f (2, 1, 1) = 2 2, 1, − 1 ⎝ 4⎠
= –0.03
Tangent plane:
2( x − 2) + 1( y − 1) − 1( z − 1) = 0, or 2 x + y − z = 4 . Δz = f (–1.98, 3.96) – f (–2.4)
= ln[(–1.98)2 (3.96)] – ln16 ≈ –0.030151
⎛1⎞
5. ∇f ( x, y ) = ⎜ ⎟ x, y ; ∇f (2, 2) = 1, 1
⎝ 2⎠ 12. Let z = f ( x, y ) = tan –1 xy;
Tangent plane: z − 2 = 1( x − 2) + 1( y − 2) , or y x
x+ y−z = 2. dz = dx + dy;
1+ x y2 2
1 + x2 y2
(−0.5)(−0.03) + (−2)(−0.01)
6. Let f ( x, y ) = xe –2 y . = = 0.0175.
1 + (4)(0.25)
∇f ( x, y ) = e−2 y , − 2 xe−2 y Δz = f (−2.03, −0.51) − f (−2, −0.5) ≈ 0.017342
∇f (1, 0) = 1, − 2
13. Let
Then 1, – 2, – 1 is normal to the surface at F ( x, y , z ) = x 2 – 2 xy – y 2 – 8 x + 4 y – z = 0;
(1, 0, 1), and the tangent plane is
∇F ( x, y, z ) = 2 x – 2 y – 8, – 2 x – 2 y + 4, – 1
1( x − 1) − 2( y − 0) − 1( z − 1) = 0 , or x − 2 y − z = 0 .
Tangent plane is horizontal if ∇F = 0, 0, k for
7. ∇f ( x, y ) = –4e3 y sin 2 x, 6e3 y cos 2 x ; any k ≠ 0 .
2x – 2y – 8 = 0 and –2x – 2y + 4 = 0 if x = 3 and
⎛π ⎞ y = –1. Then z = –14. There is a horizontal
∇f ⎜ , 0 ⎟ = –2 3, – 3
⎝3 ⎠ tangent plane at (3, –1, –14).
⎛ π⎞
Tangent plane: z + 1 = –2 3 ⎜ x – ⎟ – 3( y – 0) , 14. 8, −3, −1 is normal to 8x – 3y – z = 0.
⎝ 3⎠
Let F ( x, y, z ) = 2 x 2 + 3 y 2 − z .
or 2 3 x + 3 y + z =
( 2 3π – 3 ). ∇F ( x, y, z ) = 4 x, 6 y, – 1 is normal to
3
z = 2 x 2 + 3 y 2 at (x, y, z). 4x = 8 and
⎛1⎞ 1 1 1 1 1
8. ∇f ( x, y ) = ⎜ ⎟ , ; ∇f (1, 4) = , 6y = –3, if x = 2 and y = – ; then
⎝2⎠ x y 2 4 2
⎛1⎞ ⎛1⎞ ⎛ 1 ⎞
Tangent plane: z − 3 = ⎜ ⎟ ( x − 1) + ⎜ ⎟ ( y − 4), z = 8.75 at ⎜ 2, – , 8.75 ⎟ .
⎝2⎠ ⎝4⎠ ⎝ 2 ⎠
1 1 3
or x + y − z = − . 15. For F ( x, y , z ) = x 2 + 4 y + z 2 = 0,
2 4 2
∇F ( x, y, z ) = 2 x, 4, 2 z = 2 x, 2, z .
9. Let z = f ( x, y ) = 2 x y ; 2 3
F(0, –1, 2) = 0, and
dz = 4 xy3 dx + 6 x 2 y 2 dy. For the points given, ∇F (0, – 1, 2) = 2 0, 2, 2 = 4 0, 1, 1 .
dx = −0.01, dy = 0.02 , For G ( x, y, z ) = x 2 + y 2 + z 2 – 6 z + 7 = 0,
dz = 4(−0.01) + 6(0.02) = 0.08. ∇G ( x , y , z ) = 2 x , 2 y , 2 z – 6 = 2 x , y , z – 3 .
Δz = f (0.99, 1.02) – f (1, 1) G(0, –1, 2) = 0, and
= 2(0.99) (1.02) − 2(1) (1) ≈ 0.08017992
2 3 2 3 ∇G (0, – 1, 2) = 2 0, – 1, – 1 = –2 0, 1, 1 .
0, 1, 1 is normal to both surfaces at
10. dz = (2 x − 5 y )dx + (−5 x + 1)dy (0, –1, 2) so the surfaces have the same tangent
= (−11)(0.03) + (−9)(−0.02) = −0.15 plane; hence, they are tangent to each other at
Δz = f (2.03, 2.98) − f (2,3) = −0.1461 (0, –1, 2).


16. (1, 1, 1) satisfies each equation, so the surfaces 20. Let f ( x, y, z ) = x – z 2 , and g ( x, y, z ) = y – z 3 .
intersect at (1, 1, 1). For
∇f ( x, y, z ) = 1, 0, − 2 z and
z = f ( x, y ) = x 2 y : ∇f ( x, y ) = 2 xy, x 2 ;
∇g ( x, y, z ) = 0, 1, – 3z 2
∇f (1, 1) = 2, 1 , so 2, 1, – 1 is normal at
(1, 1, 1). ∇f (1, 1, 1) = 1, 0, – 2 and
For f ( x, y, z ) = x 2 − 4 y + 3 = 0; ∇g (1, 1, 1) = 0, 1, – 3
∇f ( x, y, z ) = 2, – 4, 0 ; 1, 0, −2 × 0,1, −3 = 2,3,1
∇f (1, 1, 1) = 2, – 4, 0 so 2, – 4, 0 is normal at Line: x = 1 + 2t , y = 1 + 3t , z = 1 + t
(1, 1, 1).
21. dS = S A dA + SW dW
2, 1, – 1 ⋅ 2, – 4, 0 = 0, so the normals, hence
W A −WdA + AdW
tangent planes, and hence the surfaces, are =− dA + dW =
perpendicular at (1, 1, 1). ( A −W ) 2
(A −W ) ( A − W )2
2

At W = 20, A = 36:
17. Let F ( x, y, z ) = x 2 + 2 y 2 + 3 z 2 – 12 = 0; −20dA + 36dW −5dA + 9dW
dS = = .
∇F ( x, y, z ) = 2 x, 2 y, 3 z is normal to the 256 64
plane. 5 dA + 9 dW 5(0.02) + 9(0.02)
Thus, dS ≤ ≤
A vector in the direction of the line, 64 64
2, 8, – 6 = 2 1, 4, – 3 , is normal to the plane. = 0.004375
x, 2 y,3 z = k 1, 4, -3 and (x, y, z) is on the
1 1
surface for points (1, 2, –1) [when k =1] and 22. V = lwh, dl = dw = , dh = , l = 72, w = 48,
2 4
(–1, –2, 1) [when k = –1].
h = 36
dV = whdl + lhdw + lwdh = 3024 in.3 (1.75 ft3)
x2 y2 z2
18. Let F ( x, y, z ) = +
= 1. +
a 2 b2 c2 23. V = πr 2 h, dV = 2πrh dr + πr 2 dh
2x 2 y 2z
∇F ( x , y , z ) = , , dV ≤ 2πrh dr + πr 2 dh ≤ 2πrh(0.02r ) + πr 2 (0.03h)
a 2 b2 c2
= 0.04πr 2 h + 0.03πr 2 h = 0.07V
x y z
∇F ( x0 , y0 , z0 ) = 2 0 , 0 , 0 Maximum error in V is 7%.
a 2 b2 c 2
The tangent plane at ( x0 , y0 , z0 ) is L
24. T = f ( L, g ) = 2π
x0 ( x – x0 ) y0 ( y – y0 ) z0 ( z – z 0 ) g
+ + = 0.
a 2
b 2
c2 dT = f L dL + f g dg
x0 x y0 y z0 z ⎛ x0 y0 z0 ⎞
2 2 2
⎛ ⎞ ⎛ ⎞
+ + –⎜ + + ⎟=0 ⎜ ⎟ ⎜ ⎟⎛
a2 b2 c2 ⎜ a 2 b2 c2 ⎟
⎝ ⎠ 1 ⎟⎛ 1 ⎞ ⎞
= 2π ⎜ dL + 2π ⎜ ⎟ ⎜ − L ⎟ dg
1
⎜ ⎜ ⎟ ⎜ g2 ⎟
L ⎟⎝ g ⎠ ⎜ ⎟
⎟⎝ ⎠
x0 x y0 y z0 z L
Therefore, + + = 1, since ⎜2
⎜ ⎟
⎟ ⎜2
⎜ ⎟
a2 b2 c2 ⎝ g⎠ ⎝ g ⎠
2
x0 2
y0 2
z0 π( gdL – Ldg )
+ + = 1. = , so
a2 b2 c2 g2
L
g
19. ∇f ( x, y, z ) = 2 9 x, 4 y , 4 z ; dT π( gdL – Ldg ) gdL – Ldg
= =
∇f (1, 2, 2) = 2 9, 8, 8 T ⎛ L ⎞⎛ 2 L ⎞ 2 gL
⎜ 2π
⎜ ⎟⎜ g
⎟⎜ ⎟
⎟
∇g ( x , y , z ) = 2 2 x , – y , 3 z ; ⎝ g ⎠⎝ g⎠
∇f (1, 2, 2) = 4 1, – 1, 3 1 ⎛ dL dg ⎞
= ⎜ – ⎟.
9, 8, 8 × 1, –1, 3 = 32, –19, –17 2⎝ L g ⎠
Line: x = 1 + 32t , y = 2 -19t , z = 2 -17t Therefore,
dT 1 ⎛ dL dg ⎞ 1
≤ ⎜ + ⎟ = (0.5% + 0.3%) = 0.4%.
T 2⎝ L g ⎠ 2


R1 R2 27. Let F(x, y, z) = xyz = k; let (a, b, c) be any point
25. Solving for R, R = , so on the surface of F.
R1 + R2
k k k
∂R R22
∂R R12 ∇F ( x, y, z ) = yz , xz , xy = , ,
= and = . x y z
∂ R1 ( R1 + R2 ) 2 ∂ R2 ( R1 + R2 )2
1 1 1
=k , ,
R2 dR1 + R1 dR2
2 2
x y z
Therefore, dR = ;
( R1 + R2 )2 1 1 1
∇F ( a , b , c ) = k , ,
2
R2 dR1 + R1
2
dR2 a b c
dR ≤ . Then at R1 = 25, An equation of the tangent plane at the point is
( R1 + R2 ) 2
⎛1⎞ ⎛1⎞ ⎛1⎞
(25)(100) ⎜ ⎟ ( x – a ) + ⎜ ⎟ ( x – b) + ⎜ ⎟ ( x – c) = 0, or
R2 = 100, dR1 = dR2 = 0.5, R = = 20 ⎝a⎠ ⎝b⎠ ⎝c⎠
25 + 100 x y z
(100)2 (0.5) + (25) 2 (0.5) + + = 3.
and dR ≤ = 0.34. a b c
(125) 2 Points of intersection of the tangent plane on the
coordinate axes are (3a, 0, 0), (0, 3b, 0), and
(0, 0, 3c).
26. Let F ( x, y , z ) = x 2 + y 2 + 2 z 2 .
The volume of the tetrahedron is
∇F ( x , y , z ) = 2 x , 2 y , 4 z ; ⎛1⎞ 1⎛1 ⎞
∇F ⎜ ⎟ (area of base)(altitude)= ⎜ 3a 3b ⎟ ( 3c )
∇F (1, 2, 1) = 2 1, 2, 2 ;
1 2 2
= , , ⎝ 3⎠ 3⎝ 2 ⎠
∇F 3 3 3 9 abc 9 k
= = (a constant).
1 2 2 2 2
Thus, u = , , is the unit vector in the
3 3 3
direction of flight, and
1 2 2
x, y, z = 1, 2, 1 + 4t , , is the location
3 3 3
of the bee along its line of flight t seconds after
takeoff. Using the parametric form of the line of
flight to substitute into the equation of the plane
yields t = 3 as the time of intersection with the
plane. Then substituting this value of t into the
equation of the line yields x = 5, y = 10, z = 9 so
the point of intersection is (5, 10, 9).

1 1 1
28. If F ( x, y , z ) = x + y + z , then ∇F ( x, y , z ) = 0.5 , , . The equation of the tangent is
x y z

1 1 1 x y z
0.5 , , ⋅ x – x0 , y – y0 , z – z0 = 0, or + + = x0 + y0 + z0 = a.
x0 y0 z0 x0 y0 z0

Intercepts are a x0 , a y0 , a z0 ; so the sum is a ( x0 + y0 + z0 = a 2 . )


29. f ( x, y ) = ( x 2 + y 2 )1/ 2 ; f (3, 4) = 5
3 4
f x ( x, y ) = x( x 2 + y 2 ) –1/ 2 ; f x (3, 4) = = 0.6 ; f y = ( x, y ) = y ( x 2 + y 2 ) –1/ 2 ; f x (3, 4) = = 0.8
5 5
16
f xx ( x, y ) = y 2 ( x 2 + y 2 ) –3 / 2 ; f x (3, 4) = = 0.128 ; f xy ( x, y ) = – xy ( x 2 + y 2 ) –3 / 2 ;
125
12
f xy (3, 4) = – = –0.096
125
9
f yy = x 2 ( x 2 + y 2 ) –3 / 2 ; f xx (3, 4) = = 0.072
125
Therefore, the second order Taylor approximation is
f ( x, y ) = 5 + 0.6( x – 3) + 0.8( y – 4) + 0.5[0.128( x – 3) 2 + 2(–0.096)( x – 3)( y – 4) + 0.072( y – 4)2 ]

a. First order Taylor approximation: f ( x, y ) = 5 + 0.6( x - 3) + 0.8( y - 4) .
Thus, f (3.1,3.9) ≈ 5 + 0.6(0.1) + 0.8(-0.1) = 4.98 .

b. f (3.1,3.9) ≈ 5 + 0.6(–0.1) + 0.8(0.1) + 0.5[0.128(0.1) 2 + 2(–0.096)(0.1)(–0.1) + 0.072(–0.1) 2 ] = 4.98196

c. f (3.1, 3.9) ≈ 4.9819675

⎛ x2 + y2 ⎞
30. f ( x, y ) = tan ⎜ ⎟ ; f (0, 0) = 0
⎜ 64 ⎟
⎝ ⎠
x ⎛ x + y2 ⎞
2
f x ( x, y ) = ⋅ sec2 ⎜ ⎟ ; f x (0, 0) = 0
32 ⎜ 64 ⎟
⎝ ⎠
y ⎛x +y
2 2⎞
f y ( x, y ) = ⋅ sec2 ⎜ ⎟ ; f y (0, 0) = 0
32 ⎜ 64 ⎟
⎝ ⎠
2x 2 ⎛ x2 + y2 ⎞ ⎛ x2 + y2 ⎞ 1 ⎛ x2 + y 2 ⎞ 1
f xx ( x, y ) = sec 2 ⎜ ⎟ tan ⎜ ⎟ + sec 2 ⎜ ⎟ ; f xx (0, 0) =
2 ⎜ 64 ⎟ ⎜ 64 ⎟ 32 ⎜ 64 ⎟ 32
32 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 y2 ⎛ x2 + y 2 ⎞ ⎛ x2 + y 2 ⎞ 1 ⎛ x2 + y2 ⎞ 1
f yy ( x, y ) = sec2 ⎜ ⎟ tan ⎜ ⎟ + sec2 ⎜ ⎟ ; f yy (0, 0) =
2 ⎜ 64 ⎟ ⎜ 64 ⎟ 32 ⎜ 64 ⎟ 32
32 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
When computed, each term of f xy ( x, y ) will contain either x or y, resulting in f xy ( 0, 0 ) = 0 . Therefore, the
second-order Taylor approximation is
1⎡ 1 1 2⎤ 1 2 1 2
f ( x, y ) = 0 + 0 ⋅ x + 0 ⋅ y + ⎢ x 2 + 2 ⋅ 0 ⋅ x ⋅ y + y = x + y
2 ⎣ 32 32 ⎥ 64
⎦ 64

a. The first-order Taylor approximation is f ( x, y ) = 0 + 0 ⋅ x + 0 ⋅ y = 0; Thus, f (0.2, −0.3) ≈ 0.

1 1
b. f (0.2, −0.3) ≈ (0.2) 2 + (−0.3)3 = 0.00203125
64 64

c. f (0.2, −0.3) ≈ 0.0020312528


1. closed bounded

2. boundary; stationary; singular

3. ∇f ( x0 , y0 ) = 0

2
4. f xx ( x0 , y0 ) f yy ( x0 , y0 ) – f xy ( x0 , y0 )

Problem Set 12.8

1. ∇f ( x, y ) = 2 x – 4, 8 y = 0, 0 at (2, 0), a stationary point.
D = f xx f yy – f xy = (2)(8) – (0)2 = 16 > 0 and f xx = 2 > 0. Local minimum at (2, 0).
2

2. ∇f ( x, y ) = 2 x − 2, 8 y + 8 = 0, 0 at (1, –1), a stationary point. D = f xx f yy – f xy = (2)(8) – (0)2 = 16 > 0 and
2

f xx = 2 > 0. Local minimum at (1, –1).

3. ∇f ( x, y ) = 8 x3 – 2 x, 6 y = 2 x(4 x 2 – 1), 6 y
= 0, 0 , at (0, 0), (0.5, 0), (−0.5, 0) all stationary points.
f xx = 24 x 2 – 2; D = f xx f yy – f xy = (24 x 2 – 2)(6) – (0)2 = 12(12 x 2 – 1).
2

At (0, 0) : D = −12 , so (0, 0) is a saddle point.
At ( 0.5, 0 ) and ( −0.5, 0 ) : D = 24 and f xx = 6 , so local minima occur at these points.

4. ∇f ( x, y ) = y 2 – 12 x, 2 xy – 6 y = 0, 0 at stationary points (0, 0), (3, −6) and (3, 6).

D = f xx f yy – f xy = (–12)(2 x – 6) – (2 y )2 = –4( y 2 + 6 x – 18), f xx = –12
2

At (0, 0): D = 72, and f xx = –12, so local maximum at (0, 0).
At (3, ±6) : D = −144, so (3, ± y ) are saddle points.

5. ∇f ( x, y ) = y, x = 0, 0 at (0, 0), a stationary point.
D = f xx f yy – f xy = (0)(0) – (1)2 = –1, so (0, 0) is a saddle point.
2

6. Let ∇f ( x, y ) = 3 x 2 – 6 y, 3 y 2 – 6 x = 0, 0 . Then 3x 2 – 6 y = 0 and 3 y 2 − 6 x = 0.
1 4
3x 2 − 6 y = 0 → 3x 2 = 6 y → x 2 = 2 y → x 4 = 4 y 2 → x = y2
4
⎛1 ⎞
⎝4 ⎠
3
4
3
4
3
( ) (
3 y 2 − 6 x = 0 → 3 ⎜ x 4 ⎟ − 6 x = 0 → x 4 − 6 x = 0 → x x3 − 8 = 0 → x ( x − 2 ) x 2 + 2 x + 4 = 0 → x = 0, x = 2
4
)
x = 0 : 3 x 2 − 6 y = 0 → 3 ( 0 ) − 6 y = 0 → −6 y = 0 → y = 0
x = 2 : 3x 2 − 6 y = 0 → 3 ( 2 ) − 6 y → 12 − 6 y = 0 → 12 = 6 y → 2 = y
2

Solving simultaneously, we obtain the solutions (0, 0) and (2, 2).
f xx = 6 x; D = f xx f yy – f xy = (6 x)(6 y ) – (–6)2 = 36 ( xy – 1) ; At (0, 0): D = −36 < 0 , so (0, 0) is a saddle point.
2

At (2, 2): D = 108 > 0 , f xx > 0, so a local minimum occurs at ( 2, 2 ) .


x 2 y − 2 xy 2 − 4
7. ∇f ( x, y ) = , = 0, 0 at (1, 2).
x2 y2
D = f xx f yy – f xy = (4 x – 3)(8 y – 3) – (1)2 = 32 x −3 y −3 − 1, f xx = 4 x −3
2

At (1, 2): D = 3 > 0 , and f xx > 0, so a local minimum at (1, 2).

8. ∇f ( x, y ) = –2 exp(– x 2 – y 2 + 4 y ) x, y – 2 = 0, 0 at (0, 2).
D = f xx f yy – f xy = exp 2(− x 2 − y 2 + 4 y )[(4 x 2 − 2)(4 y 2 − 16 y + 14) − (4 xy − 8 x) 2 ],
2

f xx = (4 x 2 – 2) exp(– x 2 – y 2 + 4 y )
At (0, 2) : D > 0 , and f xx < 0, so local maximum at (0, 2).

9. Let ∇f ( x, y ) 13. ∇f ( x, y ) = 2 x, – 2 y = 0, 0 at (0, 0).
= – sin x – sin( x + y ), – sin y – sin( x + y ) = 0, 0 D = f xx f yy – f xy = (2)(–2) – (0)2 < 0, so (0, 0) is
2

⎛ – sin x – sin( x + y ) = 0 ⎞ a saddle point. A parametric representation of the
Then ⎜ ⎟ . Therefore,
⎝ sin y + sin( x + y ) = 0 ⎠ boundary of S is x = cos t, y = sin t, t in
π [0, 2π ] .
sin x = sin y , so x = y = . However, these values
4 f ( x, y ) = f ( x(t ), y (t )) = cos 2 t – sin 2 t + 1
satisfy neither equation. Therefore, the gradient is = cos 2t – 1
defined but never zero in its domain, and the cos 2t – 1 is maximum if cos 2t = 1, which occurs
boundary of the domain is outside the domain, so for t = 0, π , 2π . The points of the curve are
there are no critical points.
(±1, 0). f (±1, 0) = 2
10. ∇f ( x, y ) = 2 x – 2a cos y, 2ax sin y = 0, 0 at f(x, y) = cos 2t – 1 is minimum if cos 2t = –1,
π 3π
⎛ π⎞ which occurs for t = , . The points of the
⎜ 0, ± ⎟ , (a, 0) 2 2
⎝ 2⎠
curve are (0, ±1). f (0, ±1) = 0 . Global minimum
D = f xx f yy – f xy = (2)(2ax cos y ) – (2a sin y )2 ,
2
of 0 at (0, ± 1); global maximum of 2 at (±1, 0).
f xx = 2
14. ∇f ( x, y ) = 2 x − 6, 2 y − 8 = 0, 0 at (3, 4),
⎛ π⎞ ⎛ π⎞
At ⎜ 0, ± ⎟ : D = –4a 2 < 0, so ⎜ 0, ± ⎟ are which is outside S, so there are no stationary
⎝ 2⎠ ⎝ 2⎠
points. There are also no singular points.
saddle points. At (a, 0): D = 4a 2 > 0 and x = cos t , y = sin t , t in [0, 2π ] is a parametric
f xx > 0, so local minimum at (a, 0). representation of the boundary of S.
f ( x, y ) = f ( x(t ), y (t ))
11. We do not need to use calculus for this one. 3x is
minimum at 0 and 4y is minimum at –1. (0, –1) is = cos 2 t – 6 cos t + sin 2 t – 8sin t + 7
in S, so 3x + 4y is minimum at (0, –1); the = 8 − 6 cos t − 8sin t = F (t )
minimum value is –4. Similarly, 3x and 4y are 4
each maximum at 1. (1, 1) is in S, so 3x + 4y is F ′(t ) = 6sin t – 8cos t = 0 if tan t = . t can be
3
maximum at (1, 1); the maximum value is 7. (Use
in the 1st or 3rd quadrants. The corresponding
calculus techniques and compare.)
⎛ 3 4⎞
points of the curve are ⎜ ± , ± ⎟ .
12. We do not need to use calculus for this one. Each ⎝ 5 5⎠
of x 2 and y 2 is minimum at 0 and (0, 0) is in S, ⎛ 3 4⎞ ⎛ 3 4⎞
f ⎜ – , – ⎟ = 18; f ⎜ , ⎟ = –2
so x 2 + y 2 is minimum at (0, 0); the minimum ⎝ 5 5⎠ ⎝5 5⎠
⎛3 4⎞
value is 0. Similarly, x 2 and y 2 are maximum at Global minimum of –2 at ⎜ , ⎟ ; global
⎝5 5⎠
x = 3 and y = 4, respectively, and (3, 4) is in S, so
⎛ 3 4⎞
x 2 + y 2 is maximum at (3, 4); the maximum maximum of 18 at ⎜ – , – ⎟ .
⎝ 5 5⎠
value is 25. (Use calculus techniques and
compare.)


15. Let x, y, z denote the numbers, so x + y + z = N. 17. Let S denote the surface area of the box with
Maximize dimensions x, y, z.
P = xyz = xy ( N – x – y ) = Nxy – x 2 y – xy 2 . S = 2xy + 2xz + 2yz and V0 = xyz , so

Let ∇P ( x, y ) = Ny – 2 xy – y 2 , Nx – x 2 – 2 xy S = 2( xy + V0 y –1 + V0 x –1 ).
Minimize f ( x, y ) = xy + V0 y –1 + V0 x –1 subject
= 0, 0 .
to x > 0, y > 0.
⎛ Ny – 2 xy – y 2 = 0 ⎞
Then ⎜ ⎟. ∇f ( x, y ) = y – V0 x –2 , x – V0 y –2 = 0, 0 at
⎜ Nx – x 2 – 2 xy = 0 ⎟
⎝ ⎠
(V01/ 3 , V01/ 3 ).
N ( x, − y ) = x − y = ( x + y )( x − y ). x = y or
2 2
D = f xx f yy – f xy = 4V02 x –3 y –3 – 1,
2
N = x + y.
Therefore, x = y (since N = x + y would mean that f xx = 2V0 x –3 .

( )
P = 0, certainly not a maximum value).
At V0 3 , V0 3 : D = 3 > 0, f xx = 2 > 0, so
1/ 1/
Then, substituting into Nx – x 2 – 2 xy = 0, we
local minimum.
obtain Nx – x 2 – 2 x 2 = 0, from which we obtain
Conclusion: The box is a cube with edge V0 3 .
1/
N
x(N – 3x) = 0, so x = (since x = 0 ⇒ P = 0).
3 18. Let L denote the sum of edge lengths for a box of
Pxx = –2 y; dimensions x, y, z. Minimize L = 4x + 4y + 4z,
D = Pxx Pyy – Pxy
2 subject to V0 = xyz.
4V0
= (−2 y )(−2 x) − ( N − 2 x − 2 y )2 L ( x, y ) = 4 x + 4 y + , x > 0, y > 0
xy
= 4 xy − ( N − 2 x − 2 y )2 Let
∇L( x, y ) = 4 x −1 y −1 x −1 ( x 2 y − V0 ), y −1 ( xy 2 − V0 )
2
N N 2N
At x = y = :D = > 0, Pxx = – < 0 (so
3 3 3
local maximum) = 0,0 .
N
If x = y = , then z = .
N Then x 2 y = V0 and xy 2 = V0 , from which it
3 3
follows that x = y. Therefore x = y = z = V0 3 .
1/
N
Conclusion: Each number is . (If the intent is 8V0
3 Lxx = ;
to find three distinct numbers, then there is no x3 y
maximum value of P that satisfies that 2
⎛ 8V ⎞ ⎛ 8V ⎞ ⎛ 4V0 ⎞
D= Lxx L yy − L2 = ⎜ 0 ⎟⎜ 0 ⎟ − ⎜
⎜ x3 y ⎟ ⎜ xy3 ⎟ ⎜ x 2 y 2 ⎟
condition.) xy ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
( )
16. Let s be the distance from the origin to (x, y, z) on
At V0 3 , V0 3 : D > 0, Lxx > 0 (so local
1/ 1/
the plane. s 2 = x 2 + y 2 + z 2 and
x + 2 y + 3z = 12. Minimize minimum).
There are no other critical points, and as
s 2 = f ( y , z ) = (12 – 2 y – 3 z )2 + y 2 + z 2 . (x, y) → boundary, L → ∞ . Hence, the optimal
∇f ( y, z ) = –48 + 12 x + 10 y, – 72 + 12 y + 20 z box is a cube with edge V0 3 .
1/

⎛ 12 18 ⎞
= 0, 0 at ⎜ , ⎟ . 19. Let S denote the area of the sides and bottom of
⎝7 7⎠
the tank with base l by w and depth h.
D = f yy f zz – f yz = 56 > 0 and f yy = 10 > 0;
2
S = lw + 2lh + 2wh and lwh = 256.
⎛ 12 18 ⎞ ⎛ 256 ⎞ ⎛ 256 ⎞
local maximum at ⎜ , ⎟ S (l , w) = lw + 2l ⎜ ⎟ + 2w ⎜ ⎟ , w > 0, l > 0.
⎝7 7⎠ ⎝ lw ⎠ ⎝ lw ⎠
504 S (l w) = w – 5121–2 , l – 512w –2 = 0, 0 at
s2 = , so the shortest distance is
49 (8, 8). h = 4 there. At (8, 8) D > 0 and S11 > 0,
s=
6 14
≈ 3.2071. so local minimum. Dimensions are 8’ × 8’ × 4’.
7


20. Let V denote the volume of the box and (x, y, z) denote its 1st octant vertex.
V = (2x)(2y)(2z) = 8xyz and 24 x 2 + y 2 + z 2 = 9.
⎡⎛ 1 ⎞ ⎤
V 2 = 64 ⎢⎜ ⎟ (9 – y 2 – z 2 ) ⎥ y 2 z 2
⎣⎝ 24 ⎠ ⎦
Maximize f ( y , z ) = (9 – y 2 – z 2 ) y 2 z 2 , y > 0,
z > 0.
2
∇f ( y, z ) = 2 yz 2 (9 – 2 y 2 – z 2 ), y 2 z (9 – y 2 – 2 z 2 = 0, 0 at ( 3, 3). x =
4
At ( )
3, 3 , D = f yy f zz – f yz > 0 and f yy < 0, so local maximum. The greatest possible volume is
2

⎛ 2⎞
8⎜
⎜ 4 ⎟
⎟ ( 3 )( 3 ) = 6 2.
⎝ ⎠

21. Let x, y, z denote the vector; let S be the sum of its components.
x 2 + y 2 + z 2 = 81, so z = (81 – x 2 – y 2 )1/ 2 .
Maximize S ( x, y ) = x + y + (81 – x 2 – y 2 )1/ 2 , 0 ≤ x 2 + y 2 ≤ 9.
Let ∇S ( x, y ) = 1 – x(81 – x 2 – y 2 ) –1/ 2 , 1 – y (81 – x 2 – y 2 ) –1/ 2 = 0, 0 .

Therefore, x = (81 – x 2 – y 2 )1/ 2 and y = (81 – x 2 – y 2 )1/ 2 . We then obtain x = y = 3 3 as the only stationary
point. For these values of x and y, z = 3 3 and S = 9 3 ≈ 15.59.
The boundary needs to be checked. It is fairly easy to check each edge of the boundary separately. The largest
18
value of S at a boundary point occurs at three places and turns out to be ≈ 12.73.
2
Conclusion: the vector is 3 3 1, 1, 1 .

22. Let P ( x, x, z ) be any point in the plane 2 x + 4 y + 3z = 12. The square of the distance between the origin and P is
d 2 = x 2 + y 2 + z 2 . Consequently, d 2 = f ( x, y ) = x 2 + y 2 + (12 − 2 x − 4 y ) 2 9. To find the critical points, set
f x ( x, y ) = 2 x + 9 (12 − 2 x − 4 y )(−2) = 0 and f y ( x, y ) = 2 y + 9 (12 − 2 x − 4 y )(−4) = 0 The resulting system of
2 2

equations is 13x + 8 y = 24 and 8 x + 25 y = 48, which leads to a critical point of( 24 , 29 ) . Since f xx ( x, y) = 26 ,
29
48
9

f yy ( x, y ) = 9 ( 29 29 ) 9 ( 29 29 )
50 , and f ( x, y ) = 16 , D 24 , 48 = 116 Since D 24 , 48 > 0 and f
xy 9 xx ( 29 29 ) ( 29 29 )
24 , 48 > 0, 24 , 48

yields a minimum distance. The point on the plane 2 x + 4 y + 3z = 12 that is closest to the origin is ( 24 , 29 , 36 )
29
48
29
and this minimum distance is approximately 2.2283.

23. Let P( x, y, z ) be any point on z = x 2 + y 2 . The square of the distance between the point (1, 2, 0) and P can be
expressed as d 2 = f ( x, y ) = ( x − 1)2 + ( y − 2)2 + z 2 . To find the critical points, set f x ( x, y )
= 4 x3 + 2 x + 4 xy 2 − 2 = 0 and f y ( x, y ) = 4 y 3 + 2 y + 4 x 2 y − 4 = 0. Multiplying the first equation by y and the
second equation by x and summing the results leads to the equation −2 y + 4 x = 0. Thus, y = 2 x . Substituting
into the first equation yields 10 x3 + x − 1 = 0, whose solution is x ≈ 0.393. Consequently, y ≈ 0.786.
f xx ( x, y ) = 2 + 12 x 2 + 4 y 2 , f yy ( x, y ) = 2 + 12 y 2 + 4 x 2 , and f xy ( x, y ) = 8 xy. The value of D for the critical point
(0.393, 0.786) is approximately 57 and since f xx (0.393, 0.786) > 0, (0.393, 0.786) yields a minimum distance.
The point on the surface z = x 2 + y 2 is (0.393, 0.786, 0.772) and this minimum distance is approximately 1.56.


24. Let (x, y, z) denote a point on the cone, and s denote the distance between (x, y, z) and (1, 2, 0).
s 2 = ( x − 1)2 + ( y − 2) 2 + z 2 and z 2 = x 2 + y 2 . Minimize s 2 = f ( x, y ) = ( x – 1)2 + ( y – 2)2 + ( x 2 + y 2 ), x, y in R.
⎛1 ⎞ ⎛1 ⎞
∇f ( x, y ) = 2 2 x − 1, 2 y − 2 = 0, 0 at ⎜ , 1⎟ . At ⎜ , 1⎟ , D > 0 and f xx > 0, so local minimum.
⎝2 ⎠ ⎝2 ⎠
5
Conclusion: Minimum distance is s = ≈ 1.5811.
2

⎛1⎞
25. A = ⎜ ⎟ [ y + ( y + 2 x sin α )]( x cos α ) and
⎝2⎠
⎛1⎞ ⎛ π⎞
2x + y = 12. Maximize A( x, α ) = 12 x cos α – 2 x 2 cos α + ⎜ ⎟ x 2 sin 2α , x in (0, 6], a in ⎜ 0, ⎟ .
⎝2⎠ ⎝ 2⎠
⎛ π⎞
A( x, α ) = 12 cos α – 4 x cos α + 2 x sin α cos α , – 12 x sin α + 2 x 2 sin α + x 2 cos 2α = 0, 0 at ⎜ 4, ⎟ .
⎝ 6⎠
⎛ π⎞
At ⎜ 4, ⎟ , D > 0 and Axx < 0, so local maximum, and A = 12 3 ≈ 20.78. At the boundary point of x = 6, we get
⎝ 6⎠
π π π 2π
α = , A = 18. Thus, the maximum occurs for width of turned-up sides = 4”, and base angle = + = .
4 2 6 3

26. The lines are skew since there are no values of s and t that simultaneously satisfy t – 1 = 3s, 2t = s + 2, and
t + 3 = 2s – 1. Minimize f, the square of the distance between points on the two lines.
f ( s, t ) = (3s – t + 1)2 + ( s + 2 – 2t )2 + (2s – 1 – t – 3)2
Let
∇f ( s, t ) = 2(3s – t + 1)(3) + 2( s – 2t + 2)(1) + 2(2 s – t – 4)(2), 2(3s – t + 1)(–1) + 2( s – 2t + 2)(–2) + 2(2 s – t – 4)(–1)
= 28s − 14t − 6, −14 s + 12t − 28 = 0, 0 .
5
Solve 28s – 14t – 6 = 0, –14s + 12t – 2 = 0, obtaining s = , t = 1.
7
D = f ss ftt – f st = (28)(12) – (–14)2 > 0; f ss = 28 > 0. (local minimum)
2

The nature of the problem indicates the global minimum occurs here.
2 2 2
⎛ 5 ⎞ ⎛ 15 ⎞ ⎛ 5 ⎞ ⎛ 25 ⎞ 875
f ⎜ , 1⎟ = ⎜ ⎟ + ⎜ ⎟ + ⎜ – ⎟ =
⎝7 ⎠ ⎝ 7 ⎠ ⎝7⎠ ⎝ 7 ⎠ 49
Conclusion: The minimum distance between the lines is 875 / 7 ≈ 4.2258.

27. Let M be the maximum value of f(x, y) on the b. x y –3x + 2y + 1
polygonal region, P. Then ax + by + (c – M) = 0 –3 0 10
is a line that either contains a vertex of P or
divides P into two subregions. In the latter case 0 5 11
ax + by + (c – M) is positive in one of the regions 2 3 1
and negative in the other. ax + by + (c – M) > 0 4 0 –11
contradicts that M is the maximum value of 1 –4 –10
ax + by + c on P. (Similar argument for
Minimum at (4, 0)
minimum.)
a. x y 2x + 3y + 4 28. x y 2x + y
–1 2 8 0 0 0
0 1 7 2 0 4
1 0 6 1 4 6
–3 0 –2 0 14 / 3 14 / 3
0 –4 –8
Maximum of 6 occurs at (1, 4 )
Maximum at ( −1, 2 )


∂f n
∂
30. a. =∑ ( yi − mxi − b )2
∂m i =1 ∂m
n
= 2∑ ( yi − mxi − b )(− xi )
i =1

( )
n
= −2∑ xi yi − mxi 2 − bxi
i =1
Setting this result equal to zero yields

( )
n
0 = −2∑ xi yi − mxi 2 − bxi
The edges of P are segments of the lines: i =1
1. y = 0
( )
n
2. 4x + y = 8 0 = ∑ xi yi − mxi 2 − bxi
3. 2x + 3y = 14, and i =1
4. x = 0 or equivalently,
n n n
29. z ( x, y ) = y 2 – x 2 ∑ xi yi = m∑ xi 2 + b∑ xi
i =1 i =1 i =1
z ( x, y ) = –2 x, 2 y = 0, 0 at (0, 0).
There are no stationary points and no singular
∂f n
∂
= ∑ ( yi − mxi − b )
points, so consider boundary points. 2
∂b i =1 ∂b
n
= 2∑ ( yi − mxi − b )(−1)
i =1
n
= −2∑ ( yi − mxi − b )
i =1
Setting this result equal to zero yields
n
0 = −2∑ ( yi − mxi − b )
i =1
On side 1:
n
y = 2x, so z = 4 x 2 – x 2 = 3x 2 0 = ∑ ( yi − mxi − b )
z ′( x) = 6 x = 0 if x = 0. i =1
Therefore, (0, 0) is a candidate. or equivalently,
n n
On side 2:
y = –4x + 6, so m∑ xi + nb = ∑ yi
i =1 i =1
z = (–4 x + 6)2 – x 2 = 15 x 2 – 48 x + 36.
z ′( x) = 30 x – 48 = 0 if x = 1.6. n n
b. nb = ∑ yi − m∑ xi
Therefore, (1.6, –0.4) is a candidate.
i =1 i =1
On side 3:
Therefore,
y = –x, so z = (– x) 2 – x 2 = 0. n n
Also, all vertices are candidates. ∑ yi − m∑ xi
i =1 i =1
b=
x y z n
0 0 0
n n n
1.6 –0.4 –2.4
2 –2 0
∑ xi yi = m∑ xi 2 + b∑ xi
i =1 i =1 i =1
1 2 3 ⎛ n n ⎞ n
Minimum value of –2.4; maximum value of 3 n
⎜ ∑ yi − m∑ xi ⎟ ∑ xi
⎜ ⎟
n
∑ xi yi = m∑ xi 2 + ⎝ i =1 i =1 ⎠ i =1
n
i =1 i =1


This simplifies into z ( x ) = 2 x 2 – 4 x + 5, x in [0, 4]. z ′( x) = 4 x – 4,
n n n
1
∑ xi yi − n ∑ xi ∑ yi so z ′( x) = 0 if x = 1. Hence, an additional critical
i =1 i =1 i =1 point is (1, 0).
m=
2 On hypotenuse: x = 4 – 4y
n
1⎛ n ⎞
∑ xi 2 − n ⎜ ∑ xi ⎟
⎜ ⎟ z ( y ) = 2(4 – 4 y ) 2 + y 2 – 4(4 – 4 y ) – 2 y + 5
i =1 ⎝ i =1 ⎠
= 33 y 2 – 50 y + 21, y in [0, 1].
∂2 f n 25
c. = 2∑ xi 2 z ′( y ) = 66 y – 50, so z ′( y ) = 0 if y =
.
∂m 2
i =1
33
⎛ 32 25 ⎞
∂ f
2 Hence, an additional critical point is ⎜ , ⎟ .
= 2n ⎝ 33 33 ⎠
∂b 2
x y z
∂2 f n
= 2∑ xi 0 0 5
∂m∂b i =1 4 0 21
Then, by Theorem C, we have
0 1 4
⎛ n 2⎞
1⎛ n ⎞
D = 4n ⎜ ∑ xi 2 − ⎜ ∑ xi ⎟ ⎟ .
1 0 3
⎜
⎜ i =1 n ⎜ i =1 ⎟ ⎟
⎝ ⎠ ⎟ 32 / 33 25 / 33 2.06
⎝ ⎠
Assuming that all the xi are not the same, we Maximum value of z is 21; it occurs at (4, 0).
Minimum value of z is about 2.06; it occurs at
∂ f2
find that D > 0 and >0. ⎛ 32 25 ⎞
∂m2 ⎜ , ⎟.
⎝ 33 33 ⎠
Thus, f (m, b) is minimized.
33. Let x and y be defined as shown in Figure 4 from
31. xi yi xi2 xi yi Section 12.8. The total cost is given by
3 2 9 6
C ( x, y ) = 400 x 2 + 502 + 200(200 − x − y )
4 3 16 12
5 4 25 20 +300 y 2 + 1002
6 4 36 24 Taking partial derivatives and setting them equal
7 5 49 35 to 0 gives
∑ 5=1 25 18 135 97 C x ( x, y ) = 200( x 2 + 502 )−1 2 (2 x) − 200 = 0
i
m(135) + b(25) = (97) and m(25) + (5)b = (18). C y ( x, y ) = 150( y 2 + 1002 )−1 2 (2 y ) − 200 = 0
Solve simultaneously and obtain m = 0.7, b = 0.1. The solution of these equations is
The least-squares line is y = 0.7x + 0.1.
50 100
x= ≈ 28.8675 and y = ≈ 89.4427
32. z = 2 x 2 + y 2 – 4 x – 2 y + 5, so 3 1.25
We now apply the second derivative test:
∇z = 4 x − 4, 2 y − 2 = 0.
∇z = 0 at (1, 1) which is outside the region. 400 x 2 + 502 − 400 x 2 x 2 + 502
C xx ( x, y ) =
Therefore, extreme values occur on the boundary. x 2 + 502
Three critical points are the vertices of the 300 y 2 + 1002 − 300 y 2 y 2 + 1002
triangle, (0, 0), (0, 1), and (4, 0). Others may C yy ( x, y ) =
occur on the interior of a side of the triangle. y 2 + 1002
C xy ( x, y ) = 0
Evaluated at x = 50 3 and y = 100 1.25 ,
D ≈ (5.196)(1.24) − 0 > 0 2

Thus, a local minimum occurs with
( )
On vertical side: x = 0
C 50 3 ,100 1.25 ≈ $79, 681
z ( y ) = y 2 – 2 y + 5, y = [0, 1]. z ′( y ) = 2 y – 2, so
We must also check the boundary. When x = 0,
z ′( y ) = 0 if y = 1. Hence, no additional critical
point. C1 ( y ) = C (0, y ) = 200(200 − y ) + 300 y 2 + 1002
On horizontal side: y = 0 and when y = 0,


C2 ( x) = C ( x, 0) = 400 x 2 + 502 + 200(200 − x) yields t = π 4 or 5 π 4.
Using the methods from Chapter 3, we find that The critical points are 3 ( 2,3 )
2 and

( −3 )
C1 reaches a minimum of about $82,361 when
2 , −3 2 .
y = 8000 and C2 reaches a minimum of about
$87,321 when x = 2500 3 . Addressing the
Since f 3 ( 2 ,3 )
2 = 10 + 6 2 and

boundary x + y = 200, we find that (
f −3 2 ,−3 )
2 = 10 − 6 2 , the
C3 ( x) = C ( x, 200 − x) = 400 x + 50 2 2
minimum value of f on x 2 + y 2 ≤ 3 is
+300 (200 − x)2 + 1002 This function reaches a 10 − 6 2 and the maximum value of f is 10 + 6 2.
minimum of about $82,214 when x ≈ 41.08 .
Thus, the minimum cost path is when 36. f ( x, y ) = x 2 + y 2 ; ∇f = 2 x , 2 y .
x = 50 3 ≈ 28.8675 ft and ∇f = 0 at (0,0).
y = 100 1.25 ≈ 89.4427 ft, which produces a
D(0, 0) = 2 ⋅ 2 − 02 = 4 > 0 and f xx (0, 0) = 2 > 0,
cost of about $79,681.
Thus, f (0, 0) = 0 is a minimum.
34. Let x and y be defined as shown in Figure 4 from In order to optimize g (t ) = f (a cos t , b sin t )
Section 12.8. The total cost is given by where 0 ≤ t ≤ 2π , we find
C ( x, y ) = 500 x 2 + 502 + 200(200 − x − y ) g ′(t ) = 2 x(−a sin t ) + 2 y (b cos t )

+100 y 2 + 1002 = 2b 2 sin t cos t − 2a 2 sin t cos t
Taking partial derivatives and setting them equal = (b 2 − a 2 ) sin 2t. Setting g ′(t ) = 0, we
to 0 gives have t = 0, π 2, π , or 3π 2. The resulting critical
C x ( x, y ) = 500( x 2 + 502 ) −1 2 (2 x) − 200 = 0 points are (a, 0) , (0, b) , (−a, 0) , and (0, −b) .
C y ( x, y ) = 100( y 2 + 1002 )−1 2 (2 y ) − 200 = 0 f (a, 0) = f (− a, 0) = a 2 ; f (0, b) = f (0, −b) = b 2 .
There is, however, no solution to C y ( x, y ) = 0 Since a > b, the maximum value of f on the given
region is a2 and the minimum value of f is 0.
Now we check the boundary. When x = 0,
C1 ( y ) = C (0, y ) = 200(200 − y ) + 100 y 2 + 1002 37. The volume of the box can be expressed as
V (l , w, h) = lwh = 2 and the surface area as
There is, however, no solution to C1′ ( y ) = 0 .
S ( l , w, h ) = 2lh + 2 wh + lw + lw . Since h = lw ,
2
When y = 0 ,
S (l , w) = w + 4 + lw + lw When cost is factored,
4
C2 ( x) = C ( x, 0) = 500 x 2 + 502 + 200(200 − x) l
1 1
C2′ ( x) = 0 yields x = 100 21 and C (l , w) = + + 0.65lw with w > 0, l > 0
w l
(
C 100 )
21, 0 ≈ $72,913 1
Cl (l , w) = − + 0.65w = 0
On the boundary x + y = 200 , we find that l2
1
C3 ( x) = C ( x, 200 − x) = 500 x 2 + 502 Cw (l , w) = − + 0.65l = 0
w2
+100 (200 − x)2 + 1002 This function reaches a Solving this system of equations leads to
minimum of about $46,961 when x ≈ 9.0016 . 0.65
Thus, the minimum cost path is when w=3 ≈ 1.1544 and l = w ≈ 1.1544 .
0.4225
x ≈ 9.0016 ft and y ≈ 190.9984 ft, which Consequently, h ≈ 1.501. Applying the second
produces a cost of about $46,961. 2
derivative test with Cll (l , w) = ,
35. f ( x, y ) = 10 + x + y l3
2
∇f = 1,1 ≠ 0; thus no interior critical Cww (l , w) = and Clw (l , w) = 0.65,
w3
points exist. Letting D ≈ 1.268 > 0. Thus, the minimum cost occurs
x = 3cos t , y = 3sin t , 0 ≤ t ≤ 2π , when the length is approximately 1.1544 feet, the
g (t ) = f (3cos t ,3sin t ) and width is approximately 1.1544 feet and the height
g ′(t ) = 3cos t − 3sin t. Setting g ′(t ) = 0 is approximately 1.501 feet.


38. The cost function in three variables is 1
C (l , w, h) = 4lw + 2lh + 2wh + 6l + 6 w + 4h, T ′( y ) = –1 – 2 y = 0 if y = – , so on the
2
60 boundary, critical points occur where y is
where lwh = 60. Substituting h = yields
lw 1
–1, – , 1.
120 240 2
C (l , w) = 4lw + + 6l + 6 w + with
w lw ⎛ 1⎞
Thus, points to consider are ⎜ 0, ⎟ , (0, –1),
l > 0 and w > 0. ⎝ 2⎠
120 240 ⎛ 3 1⎞ ⎛ 3 1 ⎞
Cl (l , w) = 4w − +6− =0 ⎜
2
wl 2 ⎜ 2 , – 2 ⎟ , ⎜ – 2 , – 2 ⎟ and (0, 1). Substituting
⎟ ⎜ ⎟
l ⎝ ⎠ ⎝ ⎠
120 240 these into T(x, y) yields that the coldest spot is
Cw (l , w) = 4l − +6− =0
w 2
lw2 ⎛ 1⎞ 1
⎜ 0, ⎟ where the temperature is – , and there
Multiplying both sides of the first equation by ⎝ 2⎠ 4
wl 2 , multiplying both sides of the second ⎛ 3 1⎞
2 is a tie for the hottest spot at ⎜ ±
⎜ 2 , – 2 ⎟ where
⎟
equation by lw , and subtracting the resulting ⎝ ⎠
equations produces −120w + 120l = 0 or l = w. 9
120 240 the temperature is .
Consequently, 4w − +6− = 0 or 4
2
w w3
2w4 + 3w3 − 60 w − 120 = 0 Using a CAS, this 40. Let x 2 , y 2 , z 2 denote the areas enclosed by the
equation yields w ≈ 3.2134 circle, and the two squares, respectively. Then
240 480 240 480 x
Cll (l , w) = + ; Cww (l , w) = + ; the radius of the circle is , and the edges of
l 3
wl 3
w3 lw3 π
240 the two squares are y and z, respectively. We
Clw (l , w) = 4 + ; Using the critical point
l 2 w2 wish to optimize A(x, y, z) = x 2 + y 2 + z 2 ,
(3.2134, 3.2134), D ≈ 131.44 > 0 ⎛ x ⎞
subject to 2π ⎜ ⎟ + 4 y + 4 z = k , or
Thus, w = l ≈ 3.2 yields a minimum. The ⎝ π⎠
minimum cost involved with making this box is
approximately $177.79. This minimum cost equivalently 2 π x + 4 y + 4 z = k , with each of x,
occurs when the length and width are y, and z nonnegative. Geometrically: we seek the
approximately 3.2 feet and the height is smallest and largest of all spheres with center at
approximately 5.8 feet. the origin and some point in common with the
triangular region indicated.
39. T ( x, y ) = 2 x 2 + y 2 – y
∇T = 4 x , 2 y – 1 = 0
1 ⎛ 1⎞
If x = 0 and y = , so ⎜ 0, ⎟ is the only interior
2 ⎝ 2⎠
critical point.

k k
Since > , the largest sphere will intersect
2 π 4
⎛ k ⎞
the region only at point ⎜ , 0, 0 ⎟ and will
⎝2 π ⎠
k
thus have radius . Thus A will be maximum
On the boundary x 2 = 1 – y 2 , so T is a function 2 π
k
of y there. if x = , y = z = 0 (all of the wire goes into
T ( y ) = 2(1 – y 2 ) + y 2 – y = 2 – y – y 2 , 2 π
the circle). The smallest sphere will be tangent to
y = [–1, 1] the triangle. The point of tangency is on the
normal line through the origin,


x, y , z = t π , 2, 2 . Substituting x = π ,
y = 2, z = 2 into the equation of the plane yields
k
the value t = , so the minimum value of
2(π + 8)
k π
A is obtained for the values of x = ,
2(π + 8)
k
y=z= . Thus the circle will have radius
π+8
⎡ k π ⎤
⎢ 2( π+8) ⎥
⎣ ⎦= k
, and the squares will each
π 2(π + 8)
k
have sides . Therefore, the circle will use
(π + 8)
πk
units and the squares will each use
(π + 8)
4k
units.
(π + 8)
[Note: sum of the three lengths is k.]

41. Without loss of generality we will assume that α ≤ β ≤ γ . We will consider it intuitively clear that for a triangle of
maximum area the center of the circle will be inside or on the boundary of the triangle; i.e., α , β , and γ are in the
interval [0,π ] . Along with α + β + γ = 2π, this implies that α + β ≥ π.

1 2
The area of an isosceles triangle with congruent sides of length r and included angle θ is r sin θ .
2
1 2 1 1
Area(ΔABC ) = r sin α + r 2 sin β + r 2 sin γ
2 2 2
1 2
= r (sin α + sin β + sin[2π − (α + β )]
2
1 2
= r [sin α + sin β − sin(α + β )]
2
Area(ΔABC ) will be maximum if (*) A(α , β ) = sin α + sin β − sin(α + β ) is maximum.
Restrictions are 0 ≤ α ≤ β ≤ π, and α + β ≥ π.
⎛π π⎞
Three critical points are the vertices of the triangular domain of A : ⎜ , ⎟ , (0, π), and (π, π). We will now search
⎝2 2⎠
for others.
ΔA(α , β ) = cos α − cos(α + β ), cos β − cos(α + β ) = 0 if
cos α = cos(α + β ) = cos β .
Therefore, cos α = cos β , so α = β [due to the restrictions stated]. Then
cos α = cos(α + α ) = cos 2α = 2 cos 2 α − 1, so cos α = 2 cos 2 α − 1.
Solve for α : 2 cos 2 α − cos α − 1 = 0; (2 cos α + 1)(cos α − 1) = 0;
1 2π
cos α = − or cos α = 1; α = or α = 0.
2 3


⎛ 2π 2π ⎞
(We are still in the case where α = β .) ⎜ , ⎟ is a new critical point, but (0, 0) is out of the domain of A.
⎝ 3 3 ⎠
There are no critical points in the interior of the domain of A.
On the β = π edge of the domain of A; A(α ) = sin α − sin(α − π) = 2sin α so A′(α ) = 2 cos α .
π ⎛π ⎞
A′(α ) = 0 if α =. ⎜ , π ⎟ is a new critical point.
2 ⎝2 ⎠
On the β = π − α edge of the domain of A:
A(α ) = sin α + sin(π − α ) − sin(2α − π) = 2sin α + sin 2α , so
A′(α ) = 2 cos α + 2 cos 2α = 2[cos α + (2 cos 2 α − 1)] = 2(2 cos α − 1)(cos α + 1) .
1 π
A′(α ) = 0 if cos α = or cos α = −1, so α = or α = π .
2 3
⎛ π 2π ⎞
⎜ , ⎟ and (π ,0) are outside the domain of A.
⎝3 3 ⎠
(The critical points are indicated on the graph of the domain of A.)

α β A
π π 2
2 2
0 π 0
π π 0
2π 2π 3 3
Maximum value of A. The triangle is equilateral.
3 3 2
π π 2
2

42. If the plane through (a, b, c) is expressed as
1 1 1
Ax + By + Cz = 1, then the intercepts are , , ; volume
A B C
⎛ 1 ⎞ ⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎛ 1 ⎞ 1
of tetrahedron is V = ⎜ ⎟ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ = .
⎝ 3 ⎠ ⎣⎝ 2 ⎠ ⎝ A ⎠ ⎝ B ⎠ ⎦ ⎝ C ⎠ 6 ABC
To maximize V subject to Aa + Bb + Cc = 1 is equivalent to
maximizing z = ABC subject to Aa + Bb + Cc = 1.
1 − aA − bB AB(1 − aA − bB)
C= , so z = .
c c
⎛1⎞ 1 1 ⎡ 1⎤
∇z = ⎜ ⎟ B − 2aAB − bB 2 , A − 2bAB − aA2 = 0 if A = , B = ⎢C = 3c ⎥ .
⎝c⎠ 3a 3b ⎣ ⎦
⎛ 1 1 ⎞
⎜ , ⎟ is the only critical point in the first quadrant. The second partials test yields that z is maximum at this
⎝ 3a 3b ⎠
1 1 1 x y z
point. The plane is x + y + z = 1, or + + = 3.
3a 3b 3c a b c
1 9abc
The volume of the first quadrant tetrahedron formed by the plane is = .
⎡6
⎣
1 1
3a 3b 3c ⎦
1 ⎤ 2 ( )( )( )
43. Local max: f(1.75, 0) = 1.15 46. Global max: f (0, 0) = 1
Global max: f(–3.8, 0) = 2.30
Global min: f (2, −2) = f (−2, 2) = e−9
44. Global max: f(0, 1) = 0.5 ≈ 0.00012341
Global min: f(0, –1) = –0.5
47. Global max: f(1.13, 0.79) = f(1.13, –0.79) = 0.53
45. Global min: f(0, 1) = f(0, –1) = –0.12 Global min: f(–1.13, 0.79) = f(–1.13, –0.79)
= –0.53


48. No global maximum or global minimum 52. Global max: f(–5.12, –4.92) = 1071
Global min: f(5.24, –4.96) = –658
49. Global max: f (3,3) = f (−3,3) ≈ 74.9225
Global min: f (1.5708, 0) = f ( −1.5708, 0) = −8 53. Global max: f(2.1, 2.1) = 3.5
Global min: f(4.2, 4.2) = –3.5
50. Global max: f(1, 43, 0) = 0.13
Global min: f(–1.82, 0) = –0.23

51. Global max: f(0.67, 0) = 5.06
Global min: f(–0.75, 0) = –3.54

54. a. 1
k (α , β ) = [80sin α + 60sin β + 48sin(2π − α − β )]
2
= 40sin α + 30sin β − 24sin(α + β )
L(α , β ) = (164 − 160 cos α )1/ 2 + (136 − 120 cos β )1/ 2
+ (100 − 96 cos(α + β ))1/ 2

b. (1.95, 2.04)

c. (2.26, 2.07)

12.9 Concepts Review 3. Let ∇f ( x, y ) = λ∇g ( x, y ), where

1. free; constrained g ( x, y ) = x 2 + y 2 – 1 = 0.
8 x - 4 y , -4 x + 2 y = λ 2 x, 2 y
2. parallel
1. 4x – 2y = λx
3. g(x, y) = 0 2. –2x + y = λy
3. x 2 + y 2 = 1
4. (2, 2)
4. 0 = λx + 2λy (From equations 1 and 2)
5. λ = 0 or x + 2y = 0 (4)
Problem Set 12.9 λ = 0: 6. y = 2x (1)
1
7. x = ± (6, 3)
1. 2 x, 2 y = λ y , x 5
2x = λy, 2y = λx, xy = 3 2
Critical points are 8. y = ± (7, 6)
5
(± ) (
3, ± 3 , f ± 3, ± 3 = 6. ) x + 2y = 0: 9. x = –2y
It is not clear whether 6 is the minimum or 1
10. y = ± (9, 3)
maximum, so take any other point on xy = 3, for 5
example (1, 3). f(1, 3) = 10, so 6 is the minimum 2
value. 11. x = (10, 9)
5
2. y, x = λ 8 x, 18 y ⎛ 1 2 ⎞ ⎛ 1 2 ⎞
Critical points: ⎜ , ⎟, ⎜ – ,– ⎟,
y = 8λx, x = 18λy, 4 x 2 + 9 y 2 = 36 ⎝ 5 5⎠ ⎝ 5 5⎠
⎛ 2 1 ⎞ ⎛ 2 1 ⎞
⎛ 3 2 ⎞ ⎛ 3 2 ⎞ ⎜ ,– ⎟, ⎜ – , ⎟
Critical points are ⎜ ,± ⎟, ⎜ – ,± ⎟. ⎝ 5 5⎠ ⎝ 5 5⎠
⎝ 2 2⎠ ⎝ 2 2⎠
f(x, y) is 0 at the first two critical points and 5 at
⎛ 3 2 ⎞ the last two. Therefore, the maximum value of
Maximum value of 3 occurs at ⎜ ± ,± ⎟.
⎝ 2 2⎠ f(x, y) is 5.


4. 2 x + 4 y, 4 x + 2 y = λ 1, -1 9. Let l and w denote the dimensions of the base, h
the depth. Maximize V(l, w, h) = lwh subject to
2x + 4y = λ, 4x + 2y = –λ, x – y = 6
0.601w + 0.20(lw + 2lh + 2wh) = 12, which
Critical point is (3, –3).
simplifies to 21w + lh + wh = 30, or
g(l, w, h) = 2lw + lh + wh – 30.
5. 2 x, 2 y, 2 z = λ 1,3, -2
Let ∇V (l , w, h) = λ∇g (l , w, h);
2x = λ, 2y = 3λ, 2z = –2λ, x + 3y – 2z = 12
wh, lh, lw = λ 2 w + h, 2l + h, l + w .
⎛ 6 18 12 ⎞
Critical point is ⎜ , , – ⎟ . 1. wh = λ(2w + h)
⎝7 7 7⎠
2. lh = λ(2l + h)
⎛ 6 18 12 ⎞ 72 3. lw = λ(l + w)
f ⎜ , , – ⎟= is the minimum.
⎝7 7 7⎠ 7 4. 2lw + lh + wh = 30
5. (w – l)h = 2λ(w – l) (1, 2)
6. Let ∇f ( x, y, z ) = λ∇g ( x, y, z ), where 6. w = l or h = 2λ
g ( x, y, z ) = 2 x 2 + y 2 – 3z = 0. w = 1:
7. l = 2λ = w (3) Note: w ≠ 0 , for then V = 0 .
4, −2,3 = λ 4 x, 2 y, −3 8. h = 4λ (7, 2)
1. 4 = 4λx 5
2. –2 = 2λy 9. λ = (7, 8, 4)
2
3. 3 = –3λ
10. l = w = 5, h = 2 5 (9, 7, 8)
4. 2 x 2 + y 2 – 3 z = 0
h = 2λ:
5. λ = –1 (3) 11. λ = 0 (2)
6. x = –1, y = 1 (5, 1, 2) 12. l = w = h = 0 (11, 1 – 3)
7. z = 1 (6, 4) (Not possible since this does not satisfy 4.)
Therefore, (–1, 1, 1) is a critical point, and
f(–1, 1, 1) = –3. (–3 is the minimum rather than ( )
5, 5, 2 5 is a critical point and
maximum since other points satisfying g = 0 have
larger values of f. For example, g(1, 1, 1) = 0, and
V ( )
5, 5, 2 5 = 10 5 ≈ 22.36 ft3 is the
f(1, 1, 1) = 5.) maximum volume (rather than the minimum
volume since, for example, g(1, 1, 14) = 30 and
7. Let l and w denote the dimensions of the base, h V(1, 1, 14) = 14 which is less than 22.36).
denote the depth. Maximize V(l, w, h) = lwh
subject to g(l, w, h) = lw + 2lh + 2wh = 48. 10. Minimize the square of the distance,
wh, lh, lw = λ w + 2h, l + 2h, 2l + 2 w f ( x, y , z ) = x 2 + y 2 + z 2 , subject to
wh = λ(w + 2h), lh = λ(l + 2h), lw = λ(2l + 2w), g ( x, y, z ) = x 2 y – z 2 + 9 = 0.
lw + 2lh + 2wh = 48
Critical point is (4, 4, 2). 2 x, 2 y, 2 z = λ 2 xy, x 2 , – 2 z
V(4, 4, 2) = 32 is the maximum. (V(11, 2, 1) = 22,
for example.) 2x = 2λxy, 2 y = λ x 2 , 2z = –2λz,

8. Minimize the square of the distance to the plane, x2 y – z 2 + 9 = 0
Critical points are (0, 0, ±3) [case x = 0];
f ( x, y , z ) = x 2 + y 2 + z 2 , subject to
x + 3y – 2z – 4 = 0. (± )
2, – 1, ± 7 [case x ≠ 0 , λ = –1]; and
2 x, 2 y, 2 z = λ 1, 3, -2
2x = λ, 2y = 3λ, 2z = –2λ, x + 3y – 2z = 4
( ±3 6 2 / 9, − 3 9 / 2, 0) [case x ≠ 0, λ ≠ −1 ].
Evaluating f at each of these eight points yields 9
⎛2 6 4⎞
Critical point is ⎜ , , – ⎟ . The nature of the (case x = 0), 10 (case x ≠ 0 , λ = –1), and
⎝7 7 7⎠
problem indicates that this will give a minimum
33 3 2
2
( )
2 9 (case x ≠ 0, λ ≠ −1 ). The latter is
rather than a maximum. The least distance to the
1/ 2 1/ 2
the smallest, so the least distance between the
⎡ ⎛2 6 4 ⎞⎤ ⎛8⎞ 3
plane is ⎢ f ⎜ , , – ⎟ ⎥ =⎜ ⎟ ≈ 1.0690. origin and the surface is 36 ≈ 2.8596.
⎣ ⎝7 7 7 ⎠⎦ ⎝7⎠ 4


11. Maximize f(x, y, z) = xyz, subject to 15. Minimize d 2 = f ( x, y, z )
2 2 2 2 2 2 2 2 2 2 2 2
g ( x, y , z ) = b c x + a c y + a b z – a b c
= ( x − 1) 2 + ( y − 2) 2 + z 2 with the constraint
=0
g ( x, y, z ) = x 2 + y 2 − z = 0;
yz , xz , xy = λ 2b 2 c 2 x, 2a 2 c 2 y, 2a 2 b 2 z
2 x − 2, 2 y − 4, 2 z = λ 2 x, 2 y, −1
yz = 2b 2 c 2 x, xz = 2a 2 c 2 y, xy = 2a 2 b 2 z , Setting up, solving each equation for λ , and
2 2 2 2 2 2 2 2 2 2 2 2
b c x +a c y +a b z = a b c substituting into equation x 2 + y 2 − z = 0
⎛ a b c ⎞ produces λ ≈ −1.5445; The resulting critical
Critical point is ⎜ , , ⎟.
⎝ 3 3 3⎠ point is approximately (0.393, 0.786, 0.772). The
nature of this problem indicates this will give a
⎛ a b c ⎞ 8abc
V⎜ , , ⎟= , which is the minimum value rather than a maximum. The
⎝ 3 3 3⎠ 3 3 minimum distance is approximately 1.5616.
maximum.
16. Minimize d 2 = f ( x, y, z )
12. Maximize V(x, y, z) = xyz, subject to
x y z = ( x − 1) 2 + ( y − 2) 2 + z 2 with the constraint
g ( x, y, z ) = + + –1 = 0. Let
a b c g ( x, y , z ) = x 2 + y 2 − z 2 = 0
∇V ( x, y, z ) = λ∇g ( x, y, z ), so 2 x − 2, 2 y − 4, 2 z = λ 2 x, 2 y, −2 z
1 1 1
yz , xz , xy = λ , , . Then λ = −1, x = 1 , y = 1, z = ± 5
; The critical points
a b c 2 2
λx λ y λz
a
=
b
=
c
(each equals xyz). are ( 1 ,1, 5
2 2 ) and ( 1 ,1, − 5
2 2 ) which both lead to
λ ≠ 0 since λ = 0 would imply x = y = z = 0 10
a minimum distance of .
which would not satisfy the constraint. 2
x y z
Thus, = = . These along with the 17. (See problem 37, section 12.8). Let the
a b c
dimensions of the box be l, w, and h . Then the
a b c cost of the box is
constraints yield x = , y = , z = .
3 3 3 .25(2hl + 2hw + lw) + .4(lw) or
The maximum value of V =
abc
. C (l , w, h) = .5hl + .5hw + .65lw .
27 We want to minimize C subject to the constraint
lhw = 2 ; set V (l , h, w) = lwh − 2 .
13. Maximize f(x,y,z) = x + y + z with the constraint
Now:
g ( x, y, z ) = x 2 + y 2 + z 2 − 81 = 0. Let ∇C (l , w, h) = (.5h + .65w)i + (.5h + .65l ) j + .5(l + w)k
∇f ( x, y, z ) = λ∇g ( x, y, z ) , so and
1,1,1 = λ 2 x, 2 y, 2 z ; Thus, x = y = z ∇V (l , w, h) = wh i + lh j + lw k
and 3x2 = 81 or x = y = z = ±3 3. Thus the Lagrange equations are
.5h + .65w = λ wh (1)
The maximum value of f is 9 3 when .5h + .65l = λ lh (2)
x, y, z = 3 3,3 3,3 3 .5(l + w) = λ lw (3)
lwh = 2 (4)
14. Minimize d 2 = f ( x, y, z ) = x 2 + y 2 + z 2 with Solving (4) for h and putting the result in (1) and
(2), we get
the constraint g ( x, y, z ) = 2 x + 4 y + 3z − 12 = 0
1 2λ
∇f ( x , y , z ) = λ ∇ g ( x , y , z ) + .65w = (5)
lw l
2 x, 2 y, 2 z = λ 2,3, 4 ; 1 2λ
+ .65l = (6)
2 x = 2λ ; 2 y = 4λ ; 2 z = 3λ leads to a critical lw w
point of ( 24 , 29 , 36 ) The nature of the problem
29
48
29
Multiply (5) by l and (6) by w to get
1
indicates this will give a minimum rather than a + .65lw = 2λ (7)
maximum value. The minimum distance is w
1
24 2 2 2
+ 29 + 36 ≈ 2.2283
48 + .65lw = 2λ (8)
29 29 l


from which we conclude that l = w . Putting this equation (3), we have
result into (3) we have ⎛ w +1⎞
2w + 2 w + 4 = λ w2 or λ = 4 ⎜ ⎟.
l = λl 2 (9) ⎝ w2 ⎠
1 Therefore, from equation (1), we have
Since V ≠ 0, l ≠ 0 and (9) tells us that l = ;
λ 120 ⎛ w + 1 ⎞ ⎛ 60 ⎞
+ 4w + 6 = 4 ⎜ ⎟ w⎜ ⎟ or
2
thus l =
1
, w=l =
1 2
= 2λ 2 .
, h= w ⎝ w2 ⎠ ⎝ w 2 ⎠
λ λ lw (multiplying through by w3 and simplifying)
Putting these results into equation (1), we
conclude 2w4 + 3w3 − 60 w − 120 = 0
⎛1⎞ ⎛1⎞ Using one of several techniques available to
.5(2λ 2 ) + .65 ⎜ ⎟ = λ ⎜ ⎟ (2λ 2 ) or solve, we conclude that w = l = 3.213 and
⎝λ⎠ ⎝λ⎠
60
⎛1⎞ h= ≈ 5.812 .
.65 ⎜ ⎟ = λ 2 . (3.213)2
⎝λ⎠
Hence: λ = 3 .65 ≈ .866 , so the minimum cost is 19. (See problem 40, section 12.8)
obtained when: Let
1 c = circumference of circle
l = w = ≈ 1.154 and h = 2λ 2 ≈ 1.5 p = perimeter of first square
λ
q = perimeter of second square
18. (See problem 38, section 12.8). Let the Then the sum of the areas is
dimensions of the box be l, w, and h . Then the c2 p2 q2 1 ⎡ c2 p2 q2 ⎤
cost of the box is A(c, p, q ) = + + = ⎢ + + ⎥
4π 16 16 4 ⎢ π ⎣ 4 4 ⎥
⎦
1(2hl + 2hw) + 4(lw) + 3(2l + 2 w) + 4h or
so we wish to maximize and minimize
C (l , w, h) = 2hl + 2hw + 4lw + 6l + 6w + 4h .
c2 p2 q2
We want to minimize C subject to the constraint A(c, p, q ) = + + subject to the
lhw = 60 ; set V (l , h, w) = lwh − 60 . π 4 4
constraint L(c, p, q) = c + p + q − k = 0 .
Now:
∇C ( l , w, h ) = ( 2h + 4 w + 6 ) i + ( 2h + 4l + 6 ) j and Now
2c p q
+ ( 2l + 2w + 4 ) k ∇A(c, p, q ) = i + j + k
π 2 2
∇V (l , w, h) = wh i + lh j + lw k ∇L ( c , p , q ) = i + j + k
Thus the Lagrange equations are so the Lagrange equations are
2h + 4 w + 6 = λ wh (1) 2c
2h + 4l + 6 = λ lh (2) =λ (1)
π
2l + 2 w + 4 = λ lw (3) p
lwh = 60 (4) =λ (2)
2
60
Solving (4) for h = and putting the result in q
=λ (3)
lw 2
(1) and (2), we get c+ p+q = k (4)
120 60λ
+ 4w + 6 = (5) Putting (1), (2) and (3) into (4) we get
lw l π 2k
120 60λ (4 + )λ = k or λ =
+ 4l + 6 = (6) 2 8+π
lw w Therefore:
Multiply (5) by l and (6) by w to get πk
120 c0 = ≈ 0.282k
+ 4lw + 6l = 60λ (7) 8+π
w 4k
120 p0 = ≈ 0.359k
+ 4lw + 6 w = 60λ (8) 8+π
l 4k
from which we conclude that q0 = ≈ 0.359k
8+π
120 120
+ 6w = + 6l or (l − w)(lw + 20) = 0 . Now A(c0 , p0 , q0 ) ≈ 0.0224k 2 while
l w
Since lw cannot be negative (= −20 ), we A(k , 0, 0) ≈ .079k 2 , so we conclude that
conclude that l = w ; putting this result into


A(c0 , p0 , q0 ) is a minimum value. There is also −a −b −c
and ∇g ( A, B, C ) = i+ j+ k . Thus
2 2
a maximum value (see problem 40, section 12.8) A B C2
but our Lagrange approach does not capture this. the Lagrange equations are
The reason is that the maximum exists because c, A −λ a
p, and q must all be ≥ 0 . Our constraint, = (1)
however, does not require this and allows
ABC A2
negative values for any or all of the variables. B −λ b
= (2)
Under these conditions, there is no global ABC B2
maximum.
C −λ c
= (3)
20. (See problem 42, section 12.8). Let P be the ABC C2
x y z
plane + + = 1 . This plane will cross the a b c
+ + =1 (4)
A B C A B C
first octant, forming a triangle, T, in P; the From (1) – (3) we have
vertices of this triangle occur where P intersects
the coordinate axes. They are: − A3 − B3 −C 3
λ ABC = = = (5)
Vx = ( A, 0, 0), V y = (0, B, 0), Vz = (0, 0, C ) . a b c
Solving in pairs we get
a. Define the vectors g = − A, B, 0 and ⎛ b⎞ ⎛ c⎞
B = ⎜ 3 ⎟ A, C = ⎜ 3 ⎟ A
⎜ a⎟ ⎜ a⎟ (6)
h = − A, 0, C . From example 3 in 11.4, we ⎝ ⎠ ⎝ ⎠
and putting these results into (4) we obtain
know the area of T is
3 3
1 1 A = a + ab 2 + ac 2
g×h = ( BC ) 2 + ( AC ) 2 + ( AB)2 .
2 2
= 3 a ⎛ a 2 + b2 + c2 ⎞
3 3 3
⎜ ⎟
⎝ ⎠
b. The height of the tetrahedron in question is Similarly, we have
the distance is the distance between (0,0,0)
3 3
and P . By example 10 in 11.3, this distance B = a 2 b + b + bc 2
is
= 3 b ⎛ a 2 + b2 + c2 ⎞
3 3 3
⎜ ⎟
1 ( ABC ) 2 ⎝ ⎠
h= =
1
+
1
+
1
( BC ) + ( AC )2 + ( AB) 2
2 3
C = a 2 c + b2 c + c
3
A2 B 2 C 2

= 3 c ⎛ a 2 + b2 + c2 ⎞
3 3 3
⎜ ⎟
c. Finally, the volume of the tetrahedron is ⎝ ⎠
1 Finally, the volume of the tetrahedron is
h(area of T ) , or 3
3 ⎛3 3 3 ⎞
3
abc ⎜ a 2 + b 2 + c 2 ⎟
1 ABC ⎝ ⎠ .
V ( A, B, C ) = ( BC )2 + ( AC )2 + ( AB )2 =
6 6 6
⎡ ( ABC ) 2 ⎤
⋅⎢ ⎥ 21. Finding critical points on the interior first:
⎢ ( BC ) 2 + ( AC ) 2 + ( AB )2 ⎥ ∂f ∂f
⎣ ⎦ =1≠ 0 = 1 ≠ 0; There are no critical
∂x ∂y
1 1
That is, V ( A, B, C ) = ( ABC ) 2 = ABC . points on the interior. Finding critical points on
6 6
the boundary: ∇f ( x, y ) = λ∇g ( x, y );
Hence we want to minimize
V ( A, B, C ) = ABC subject to the constraint 1,1 = λ 2 x, 2 y ; The solution to the system
a b c 1 = λ ⋅ 2 x, 1 = λ ⋅ 2 y , x 2 + y 2 = 1 is λ = ± 1 ,
+ + = 1 ; define 2
A B C
x=± 1 , y=± 1 The four critical points are
a b c 2 2
g ( A, B, C ) = + + − 1 .

Now
A B C
( 1 ,± 1
2 ) and ( − , ± ).
2
1
2
1
2

∇V ( A, B, C ) =
A
i+
B
j+
C
k f ( , ) = 10 + 2 is the maximum value.
1 1
2 2
ABC ABC ABC
f ( − , − ) = 10 − 2 is the minimum value.
1 1
2 2


22. Finding critical points on the interior:
∂f
= 1 − y = 0 ⇒ y = 1;
{
boundary S = ( x, y ): x4
2
+
y2
9 }
= 1 ; this is done

∂x using Lagrange multipliers. Let
∂f x2 y 2
= 1 − x = 0 ⇒ x = 1; The only critical point on g ( x, y ) = + − 1 ; then
∂y 4 9
the interior is c1 = (1,1). Finding critical points 1 −2 xy
∇f ( x, y ) = i+ j and
2
on the boundary: Solve the system of equations 1+ y (1 + y 2 )2
1 − y = λ ⋅ 2 x; 1 − x = λ ⋅ 2 y; x 2 + y 2 = 9 x 2y
∇g ( x , y ) = i + j
Using substitution, it can be found that the 2 9
critical points on the boundary are The Lagrange equations are
⎛ 3 3 ⎞ ⎛ 3 3 ⎞ 1 λx
c2 = ⎜ , ⎟ , c3 = ⎜ − ,− ⎟, = (1)
⎝ 2 2⎠ ⎝ 2 2⎠ 2 2
1+ y
c4 = (2.56155, -1,56155), −2 xy 2λ y
= (2)
c5 = (-1.56155, 2.56155) 2 2
(1 + y ) 9
The maximum value of 5 is obtained substituting
9 x 2 + 4 y 2 = 36 (3)
either c4 or c5 into f. The minimum value of
Putting (1) into (2) yields
about -8.7426 is obtained by substituting c3 into
−λ 2 x3 y 2λ y
f. = (4)
2 9
23. Finding critical points on the interior: One solution to (4) is y = 0 which yields, from
∂f ∂f (3), x = ±2 . Thus (2, 0) and (−2, 0) are
= 2 x + 3 − y = 0; = 2y − x = 0
∂x ∂y candidates for optimization points.
The solution to this system is the only critical If y ≠ 0 , (4) can be reduced to
point on the interior, c1 = (-2,-1). − λ 2 x 3 2λ
Critical points on the boundary will come from = (5)
2 9
the solutions to the following system of
−4
equations: so that λ = . Putting this result into (1)
2 x + 3 − y = λ ⋅ 2 x, 2 y − x = λ ⋅ 2 y , 9 x3
1 2
x 2 + y 2 = 9 . From the solutions to this system, yields =− , which has no solutions
2
the critical points are c2 = (0,3), 1+ y 9 x2
(left side always +, right side always -).
⎛3 3 3⎞ ⎛ 3 3 3⎞ Therefore the only two candidates for max/min
c3 = ⎜⎜ 2 , − ⎟ , c4 = ⎜ −
⎜ 2 ,− 2 ⎟
⎝ 2⎟⎠ ⎝
⎟
⎠ are (2, 0) and (−2, 0) . Since f (2, 0) = 2 and
f (c1 ) = −3, f (c2 ) = 9, f (c3 ) ≈ 20.6913, f (−2, 0) = −2 we conclude that the max value of
f (c4 ) ≈ −2.6913 The max value of f is f on S is 2 and the min value is −2 .
≈ 20.6913 and the min value is -3.
∂f ∂f
= = 2(1 + x + y ) = 0 ⇒ x + y = −1
{ }
2 2
25.
24. f ( x, y ) =
x
on the set S = ( x, ≤1 y ): x4 + y9 ∂x ∂y
1 + y2 There is no minimum or maximum value on the
We first find the max and min for f on the set interior since there are an infinite number of

{ }
2
critical points. The critical points on the
y2
S = ( x, y ): x4 +
9
< 1 using the methods of boundary will come from the solutions to the
following system of equations:
section 12.8: 1
1 −2 xy 2(1 + x + y ) = λ ⋅ x
∇f ( x, y ) = i+ j so setting 2
2
1+ y (1 + y 2 )2 1
2(1 + x + y ) = λ ⋅ y
1 8
∇f ( x, y ) = 0 we have = 0 (impossible).
1 + y2 Solving these two equations for λ leads to
y = − x − 1 or y = 4 x . Together with the
Thus f has no max or min on S .
x2 y2
We now look for the max and min of f on the constraint + − 1 = 0 leads to the critical
4 16


⎛ −1 − 2 19 −4 + 2 19 ⎞ ⎛θ ⎞
points on the boundary: ⎜ , ⎟, 2r sin ⎜ ⎟ . Then we wish to maximize
⎜ 5 5 ⎟ ⎝2⎠
⎝ ⎠
⎡ ⎛α ⎞ ⎛β⎞ ⎛ γ ⎞⎤
⎛ −1 + 2 19 −4 − 2 19 ⎞ ⎛ 2 8 ⎞ P (α , β , γ ) = 2r ⎢sin ⎜ ⎟ + sin ⎜ ⎟ + sin ⎜ ⎟ ⎥
⎜ , ⎟ , ⎜− ,− ⎟ and ⎣ ⎝2⎠ ⎝2⎠ ⎝ 2 ⎠⎦
⎜ 5 5 ⎟ ⎝ 5 5⎠
⎝ ⎠ subject to g(α, β, γ) = α + β + γ − 2π = 0 = 0.
⎛ 2 8 ⎞
⎜ , ⎟ . Respectively, the maximum value is ⎛α ⎞ ⎛β⎞ ⎛γ ⎞
Let r cos ⎜ ⎟ , cos ⎜ ⎟ , cos ⎜ ⎟ = λ 1, 1, 1 .
⎝ 5 5⎠ ⎝2⎠ ⎝2⎠ ⎝2⎠
≈ 29.9443 and the minimum value is 0.
⎛α ⎞ ⎛β⎞ ⎛γ ⎞
Then λ = r cos ⎜ ⎟ = r cos ⎜ ⎟ = r cos ⎜ ⎟ , so
26. It is clear that the maximum will occur for a ⎝2⎠ ⎝2⎠ ⎝2⎠
triangle which contains the center of the circle. ⎛ α β γ ⎞
(With this observation in mind, there are α = β = γ ⎜ since + + = π ⎟ .
⎝ 2 2 2 ⎠
additional constraints: 0 < α < π , 0 < β < π ,
2π 2π
0 > γ < π .) 3αa = 2 π , so α = ; then β = γ = .
Note that in an isosceles triangle, the side 3 3
opposite the angle θ which is between the
congruent sides of length r has length

27. Let α + β + γ = 1, α > 0, β > 0, and γ > 0.
Maximize P ( x, y, z ) = kxα y β zγ , subject to g(x, y, z) = ax + by + cz – d = 0.
Let ∇P ( x, y, z ) = λ∇g ( x, y, z ). Then kα xα –1 y β zγ , k β xα y β –1 zγ , kγ xα y β zγ –1 = λ a, b, c .
λ ax λ by λ cz
Therefore, = = (since each equals kxα y β zγ ).
α β γ
λ ≠ 0 since λ = 0 would imply x = y = z = 0 which would imply P = 0.
ax by cz
Therefore, = = (*).
α β γ
⎛ ax ⎞ ⎛ by ⎞ ⎛ cz ⎞
The constraints ax + by + cz = d in the form α ⎜ ⎟ + β ⎜ ⎟ + γ ⎜ ⎟ = d becomes
⎝α ⎠ ⎝β ⎠ ⎝γ ⎠
⎛ ax ⎞ ⎛ ax ⎞ ⎛ ax ⎞
α ⎜ ⎟ + β ⎜ ⎟ + γ ⎜ ⎟ = d , using (*).
⎝α ⎠ ⎝α ⎠ ⎝α ⎠
⎛ ax ⎞ ax
Then (α + β + γ ) ⎜ ⎟ = d , or = d (since α + β + γ = 1).
⎝ α ⎠ α
αd βd γd
x= (**); y = and z = then following using (*) and (**).
a b c
Since there is only one interior critical point, and since P is 0 on the boundary, P is maximum when
αd βd γd
x= , y= ,z= .
a b c

28. Let (x, y, z) denote a point of intersection. Let 5. 2x – y + 3z = 28
f(x, y, z) be the square of the distance to the 6. 3λ + 4μ = 16 (1, 2, 3, 4)
origin. Minimize f ( x, y , z ) = x 2 + y 2 + z 2 7. 2λ + 7μ = 28 (1, 2, 3, 5)
8. λ = 0, μ = 4 (6, 7)
subject to g ( x, y, z ) = x + y + z - 8 = 0 and
9. x = 4, y = –2, z = 6 (8, 1-3)
h(x, y, z) = 2x – y + 3z – 28 = 0. f(4, –2, 6) = 56, and the nature of the problem
Let ∇f ( x, y, z ) = λ∇g ( x, y, z ) + μ∇h( x, y, z ). indicates this is the minimum rather than the
2 x, 2 y, 2 z = λ 1,1,1 + μ 2, -1, 3 maximum.
1. 2x = λ + 2μ Conclusion: The least distance is 56 ≈ 7.4833.
2. 2y = λ – μ
3. 2z = λ + 3μ
4. x + y + z = 8


29. -1, 2, 2 = λ 2 x, 2 y, 0 + μ 0,1, 2 The function to be maximized in a hyperplane
with positive coefficients and constant (so
–1 = 2λx, 2 = 2λy + μ, 2 = 2μ, x 2 + y 2 = 2, intercepts on all axes are positive), and the
y + 2z = 1 constraint is a hypersphere of radius 1, so the
Critical points are (–1, 1, 0) and (1, –1, 1). maximum will occur where each xi is positive.
f(–1, 1, 0) = 3, the maximum value; There is only one such critical point, the one
f(1, –1, 1) = –1, the minimum value. obtained from the above by taking the principal
square root to solve for xi .
30. a. Maximize
w( x1 , x2 , …, xn ) = x1 x2 , …, xn , ( xi > 0) Then the maximum value of w is
⎛ a ⎞ ⎛ a ⎞ ⎛ a ⎞ A
subject to the constraint a1 ⎜ 1 ⎟ + a2 ⎜ 2 ⎟ +…+ an ⎜ n ⎟ = = A
g ( x1 , x2 , …, xn ) = x1 + x2 +…+ xn – 1 = 0. Le ⎝ A⎠ ⎝ A⎠ ⎝ A⎠ A
2 2 2
t ∇w( x1 , x2 , …, xn ) = λ∇g ( x1 , x2 , …, xn ). where A = a1 + a2 + … + an .
x2 … xn , x1 x3 … xn , x1 … xn –1 = λ 1, 1, …, 1 .
32. Max: f(–0.71, 0.71) = f(–0.71, – 0.71) = 0.71
Therefore, λ x1 = λ x2 =…= λ xn (since each
equals x1 x2 … xn ). Then x1 = x2 = … = xn . 33. Min: f(4, 0) = –4
(If λ = 0, some xi = 0, so w = 0.) 34. Max: f(1.41, 1.41) = f(–1.41, –1.44) = 0.037
1
Therefore, nxi = 1; xi = . 35. Min: f(0, 3) = f(0, –3) = –0.99
n
n
⎛1⎞
The maximum value of w is ⎜ ⎟ , and 12.10 Chapter Review
⎝n⎠
1
occurs when each xi = . Concepts Test
n
n 1. True: Except for the trivial case of z = 0,
⎛1⎞ which gives a point.
b. From part a we have that x1 x2 … xn ≤ ⎜ ⎟ .
⎝n⎠
1 xy
Therefore, n x1 x2 … xn ≤ . 2. False: Use f(0, 0) = 0; f ( x, y ) =
n x + y2 2

ai a elsewhere for counterexample.
If xi = = i for each i, then
a1 + … + an A
3. True: Since g ′(0) = f x (0, 0)
a a a 1 A
n 1 2 … n ≤ , so n a a … a ≤ , or
1 2 n
A A A n n 4. True: It is the limit along the path, y = x.
n a a … a ≤ a1 + a2 + … an . 5. True: Use “Continuity of a Product”
1 2 n
n Theorem.
31. Let a1 , a2 , … an = λ 2 x1 , 2 x2 , …, 2 xn . 6. True: Straight forward calculation of partial
Therefore, ai = 2λ xi , for each i = 1, 2, …, n derivatives
(since λ = 0 implies ai = 0, contrary to the 7. False: See Problem 25, Section 12.4.
hypothesis).
xi x j ⎛ 1 ⎞ 8. False: It is perpendicular to the level curves
= for all i, j ⎜ since each equals ⎟. of f. The gradient of
ai a j ⎝ 2λ ⎠
F(x, y, z) = f(x, y) – z is perpendicular
The constraint equation can be expressed to the graph of z = f(x, y).
2 2 2
2⎛x ⎞ 2⎛x ⎞ 2⎛x ⎞
a1 ⎜ 1 ⎟ + a2 ⎜ 2 ⎟ + … + an ⎜ n ⎟ = 1. 9. True: Since 0, 0, −1 is normal to the
⎝ a1 ⎠ ⎝ a2 ⎠ ⎝ an ⎠
tangent plane
2
Therefore, ( 2
a1 2
+ a2 + 2
… + an ) ⎛ x1 ⎞
⎜ ⎟ = 1.
⎝ a1 ⎠ 10. False: C ex : For the cylindrical surface
2 f ( x, y ) = y 3 , f(p) = 0 for every p on
a1
2
x1 = ; similar for each other xi2 . the x-axis, but f(p) is not an extreme
2 2
a1 + … + an value.


11. True: It will point in the direction of greatest b. 2x − y −1 ≥ 0
increase of heat, and at the origin,
∇T (0, 0) = 1, 0 is that direction.

12. True: It is nonnegative for all x, y, and it has
a value of 0 at (0, 0).

13. True: Along the x-axis, f ( x, 0) → ±∞ as
x → ±∞.

14. False: Du f ( x, y ) = 4, 4 ⋅ u ≤ 4 2
⎛ 1 ⎞ ⎞ 2. x + y 2 = k
(equality if u = ⎜ ⎟ 1,1 ⎟
⎝ 2⎠ ⎠ x – k = – y2
15. True: – Du f ( x, y ) = –[∇f ( x, y ) ⋅ u]
= ∇f ( x, y ) ⋅ (– u) = D– u f ( x, y )

16. True: The set (call it S, a line segment)
contains all of its boundary points
because for every point P not in S
(i.e., not on the line segment), there is
an open neighborhood of P (i.e., a
circle with P as center) that contains
no point of S. 3. f x ( x, y ) = 12 x3 y 2 + 14 xy 7
17. True: By the Min-Max Existence Theorem f xx ( x, y ) = 36 x 2 y 2 + 14 y 7

18. False: ( x0 , y0 ) could be a singular point. f xy ( x, y ) = 24 x3 y + 98 xy 6

⎛π ⎞ ⎛π⎞ 4. f x ( x, y ) = –2 cos x sin x = – sin 2 x
19. False: f ⎜ , 1⎟ = sin ⎜ ⎟ = 1, the maximum
⎝2 ⎠ ⎝2⎠ f xx ( x, y ) = –2 x cos 2 x
value of f, and (π / 2, 1) is in the set. f xy ( x, y ) = 0

20. False: The same function used in Problem 2 5. f x ( x, y ) = e – y sec2 x
provides a counterexample.
f xx ( x, y ) = 2e – y sec 2 x tan x
f xy ( x, y ) = – e – y sec2 x
Sample Test Problems

1. a. x 2 + 4 y 2 − 100 ≥ 0 6. f x ( x, y ) = – e – x sin y

x2 y 2 f xx ( x, y ) = e – x sin y
+ ≥1
100 25 f xy ( x, y ) = – e – x cos y

7. Fy ( x, y ) = 30 x3 y5 – 7 xy 6
Fyy ( x, y ) = 150 x3 y 4 – 42 xy 5

Fyyx ( x, y ) = 450 x 2 y 4 – 42 y 5

8. f x ( x, y, z ) = y 3 – 10 xyz 4
f y ( x, y, z ) = 3 xy 2 – 5 x 2 z 4

f z ( x, y, z ) = −20 x 2 yz 3
Therefore, f x (2, – 1, 1) = 19 ;
f y (2, – 1, 1) = –14 ; f z (2, – 1, 1) = 80

794 Section 12.10 Instructor's Resource Manual

y 2 16. In the direction of ∇f (1, 2) = 4 9, 4
9. z y ( x, y ) = ; z y (2, 2) = = 1
2 2
10. Everywhere in the plane except on the parabola x2 x2 y 2
17. a. f(4, 1) = 9, so + y 2 = 9, or + = 1.
x 2 = y. 2 18 9
b. ∇f ( x, y ) = x, 2 y , so f (4,1) = 4, 2 .
x–x
11. No. On the path y = x, lim = 0. On the
x →0 x + x c.
x–0
path y = 0, lim = 1.
x →0 x + 0

x2 – 2 y 4–4
12. a. lim = =0
( x, y )→(2, 2) x 2 + 2y 4+4

b. Does not exist since the numerator lends to 4
and the denominator to 0.
18. Fx = Fu u x + Fv vx
( x 2 + 2 y 2 )( x 2 – 2 y 2 )
c. lim v y u 1
( x, y )→(0, 0) x2 + 2 y 2 = +
2 2 2 2
1 + u v 2 xy 1+ u v 2 x
= lim ( x2 – y 2 ) = 0
( x, y )→(0, 0) v y +u
=
2(1 + u 2 v 2 ) x
13. a. ∇f ( x, y, z ) = 2 xyz 3 , x 2 z 3 , 3 x 2 yz 2 Fy = Fu u y + Fv v y
f (1, 2, −1) = −4, −1, 6 v x u −1
= +
2 2 2 2
b. ∇f ( x , y , z ) 1 + u v 2 xy 1+ u v 2 y
v x –u
= y 2 z cos xz , 2 y sin xz , xy 2 cos xz =
2(1 + u 2 v 2 ) y
∇f (1, 2, – 1) = –4 cos(1), sin(1), – cos(1)
≈ −2.1612, −3.3659, 2.1612 ⎛1⎞ ⎛ u ⎞
19. f x = fu u x + fv u y = ⎜ ⎟ (2 x) + ⎜ – ⎟ ( yz )
⎝ v⎠ ⎝ v2 ⎠
14. Du f ( x, y ) = 3 y (1 + 9 x 2 y 2 ) –1 , 3x(1 + 9 x 2 y 2 ) –1 ⋅ u
= x –2 y –1 z –1 ( x 2 + 3 y – 4 z )

Du f (4, 2) =
6 12
, ⋅
3
,−
1
=
(3 3−6 ) ⎛1⎞ ⎛ u ⎞
f y = fu u y + f v v y = ⎜ ⎟ (–3) + ⎜ – ⎟ ( xz )
577 577 2 2 577 ⎝v⎠ ⎝ v2 ⎠
≈ −0.001393 = − x −1 y − 2 z − 1 ( x 2 + 4 z )
⎛1⎞ ⎛ u ⎞
15. z = f ( x, y ) = x 2 + y 2 f z = fu u z + f v vz = ⎜ ⎟ (4) + ⎜ – ⎟ ( xy )
⎝v⎠ ⎝ v2 ⎠
1, – 3, 0 is horizontal and is normal to the
= x −1 y −1 z −2 (3 y − x 2 )
vertical plane that is given. By inspection,
3, 1, 0 is also a horizontal vector and is

perpendicular to 1, – 3, 0 and therefore is

3 1
parallel to the vertical plane. Then u = ,
2 2
is the corresponding 2-dimensional unit vector.
Du f ( x, y ) = ∇f ( x, y ) ⋅ u
3 1
= 2 x, 2 y ⋅ , = 3x + y
2 2
Du f (1, 2) = 3 + 2 ≈ 3.7321 is the slope of the
tangent to the curve.


dF dF dx dF dy
20. = +
dt dx dt dy dt
= (3 x 2 − y 2 )(−6sin 3t ) + (−2 xy − 4 y 3 )(3cos t )
⎛ dF ⎞
t = 0 ⇒ x = 2 and y = 0, so ⎜ ⎟ = 0.
⎝ dt ⎠ t =0

21. Ft = Fx xt + Fy yt + Fz zt

⎛ 10 xy ⎞ ⎛ 3t1/ 2 ⎞ ⎛ 5x2 ⎞ ⎛ 1 ⎞ ⎛ 15 x 2 y ⎞ 3t
=⎜ ⎟⎜ ⎟+⎜ 3 ⎟ ⎜ ⎟ + ⎜ – 4 ⎟ (3e )
⎝ z3 ⎠ ⎜ 2⎝
⎟ ⎜ z
⎠ ⎝
⎟⎝ t ⎠ ⎜
⎠ ⎝ z ⎟ ⎠
15 xy t 5x2 45 x 2 ye3t
= + –
z3 z 3t z4

dc db dα
22. = 3, = –2, = 0.1
dt dt dt
⎛1⎞
Area = A(b, c, α ) = ⎜ ⎟ c(b sin α )
⎝2⎠
dA ⎡⎛ b ⎞ ⎛ dc ⎞ ⎛ c ⎞ ⎛ db ⎞ ⎛ b ⎞ ⎛ dα ⎞ ⎤
= ⎜ ⎟ (sin α ) ⎜ ⎟ + ⎜ ⎟ (sin α ) ⎜ ⎟ + ⎜ ⎟ (bc cos α ) ⎜ ⎟⎥
dt ⎢⎝ 2 ⎠
⎣ ⎝ dt ⎠ ⎝ 2 ⎠ ⎝ dt ⎠ ⎝ 2 ⎠ ⎝ dt ⎠ ⎦

⎛ dA ⎞
=
( 7 + 4 3 ) ≈ 6.9641 in.2/s
⎜ ⎟⎛ π⎞
⎝ dt ⎠ ⎜ 8, 10, ⎟
2
⎝ 6⎠

23. Let F ( x, y , z ) = 9 x 2 + 4 y 2 + 9 z 2 – 34 = 0
∇F ( x, y, z ) = 18 x, 8 y , 18 z , so ∇f (1, 2, – 1) = 2 9, 8, – 9 .
Tangent plane is 9(x – 1) + 8(y – 2) – 9(z + 1) = 0, or 9x + 8y – 9z = 34.

24. V = πr 2 h; dV = Vr dr + Vh dh = 2πrh dr + πr 2 dh
If r = 10, dr ≤ 0.02, h = 6, dh = 0.01, then
dV ≤ 2πrh dr + πr 2 dh ≤ 2 π (10)(6)(0.02) + π (100)(0.01) = 3.4 π
V(10, 6) = π (100)(6) = 600 π
Volume is 600 π ± 3.4 π ≈ 1884.96 ± 10.68

25. df = y 2 (1 + z 2 ) –1 dx + 2 xy (1 + z 2 ) –1 dy – 2 xy 2 z (1 + z 2 ) –2 dz
If x = 1, y = 2, z = 2, dx = 0.01, dy = –0.02, dz = 0.03, then df = –0.0272.
Therefore, f(1.01, 1.98, 2.03) ≈ f(1, 2, 2) + df = 0.8 – 0.0272 = 0.7728

26. ∇f ( x, y ) = 2 xy – 6 x, x 2 – 12 y = 0, 0
at (0, 0) and (±6, 3).
D = f xx f yy – f xy = (2 y – 6)(–12) – (2 x)2
2

= 4(18 − 6 y − x 2 ); f xx = 2( y – 3)
At (0, 0): D = 72 > 0 and f xx < 0, so local maximum at (0, 0).
At (±6, 3): D < 0, so (±6, 3) are saddle points.


27. Let (x, y, z) denote the coordinates of the 1st Review and Preview Problems
octant vertex of the box. Maximize
f(x, y, z) = xyz subject to 1.
2 2 2
g ( x, y, z ) = 36 x + 4 y + 9 z – 36 = 0
(where x, y, z > 0 and the box’s volume is
V(x, y, z) = f(x, y, z).
Let ∇f ( x, y, z ) = λ∇g ( x, y, z ).
yz , xz , xy 8 = λ 72 x,8 y,18 z
1. 8yz = 72λx
2. 8xz = 8λy
3. 8xy = 18λz
4. 36 x 2 + 4 y 2 + 9 z 2 = 36
yz 72λ x
5. = , so y 2 = 9 x 2 . (1, 2)
xz 8λ y 2.
yz 72λ x
6. = , so z 2 = 4 x 2 . (1, 3)
xz 18λ y
1
7. 36 x 2 + 36 x 2 + 36 x 2 = 36, so x = .
3
(5, 6, 4)
3 2
8. y = ,z= (7, 5, 6)
3 3
⎛ 1 3 2 ⎞ ⎛ 1 ⎞⎛ 3 ⎞⎛ 2 ⎞
V⎜ , , ⎟ = 8⎜ ⎟⎜ ⎟⎜ ⎟
⎝ 3 3 3⎠ ⎝ 3 ⎠⎝ 3 ⎠⎝ 3 ⎠
16
= ≈ 9.2376
3
The nature of the problem indicates that the
critical point yields a maximum value rather than 3.
a minimum value.

28. y , x = λ 2 x, 2 y
y = 2λx, x = 2λy, x 2 + y 2 = 1
⎛ 1 1 ⎞
Critical points are ⎜ ,± ⎟ and
⎝ 2 2⎠
⎛ 1 1 ⎞ 1
⎜– ,± ⎟ . Maximum of at
⎝ 2 2⎠ 2
⎛ 1 1 ⎞ 1
⎜± ,± ⎟ ; minimum of – at
⎝ 2 2⎠ 2 4.
⎛ 1 1 ⎞
⎜± , ⎟.
⎝ 2 2⎠

29. Maximize V (r , h) = πr 2 h, subject to
S (r , h) = 2πr 2 + 2πrh – 24π = 0.
2π rh, π r 2 = λ 4π r + 2π h, 2π r

rh = λ(2r + h), r = 2λ, r 2 + rh = 12
Critical point is (2, 4). The nature of the problem
indicates that the critical point yields a maximum
value rather than a minimum value. Conclusion:
The dimensions are radius of 2 and height of 4.

Instructor’s Resource Manual Review and Preview 797

5. 9.

6.
10.

7.

11.

8.

798 Review and Preview Instructor’s Resource Manual

12. 2
⎡ 1 1 ⎤
∫ ( a + bx + c x ) dx = ⎢ax + 2 bx
2
18. 2 2 2
+ c 2 x3 ⎥
0
⎣ 3 ⎦0
8
= 2a + 2b + c 2
3
π
π ⎡1 1 ⎤ π
19. ∫0
sin 2 x dx = ⎢ x − sin 2 x ⎥ =
⎣2 4 ⎦0 2

3/ 4
3/ 4 1 1 x +1 1 ⎛ 21 ⎞
20. ∫1/ 4 1 − x2
dx = ln = ln ⎜ ⎟
2 x − 1 1/ 4 2 ⎝ 5 ⎠

3/ 4 1 1 3/ 4 1
13.
21. ∫ x =1/ 4 1+ u
du = ln 1 + x 2
2 1/ 4
= ln 2
2

4 1 4
22. ∫ x=0 1+ u 2
du = ⎡ tan −1 e x ⎤ ≈ 0.7671
⎣ ⎦0

u = 4r 2 + 1; du = 8r dr
3 1 3
∫0 r 4r 2 + 1 dr =
8 ∫r =0
23. u du
3
−1 + 373 / 2
⎡1 2
( )
3/ 2 ⎤
= ⎢ i 4r 2 + 1 ⎥ =
⎣8 3 ⎦0 12

24. u = a 2 − r 2 ; du = −2r dr
a/2 ar a a/2 1
∫0 dr = − ∫
2 r =0 u
du
a2 − r 2
14. a ⎡1 2 2 ⎤
=− ⎢ a −r ⎥
a/2
=
(
a2 2 − 3 )
2 ⎣2 ⎦0 8

π / 2⎛ 1 1 ⎞
25. ∫0 ⎜ 2 + 2 cos 2θ ⎟ dθ
⎝ ⎠
π /2
⎡1 1 ⎤ π
= ⎢ θ + sin 2θ ⎥ =
⎣2 4 ⎦0 4

2
π / 2⎛ 1 1 ⎞
26. ∫0 ⎜2
⎝
+
2
cos 2θ ⎟ dθ
⎠
2
π / 2⎛ 1 1 1 ⎞
=∫ ⎜4 + cos 2θ + cos 2 2θ ⎟ dθ
0
⎝ 2 4 ⎠
1
∫e
−2 x
15. dx = − e −2 x + C π / 2⎛ 3 1 1 ⎞
2 =∫ ⎜ 8 + 2 cos 2θ + 8 cos 4θ ⎟ dθ
0 ⎝ ⎠
1 1 π /2
⎡3 1 1 ⎤ 3π
∫ xe dx = − xe −2 x + ∫ e−2 x dx + C
−2 x
16. = ⎢ θ + sin 2θ + sin 4θ ⎥ =
2 2 ⎣8 4 32 ⎦0 16
1 −2 x 1 − 2 x
= − xe − e + C
2 4
27. 2π ⎛ a 2 − b 2 − a 2 − c 2 ⎞
⎜ ⎟
a/2
⎝ ⎠
a/2 ⎛ xπ ⎞ a ⎛ xπ ⎞ 2a θ is not part of the integrand.
17. ∫− a / 2 cos ⎜ a
⎝
⎟dx = sin ⎜
⎠ π ⎟ =
⎝ a ⎠− a / 2 π

Instructor’s Resource Manual Review and Preview 799

28. The area is an equilateral triangle of length 2.
1 6 3
A= 2 =
2 2 2

29. The solid is half of a right circular cylinder of
radius 3 and height 8.
1 π
V = π r 2 h = ( 9 )( 8 ) = 36π
2 2

30. The solid is a sphere of radius 7.
4 4π 3 1372π
V = π r3 = 7 = ≈ 1436.8 31.
3 3 3
The solid looks similar to a football.
π
π ⎡1 1 ⎤ π2
V = π ∫ sin 2 x dx = π ⎢ x − sin 2 x ⎥ =
0
⎣2 4 ⎦0 2

32. The solid is a right circular cylinder of radius 7
and height 100.
V = π r 2 h = 4900π

33. The solid is half an elliptic paraboloid.
In the xz-plane, we can consider rotating the
graph of z = 9 − x 2 around the z-axis for
0 ≤ x ≤ 3 . Using the Shell Method, we would get
3
(
V = 2π ∫ x 9 − x 2 dx
0
)
3
⎡ 9 x2 x4 ⎤ ⎡ 81 81 ⎤ 81π
= 2π ⎢ − ⎥ = 2π ⎢ − ⎥ =
⎢ 2
⎣ 4⎥ ⎦0 ⎣2 4⎦ 2

34. The solid is half of a hollow sphere of radius 1
inside half of a solid sphere of radius 4.
1⎛4 4 ⎞
V = ⎜ π 43 − π 13 ⎟ = 42π
2⎝3 3 ⎠

800 Review and Preview Instructor’s Resource Manual

Sol Purcell Ingles

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